Alta - Chapter 7 - The Central Limit Theorem - Part 1

The lengths of text messages have an unknown distribution with mean 32 and standard deviation 4 characters. A sample, with size n=49, is randomly drawn from the population and the sum is taken. What is the probability that the sum is between 1585 and 1602 characters?

$0.1596$0.1596​ The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution times the number of samples, so the mean is (32)(49)=1568. The standard deviation is equal to the original standard deviation multiplied by the square root of the sample size. So, the standard deviation is (4)(49−−√)≈28.000. To find the probability using the Standard Normal Table, we find that the z-scores for the two values, 1585 and 1602, are 0.61 and 1.21 respectively, using the formula z=x−μσ. Using the Standard Normal Table, the area to the left of z=0.61 is 0.7291, and the area to the left of z=1.21 is 0.8869. 0.8869−0.7291=0.1578, so the probability is about 16%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(1585, 1602, 1568, 449−−√), which gives us a result of 0.1596.

Suppose student test scores for a nationwide standardized test have an unknown distribution with mean 259 and standard deviation 33 points. A sample of size n=38 is randomly taken from the population and the mean is taken. What is the probability that the resulting mean is more than 264.3 points?

$0.1611$0.1611​ The Central Limit Theorem for Means states that the mean of the normal distribution of means is equal to the mean of the original distribution. The standard deviation is equal to the original standard deviation divided by the square root of the sample size. So, the mean of this mean distribution is 259 and the standard deviation is 3338√≈5.353. To find the probability using the Standard Normal Table, we find that the z-score for the value 264.3 is 0.990, using the formula z=x−μσ. Using the Standard Normal Table, the area to the left of z=0.990 is 0.8389. Remember that the area to the right of z=0.990 will be the complement of this, 1−0.8389=0.1611. So the probability is about 16%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(264.3, 1010, 259, 33/38−−√), which gives us a result of 0.1611.

Suppose weights of running shoes have an unknown distribution with mean 11 and standard deviation 2 ounces. A sample of size n=40 is randomly taken from the population and the mean is taken. What is the probability that the resulting mean is more than 11.3 ounces?

$0.1714$0.1714​ The Central Limit Theorem for Means states that the mean of the normal distribution of means is equal to the mean of the original distribution. The standard deviation is equal to the original standard deviation divided by the square root of the sample size. So, the mean of this mean distribution is 11 and the standard deviation is 240√≈0.316. To find the probability using the Standard Normal Table, we find that the z-score for the value 11.3 is 0.949, using the formula z=x−μσ. Using the Standard Normal Table, the area to the left of z=0.949 is 0.8289. Remember that the area to the right of z=0.949 will be the complement of this, 1−0.8289=0.1711. So the probability is about 17%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(11.3, 1010, 11, 2/40−−√), which gives us a result of 0.1714.

What is the probability that the sample mean for a sample of size 30 will be at least 45? Use the results from above in your calculation and round your answer to the nearest tenth of a percent.

$0.3$0.3​ From above, we know the sampling distribution is normally distributed with mean 40 and standard deviation 1.8. The problem is asking for the probability the sample mean will be at least 45. So, we need to compute P(x¯≥45). To compute this, we will compute the z−score of 45. P(x¯≥45)=P(Z≥45−401.8)=P(Z≥2.777)=1−P(Z<2.777)=1−0.997=0.003.

The pages per book in a library have an unknown distribution with mean 316 and standard deviation 20 pages. A sample, with size n=40, is randomly drawn from the population and the values are added together. Using the Central Limit Theorem for Sums, what is the mean for the sample sum distribution?

$12640\text{ pages}$12640 pages​ The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution multiplied by the sample size ((n)(μX)). So the mean of the sample sum distribution is (n)(μX)=(40)(316)=12640

A doctor is concerned that patients in her office are consuming an excessive amount of red meat, which can be high in saturated fat. According to a recent study, the yearly consumption of red meat per person follows an approximately normal distribution with a mean of 129 pounds and a standard deviation of 36 pounds. If the doctor randomly samples patients in her office to analyze their consumption of red meat and gets a standard error of 9 pounds, how many patients did she sample?

$16\ \text{patients}$16 patients​ The standard error, or standard deviation of the sampling distribution, is σx¯=σn−−√ . 9=36n−−√ n−−√=369=4 n=16 Therefore, a sample size of 16 patients results in a standard error of 9 pounds.

The length, in words, of the essays written for a contest have an unknown distribution with mean 1940 and standard deviation 394 words. A sample, with size n=45, was randomly drawn from the population. Using the Central Limit Theorem for Means, what is the mean for the sample mean distribution?

$1940$1940​ The Central Limit Theorem for Means states that the mean of the normal distribution of sample means is equal to the mean of the original distribution, 1940.

A financial analyst is interested in learning more about how her account holders save for retirement. The percent of a retirement portfolio that account holders place in bonds follows a nearly normal distribution with a mean of 25 percent and a standard deviation of 5 percent. What sample size of account holders is needed to a get a standard error of 0.1 percent?

$2500\text{ account holders}$2500 account holders​ The standard error, or standard deviation of the sampling distribution, is σx¯=σn−−√ . 0.1=5n−−√ n−−√=50.1=50 n=2500

The lengths, in inches, of adult corn snakes have an unknown distribution with mean 57 and standard deviation 10 inches. A sample, with size n=49, is randomly drawn from the population and the values are added together. Using the Central Limit Theorem for Sums, what is the mean for the sample sum distribution?

$2793\text{ inches}$2793 inches​ The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution multiplied by the sample size ((n)(μX)). So the mean of the sample sum distribution is (n)(μX)=(49)(57)=2793

The ads per magazine have an unknown distribution with mean 290 and standard deviation 24 ads. A sample, with size n=55, was randomly drawn from the population. Using the Central Limit Theorem for Means, what is the mean for the sample mean distribution?

$290$290​ The Central Limit Theorem for Means states that the mean of the normal distribution of sample means is equal to the mean of the original distribution, 290.

Suppose speeds of vehicles traveling on a highway have an unknown distribution with mean 71 and standard deviation 5 miles per hour. A sample of size n=72 is randomly taken from the population and the sum of the values is computed. Using the Central Limit Theorem for Sums, what is the mean for the sample sum distribution?

$5,112\text{ miles per hour}$5,112 miles per hour​ The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution multiplied by the sample size ((n)(μX)). So the mean of the sample sum distribution is (n)(μX)=(72)(71)=5112

Suppose the birthweights of infants born in 2015 have a μ of 3056 grams and the σ is 514 grams. A sample of size n=64 is randomly taken from the population and the mean is taken. Using the Central Limit Theorem for Means, what is the standard deviation for the sample mean distribution? Round to the nearest hundredth.

$64.25$64.25​ The standard deviation for the sample mean distribution is calculated by dividing the standard deviation by the square root of the sample size. 51464−−√=5148=64.25

A businesswoman wants to open a coffee stand across the street from a competing coffee company. She notices that the competing company has an average of 170 customers each day, with a standard deviation of 45 customers. Suppose she takes a random sample of 31 days. Identify the following to help her decide whether to open her coffee stand, rounding to the nearest whole number when necessary:

$\mu=$μ=​ $$170 customers per day $\sigma=$σ=​ $$45 customers per day $n=$n=​ $$31 $\mu_{\overline{x}}=$μx​=​ $$170 customers per day $\sigma_{\overline{x}}=$σx​=​ $$8 customers per day 1$170$170​ 2$45$45​ 3$31$31​ 4$170$170​ 5$8$8​ We are given population mean μ=170 and population standard deviation σ=45, and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=31. By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=170 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=4531−−√≈8

The owners of a baseball team are building a new baseball field for their team and must determine the number of seats to include. The average game is attended by 6,500 fans, with a standard deviation of 450 people. Suppose a random sample of 35 games is selected to help the owners decide the number of seats to include. Identify each of the following and be sure to round to the nearest whole number:

$\mu=$μ=​ $$6500 $\sigma=$σ=​ $$450 $n=$n=​ $$35 $\mu_{\overline{x}}=$μx​=​ $$6500 $\sigma_{\overline{x}}=$σx​=​ $$76 1$6500$6500​ 2$450$450​ 3$35$35​ 4$6500$6500​ 5$76$76​ We are given population mean μ=6,500 and population standard deviation σ=450, and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=10. By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=6,500 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=45035−−√≈76

8,900 income freshman to a university take an entrance exam, with the top 10% being offered admission to the university's honors program. The average test score is 78%, with a standard deviation of 11%. What must a student score to be invited to join the honors program if only 65 scores are looked at to determine the honors admissions requirement? Use the z-table below:

$z$z​ -score = $$1.28 $\sigma_{\overline{x}}$σx​​ = $$1.36% $\overline{x}$x​ = $$79.74% 1$1.28$1.28​ 2$1.36$1.36​ 3$79.74$79.74​ The top 10% is the complement of the bottom 90%. Looking at the standard normal probabilities table the z-score value that corresponds to 90% would be z=1.28 . σx¯ would be equal to the standard deviation divided by the square root of n. σx¯=11%65−−√=1.36 By plugging all the numbers into the formula z=x¯−μσx¯ we find that 1.28=x¯−781.36 1.74=x¯−78 79.74=x¯ So, a student scoring at least 79.74% should be invited to join the honors college since they will be in the top 10%. FEEDBACK

The lengths of text messages have an unknown distribution with mean 29 and standard deviation 3 characters. A sample, with size n=65, was randomly drawn from the population. Using the Central Limit Theorem for Means, what is the standard deviation for the sample mean distribution?

.37

The heights, in inches, of male orangutans have an unknown distribution with mean 51 and standard deviation 6 inches. A sample, with size n=46, is randomly drawn from the population and the mean is taken. What is the probability that the sum is between 2379 and 2401 inches?

0.120 The Central Limit Theorem for Sums states that the mean of the normal distribution of means is equal to the mean of the original distribution times the number of samples, so the mean is (51)(46)=2346. The standard deviation is equal to the original standard deviation multiplied by the square root of the sample size. So, the standard deviation is (6)(46−−√)≈40.694. To find the probability using the Standard Normal Table, we find that the z-scores for the two values, 2379 and 2401, are 0.81 and 1.35 respectively, using the formula z=x−μσ. Using the Standard Normal Table, the area to the left of z=0.81 is 0.791, and the area to the left of z=1.35 is 0.9115. 0.9115−0.791=0.1205, so the probability is about 12%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(2379, 2401, 2346, 6(46−−√)), which gives us a result of 0.1204.

The heights of dogs, in inches, in a city have an unknown distribution with mean 26 and standard deviation 4 inches. A sample, with size n=42, is randomly drawn from the population and the sum is taken. What is the probability that the sum is more than 1075 inches?

0.745 The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution times the number of samples, so the mean is (26)(42)=1092. The standard deviation is equal to the original standard deviation multiplied by the square root of the sample size. So, the standard deviation is (4)(42−−√)≈25.923. To find the probability using the Standard Normal Table, we find that the z-score for the value 1075 is −0.656, using the formula z=x−μσ. Using the Standard Normal Table, the area to the left of z=−0.656 is 0.2546. Remember that the area to the right of z=−0.656 will be the complement of this, 1−0.2546=0.7454. So the probability is about 75%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(1075, 1010, 1092, (4)(42−−√)), which gives us a result of 0.7440.

Students at a university frequent the university cafeteria an average of 30 times each week, with a standard deviation of 5 visits, distributed normally. The dean of the university wants to know how many visits separate the bottom 35% from the top 65% visits in a sampling distribution of 100 students in order to make better decisions about the staffing of the university cafeteria. Use the z-table below:

1$-0.39$−0.39​ 2$29.81$29.81​ We know the separating percentage, so we can find the separating length by working backward (from percentage to the z-score in a standard normal distribution that separates the lowest 35% from the highest 65%). 35% in a standard normal distribution corresponds to a z-score of −0.39. The mean of the sampling distribution is also 30 visits, since the distribution is normal. The standard deviation of the sampling distribution of the means is σx¯¯¯=σn−−√=5100−−−√=0.5. Using the formula z=x¯¯¯−μx¯σx¯ to find the sample mean x¯¯¯, we get −0.39x¯¯¯=≈x¯¯¯−300.529.81 If the area under the sampling distribution is 0.35 units, the value that separates this area from the other 0.65 units is approximately 29.81 visits.

A company is offering 401k matching for its employees who stay with the company for more than 10 years. The company's CFO finds that the average retirement account holds $490,000, with a standard deviation of $55,000, distributed normally. What amount of money separates the lowest 20% of the means of retirement accounts from the highest 80% in a sampling of 80 employees? Use the z-table below:

1$-0.85$−0.85​ 2$484773.19$484773.19​ We know the separating percentage, so we can find the separating length by working backward (from percentage to the z-score in a standard normal distribution that separates the lowest 20% from the highest 80%). 20% in a standard normal distribution corresponds to a z-score of −0.85. The mean of the sampling distribution is also $490,000 since the distribution is normal. The standard deviation of the sampling distribution of the means is σx¯¯¯=σn−−√=$55,00080−−√=$6,149.19 Using the formula z=x¯¯¯−μx¯σx¯ to find the sample mean x¯¯¯, we get −0.85x¯¯¯=≈x¯¯¯−$490,000$6,149.19$484,773.19 If the area under the sampling distribution is 0.20 units, the value that separates this area from the other 0.80 units is approximately $484,773.19.

A doctor is researching the effects of "too much text messaging" on a group of teenagers. He has already shown that the average number of text messages sent per day by teenagers is 312 with a standard deviation of 125. He wants to know how many text messages separate the the lowest 15% from the highest 85% in a sampling distribution of 144 teenagers. Use the z-table below:

1$-1.04$−1.04​ 2$301$301​ We know the separating percentage, so we can find the separating length by working backward (from percentage to the z-score in a standard normal distribution that separates the lowest 15% from the highest 85%). 15% in a standard normal distribution corresponds to a z-score of −1.04. The mean of the sampling distribution is also 312 text messages. The standard deviation of the sampling distribution of the means is σx¯¯¯=σn−−√=125144−−−√=10.42. Using the formula z=x¯¯¯−μx¯σx¯ to find the sample mean x¯¯¯, we get −1.04x¯¯¯=≈x¯¯¯−31210.42301 If the area under the sampling distribution is 0.15 units, the value that separates this area from the other 0.85 units is approximately 301 text messages.

8,900 incoming freshman to a university take an entrance exam, with the top 10% being offered admission to the university's honors program. The average test score is 78%, with a standard deviation of 11%. Use Excel to find what the student must score to be invited to join the honors program if only 65 scores are looked at to determine the honors admissions requirement. Round the σx¯ and x¯ to two decimal places.

1$1.36$1.36​ 2$79.74$79.74​ 1. We can use the Central Limit Theorem here since the sample size, n=65, is greater than 30. 2. We know that μ=0.78, so μx¯=0.78 by the Central Limit Theorem. We can calculateσx¯ as: σx¯=0.1165√≈0.0136. To express this as a percentage, we multiply by 100 and get 1.36%. 3. Open Excel and click on any cell. We can either set up the spreadsheet with the information given, or we can continue with the =NORM.INV function on its own. 4. Type =NORM.INV(. Then enter the probability, 0.90, the mean .78, and the standard_deviation, 0.0136. Hit enter. You get 0.7974. To express this as a percentage, multiply by 100. So, a student scoring at least 79.74% should be invited to join the honors college since they will be in the top 10%.

The closing stock prices for a particular social media company follows an unknown distribution with a mean of $150 and a standard deviation of $25. An investor is looking to find the likelihood of the closing stock price falling above the average. After randomly selecting n=52 closing stock prices from the social media company, what is the probability that the sample mean is between $155 and $160? You may use a calculator or the portion of the z-table given below. Round your answer to three decimal places if necessary.

1$150$150​ 2$3.467$3.467​ 3$0.073$0.073​ The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution. The standard deviation is equal to the standard deviation of the population divided by the square root of the sample size. So, the mean of this sampling distribution of the means with sample size 52 is $150 and the standard deviation is 2552√≈$3.467. Next two z-scores would need to be calculated: z1=x¯−μx¯σx¯=160−1503.467=2.88 z2=x¯−μx¯σx¯=155−1503.467=1.44 Using the Standard Normal Table, the area to the left of 2.88 is approximately 0.998, and the area to the left of 1.44 is approximately 0.925. Therefore, the probability that the sample mean will be between $155 and $160 is 0.998−0.925=0.073.

According to the latest real estate report from a suburban town, the mean number of home purchases during last quarter was 175 homes with a population standard deviation of 6 homes. A real estate agent believes that the recent increase of mortgage interest rates is causing a decline in home purchases. To test this, the real estate agent randomly selects 60 recent home purchases. Use Excel to find the probability that the mean of this sample of home purchases is between 173 and 174 homes. Round your answer to three decimal places.

1$175$175​ 2$0.775$0.775​ 3$0.093$0.093​ 1. We can use the Central Limit Theorem for Means here as the sample size of 60 is greater than 30. 2. The following information is given: The mean of the population is μ=175, so we know μx¯=175 by the CLT; The standard deviation of the population σ=6, so we know that σx¯=660√=0.775; 3. Open Excel and click on any cell. 4. Note that because this is a between probability we need to find two values and then the difference between them. First, type =NORM.DIST(173, 175, 0.775, TRUE). This tells us that P(x¯≤173)≈0.0049. Then, type =NORM.DIST(174, 175, 0.775, TRUE). *Note, when using Excel, it's helpful to reference cells in the formula instead of typing in numbers by hand so as to utilize exact values in the calculation instead of rounded numbers. This tells us that P(x¯≤174)≈0.098. The difference between these two values gives us P(173≤x¯≤174)=0.0934. When rounded to three decimal places the probability is 0.093. Therefore, the probability that the sample mean home purchases will be between 173 homes and 174 homes is 0.093.

According to the latest real estate report from a suburban town, the mean number of home purchases during last quarter was 175 homes with a population standard deviation of 6 homes. A real estate agent believes that the recent increase of mortgage interest rates is causing a decline in home purchases. To test this, the real estate agent randomly selects 60 recent home purchases. What is the probability that the mean of this sample of home purchases is between 173 and 174 homes? You may use a calculator or the portion of the z -table given below. Round your answer to three decimal places if necessary.

1$175$175​ 2$0.775$0.775​ 3$0.094$0.094​ The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution. The standard deviation is equal to the standard deviation of the population divided by the square root of the sample size. So, the mean of this sampling distribution of the means with sample size 60 is 175 homes and the standard deviation is 660√≈0.775 homes. Next two z-scores would need to be calculated: z1=x¯−μx¯σx¯=174−1750.775=−1.29 z2=x¯−μx¯σx¯=173−1750.775=−2.58 Using the Standard Normal Table, the area to the left of −1.29 is approximately 0.099, and the area to the left of −2.58 is approximately 0.005. Therefore, the probability that the sample mean home purchases will be between 173 homes and 174 homes is 0.099−0.005=0.094.

Assume that the amount of money professional gamers spend on tech devices per year has an unknown distribution with a mean of 1899 dollars and a standard deviation of 120 dollars. A gaming website claims most gamers spend more per year on tech devices. If a statistician, who works for the gaming website, randomly selects 36 gamers, use Excel to find the probability that the average spent per year on tech devices is greater than 1920 dollars. Round your answers to three decimal places

1$1899$1899​ 2$20$20​ 3$0.147$0.147​ 1. We can use the Central Limit Theorem for Means here as the sample size of 36 is greater than 30. 2. The following information is given: The mean of the population is μ=1899, so we know μx¯=1899 by the CLT; The standard deviation of the population σ=120, so we know that σx¯=12036√=20; 3. Open Excel and click on any cell. 4. Note that because this is a greater than probability we need to use the complement rule. First, type =NORM.DIST(1920, 1899, 20, TRUE). *Note, when using Excel, it's helpful to reference cells in the formula instead of typing in numbers by hand so as to utilize exact values in the calculation instead of rounded numbers. This tells us that P(x¯≤1920)=0.853. Then, use the complement rule to find P(x¯>1920)=1−0.853=0.147. Therefore, the probability that the average of the sample will be larger than 1920 dollars is 0.147.

A bank is reviewing its risk management policies with regards to mortgages. To minimize the risk of lending, the bank wants to compare the typical mortgage owed by their clients against other homebuyers. The average mortgage owed by Americans is $306,500, with a standard deviation of $24,500. Suppose a random sample of 150 Americans is selected. Identify each of the following, rounding your answers to the nearest cent when appropriate:

1$306500 2$24500 3$150 4$306500​ 5$2000.42 We are given population mean μ=$306,500 and population standard deviation σ=$24,500, and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=150. By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=$306,500 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=$24,500150−−−√=$2,000.42

The average home price in Massachusetts is $410,000, with a standard deviation of $65,000. A city planner is studying the distribution across the state of home prices in the bottom 10%. Use Excel to find the price that will indicate that a home is in the bottom 10% if the city planner examines 200 homes around the state. Round σx¯ and x¯ to two decimal places.

1$4596.19$4596.19​ 2$404109.74$404109.74​ 1. We can use the Central Limit Theorem here since the sample size, n=200, is greater than 30. 2. We know that μ=410,000, so μx¯=410,000 by the Central Limit Theorem. We can calculate σx¯ as: σx¯=65,000200√≈4,596.19. We are looking for the price that indicates a home is priced in the bottom 10%. 3. Open Excel and click on any cell. We can either set up the spreadsheet with the information given, or we can continue with the =NORM.INV function on its own. 4. Type =NORM.INV(. Then enter the probability, 0.10, the mean 410,000, and the standard_deviation, 4,596.19. Hit enter. You get 404,109.74. The city planner should examine homes that are priced less than $404,109.74 in order to consider the homes priced in the bottom 10% of all homes in Massachusetts.

A company is offering 401k matching retirement plan for its employees who stay with the company for more than 10 years. The company's CFO finds that the average retirement account holds $490,000, with a standard deviation of $55,000, distributed normally. Use Excel to calculate the amount of money that separates the lowest 20% of the means of retirement accounts from the highest 80% in a sampling of 80 employees.

1$484824.71$484824.71​ 1. We can use the Central Limit Theorem here since the sample size, n=80, is greater than 30. 2. We know that μ=490,000, so μx¯=490,000 by the Central Limit Theorem. We can calculate σx¯ as: σx¯=55,00080√≈6,149.187. 3. Open Excel and click on any cell. We can either set up the spreadsheet with the information given, or we can continue with the =NORM.INV function on its own. 4. Type =NORM.INV(. Then enter the probability, 0.20, the mean 490,000, and the standard_deviation, 6,149.187. Hit enter. You get 484,824.713. If the area under the sampling distribution is 0.20 units, the value that separates this area from the other 0.80 units is approximately $484,824.71.

The average one-bedroom apartment rents for $2,000 a month in Washington, D.C., with a standard deviation of $400. A real estate agent is looking to find a one-bedroom apartment for a client that is in the top 15% of all rentals in the city. Use Excel to calculate how much a client should expect to pay for a one-bedroom apartment in the top 15% of rentals if the agent only surveys 50 apartments in the city. Round σx¯ and x¯ to two decimal places.

1$56.57$56.57​ 2$2058.63$2058.63​ 1. We can use the Central Limit Theorem here since the sample size, n=50, is greater than 30. 2. We know that μ=2000, so μx¯=2000 by the Central Limit Theorem. We can calculateσx¯ as: σx¯=40050√≈56.569. We are looking for how much a client should expect to pay for a one-bedroom apartment in the top 15% of rentals. 3. Open Excel and click on any cell. We can either set up the spreadsheet with the information given, or we can continue with the =NORM.INV function on its own. 4. Type =NORM.INV(. Then enter the probability, 0.85, the mean 2000, and the standard_deviation, 56.569. Hit enter. You get 2,058.63. The client should expect to pay at least $2,058.63 for an apartment in the top 15% of all rentals in the city.

The height of seaweed of all plants in a body of water are normally distributed with a mean of 10 cm and a standard deviation of 2 cm. Use Excel to find which length separates the lowest 30% of the means of the plant heights in a sampling distribution of sample size 15 from the highest 70%? Round your answer to the nearest hundredth.

1$9.73$9.73​ 1. We can use the Central Limit Theorem here since the population is normally distributed and the population standard deviation is known. 2. We know that μ=10, so μx¯=10 by the Central Limit Theorem. We can calculateσx¯ as: σx¯=215√=0.52. We are looking for the length separates the lowest 30% of the means of the plant heights in a sampling distribution of sample size 15 from the highest 70%. Remember that the =NORM.INV shows us left-tail values. So, the we will use the lowest 30% as it is a left-tail value. 3. Open Excel and click on any cell. We can either set up the spreadsheet with the information given, or we can continue with the =NORM.INV function on its own. 4. Type =NORM.INV. Then enter the probability, 0.30, the mean 10, and the standard_deviation, 0.52. Hit enter. If the area under the sampling distribution is 0.3 units, the value that separates this area from the other 0.7 units is approximately 9.73 cm.

The heights of seasonal pine saplings have an unknown distribution with mean 273 and standard deviation 15 millimeters. A sample, with size n=75, is randomly drawn from the population and the sum of the values is taken. Using the Central Limit Theorem for Sums, what is the standard deviation for the sample sum distribution?

129.90 The Central Limit Theorem for Sums states the standard deviation of the normal distribution of sample sums is equal to the original distribution's standard deviation multiplied by the square root of the sample size, (σX)(n−−√). The original standard deviation is 15, and the sample size is 75. So, the standard deviation of the distribution of sample sums is:(σX)(n−−√)=(15)(75−−√)≈129.90

The finishing time for cyclists in a race have an unknown distribution with mean 144 and standard deviation 12 minutes. A sample, with size n=46, is randomly drawn from the population and the sum is taken. What is the probability that the sum is between 6643 and 6676 minutes?

15%

The College Board claimed in 2016 the mean SAT score was 1083 with standard deviation 193 points. A sample school, with size n=68, is randomly drawn from the population and the sum of the values is taken. Using the Central Limit Theorem for Sums, what is the standard deviation for the sample sum distribution?

1591.5 The Central Limit Theorem for Sums states the standard deviation of the normal distribution of sample sums is equal to the original distribution's standard deviation multiplied by the square root of the sample size, (σX)(n−−√). The original standard deviation is 193, and the sample size is 68. So, the standard deviation of the distribution of sample sums is: (σX)(n−−√)=(193)(68−−√)≈1591.5

The pages per book in a library have an unknown distribution with mean 319 and standard deviation 22 pages. A sample, with size n=62, was randomly drawn from the population. Using the Central Limit Theorem for Means, what is the standard deviation for the sample mean distribution?

2.79 The Central Limit Theorem for Means states the standard deviation of the normal distribution of sample means is equal to the original distribution's standard deviation divided by the square root of the sample size, σXn√. The original standard deviation is 22, and the sample size is 62. So, the standard deviation of the distribution of sample means is: σXn−−√=2262−−√≈2.79

Suppose weights, in pounds, of dogs in a city have an unknown distribution with mean 27 and standard deviation 4 pounds. A sample of size n=42 is randomly taken from the population and the sum of the values is taken. Using the Central Limit Theorem for Sums, what is the standard deviation for the sample sum distribution?

25.92 The Central Limit Theorem for Sums states the standard deviation of the normal distribution of sample sums is equal to the original distribution's standard deviation multiplied by the square root of the sample size, (σX)(n−−√). The original standard deviation is 4, and the sample size is 42. So, the standard deviation of the distribution of sample sums is:(σX)(n−−√)=(4)(42−−√)≈25.92

Suppose lengths of text messages have an unknown distribution with mean 30 and standard deviation 4 characters. A sample of size n=61 is randomly taken from the population and the sum of the values is taken. Using the Central Limit Theorem for Sums, what is the standard deviation for the sample sum distribution?

31.24 The Central Limit Theorem for Sums states the standard deviation of the normal distribution of sample sums is equal to the original distribution's standard deviation multiplied by the square root of the sample size, (σX)(n−−√). The original standard deviation is 4, and the sample size is 61. So, the standard deviation of the distribution of sample sums is:(σX)(n−−√)=(4)(61−−√)≈31.24

A psychologist researching new communication methods believes that is unusual for teenagers to send less than 275 texts each day. Assume that the daily number of texts sent by teenagers follows an unknown distribution with a mean of 300 texts and a standard deviation of 59 texts. The psychologist randomly selects 45 teenagers and records the number of texts they send on a particular day. What is the probability that the sample mean is less than 275 texts? Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find μx¯ and σx¯; Calculate the z-score for x¯=275 and move the slider along the x-axis to the appropriate z-score; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve.

561$300$300​ 2$8.795$8.795​ 3$0.002$0.002​ The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution, μx¯=μ. The standard deviation is equal to the standard deviation of the population divided by the square root of the sample size, σx¯=σn√. So, the mean of this sampling distribution of the means with sample size n=45 is μx¯=μ=300 texts and the standard deviation is σx¯=σn−−√=5945−−√=8.795 The z-score for 275 using the formula z=x¯¯¯−μx¯σx¯ is z=275−3008.795=−2.84 Using the graph: Select the correct probability area graph: the one that has the purple area in the left tail represents a less than probability; The Central Limit Theorem applies: μx¯=300 and σx¯=8.795; The z-score for x¯=275 is z=−2.84, so we move the black slider along the x-axis to z=−2.84; The purple area under the curve is: 0.0023. This represents the probability that the mean of the sample is less than 275. Rounding to three decimal places, we have P(x¯<275)=0.002. That is, the probability that the sample mean will be less than 275 texts is approximately 0.002.

The number of square feet per house has an unknown distribution with mean 1490 and standard deviation 113 square feet. A sample, with size n=56, is randomly drawn from the population and the sum of the values is taken. Using the Central Limit Theorem for Sums, what is the standard deviation for the sample sum distribution?

845.61 The Central Limit Theorem for Sums states the standard deviation of the normal distribution of sample sums is equal to the original distribution's standard deviation multiplied by the square root of the sample size, (σX)(n−−√). The original standard deviation is 113, and the sample size is 56. So, the standard deviation of the distribution of sample sums is:(σX)(n−−√)=(113)(56−−√)≈845.61

The Sampling Distribution of the Means

A Sampling Distribution is a form of a probability distribution. It is created using sample statistics based on all possible samples of the same size, from the same population. The sampling distribution of the means can be obtained as follows: Obtain a random sample of size n. For this sample, calculate the sample mean (x¯) for the variable under study. Repeat this sampling for all possible random samples of size n. Then the distribution of all of the x¯'s collected from these samples will form the sampling distribution of the sample means. For example, suppose the test scores of 5 students are 12, 16, 13, 9, and 7. The table lists all possible combinations of 2 students (the sample size), along with the average test score (x¯), for those two students We can now analyze the distribution for these sample means by organizing them into a frequency distribution table with a class width of 2, and constructing a histogram using the relative frequencies: ** Note - Remember that Class Limits are the interval that each class spans. Class 1 above will contain all the values from 7 to 8.9 inclusively, and the class width is maximum−minimumDesirednumberofclasses .

Assume that the heights of male orangutans have an unknown distribution with a mean of 55 inches and a standard deviation of 5 inches. A sample of size n=39 is randomly taken from the population. What is the probability that the sample mean is in between 55.2 inches and 55.4 inches? You may use a calculator or the portion of the z-table given below. Round your answer to three decimal places.

Correct answer: 0.093 The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution. The standard deviation is equal to the standard deviation of the population divided by the square root of the sample size. So, the mean of this sampling distribution of the means with sample size 39 is 55 inches and the standard deviation is 539√≈0.801 inches. To find the probability with the help of a Standard Normal Table, we find that the z-scores for the two values, 55.2 and 55.4, using the formula z=x−μσ. The z-score for 55.2 is 0.25 and the z-score for 55.4 is 0.5. Using the Standard Normal Table, the area to the left of 0.25 is 0.5987, and the area to the left of 0.5 is 0.6915. As 0.6915−0.5987=0.0928, the probability that the sample mean is in between 55.2 inches and 55.4 inches is approximately 9%. If you are using a calculator, input the parameters into the normalcdf( ) function as follows: normalcdf(55.2,55.4,55,539√) as the format for normalcdf function is normalcdf(lower value, upper value, μ, σn√). After pressing enter, the calculator displays a value closer to 0.0928.

The number of square feet per house have an unknown distribution with mean 1630 and standard deviation 149 square feet. A sample, with size n=36, is randomly drawn from the population and the sum is taken. What is the probability that the sum is between 58260 and 58492 square feet?

Correct answer: 0.098 The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution times the number of samples, so the mean is (1630)(36)=58680. The standard deviation is equal to the original standard deviation multiplied by the square root of the sample size. So, the standard deviation is (149)(36−−√)≈894.000. To find the probability using the Standard Normal Table, we find that the z-scores for the two values, 58260 and 58492, are −0.47 and −0.21 respectively, using the formula z=x−μσ. Using the Standard Normal Table, the area to the left of z=−0.47 is 0.319, and the area to the left of z=−0.21 is 0.417. 0.417−0.319=0.098, so the probability is about 10%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(58260, 58492, 58680, 149⋅36−−√), which gives us a result of 0.0975.

Question The mean age of all students at a certain college is 22 years and the standard deviation is 2 years. What is the probability that the mean age of a randomly selected sample of 100 students will be less than 21.8 years? Round your answer to three decimal places. Do not include units in your answer.

Correct answer: 0.159 The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution. The standard deviation is equal to the standard deviation of the population divided by the square root of the sample size. So, the mean of this sampling distribution of the means with sample size 100 is 22 years and the standard deviation is 2100√=0.2 years. The z-score for 21.8 using the formula z=x−μσ is −1. Using the Standard Normal Table, the area to the left of −1 is approximately 0.159. Therefore, the probability that the sample mean will be less than 21.8 years is approximately 0.159.

The finishing times, in minutes, for running marathons have an unknown distribution with mean 288 and standard deviation 40 minutes. A sample, with size n=44, is randomly drawn from the population and the mean is taken. What is the probability that the mean is more than 293.8 minutes?

Correct answer: 0.168 The Central Limit Theorem for Means states that the mean of the normal distribution of means is equal to the mean of the original distribution. The standard deviation is equal to the original standard deviation divided by the square root of the sample size. So, the mean of this mean distribution is 288 and the standard deviation is 4044√≈6.030. To find the probability using the Standard Normal Table, we find that the z-score for the value 293.8 is 0.962, using the formula z=x−μσ. Using the Standard Normal Table, the area to the left of z=0.962 is 0.8315. Remember that the area to the right of z=0.962 will be the complement of this, 1−0.8315=0.1685. So the probability is about 17%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(293.8, 1010, 288, 40/44−−√), which gives us a result of 0.1681.

Question Assume that the student test scores for a nationwide standardized test have an unknown distribution but have a mean of 243 and a standard deviation of 34. A sample of size n=40 is randomly taken from the population. What is the probability that the sum of the scores in this sample is less than 9522?

Correct answer: 0.179 The Central Limit Theorem for Sums states that the mean of any sampling distribution of the sums is equal to the mean of the population distribution multiplied by the sample size. The standard deviation is equal to the standard deviation of the population multiplied by the square root of the sample size. So, the mean of this sampling distribution of the sums with sample size 40 is (243)(40)=9720 and the standard deviation is (34)(40−−√)≈215.035. The z-score for 9522 using the formula z=x−μσ is approximately −0.92. Using the Standard Normal Table, the area to the left of −0.92 is approximately 0.1788. Therefore, the probability that the sum of the scores in this sample is less than 9522 is approximately 0.179.

The length, in words, of the essays written for a contest have an unknown distribution with mean 1830 and standard deviation 311 words. A sample, with size n=42, is randomly drawn from the population and the sum is taken. What is the probability that the sum is more than 75812 words?

Correct answer: 0.699 The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution times the number of samples, so the mean is (1830)(42)=76860. The standard deviation is equal to the original standard deviation multiplied by the square root of the sample size. So, the standard deviation is (311)(42−−√)≈2015.510. To find the probability using the Standard Normal Table, we find that the z-score for the value 75812 is −0.520, using the formula z=x−μσ. Using the Standard Normal Table, the area to the left of z=−0.520 is 0.3015. Remember that the area to the right of z=−0.520 will be the complement of this, 1−0.3015=0.6985. So the probability is about 70%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(75812, 1010, 76860, (311)(42−−√)), which gives us a result of 0.6985.

Suppose test scores for a statewide real estate licensing test have an unknown distribution with mean 230 and standard deviation 34 points. A sample of size n=45 is randomly taken from the population and the mean is taken. Using the Central Limit Theorem for Means, what is the standard deviation for the sample mean distribution?

Correct answer: 5.07 The Central Limit Theorem for Means states the standard deviation of the normal distribution of sample means is equal to the original distribution's standard deviation divided by the square root of the sample size, σXn√. The original standard deviation is 34, and the sample size is 45. So, the standard deviation of the distribution of sample means is: σXn−−√=3445−−√≈5.07

At a financial services company, the mean percent allocation for bonds in a retirement portfolio is 26.9 percent with a population standard deviation of 3.6 percent. A random sample of 35 retirement plan participants was taken and the probability that the mean bond percent for the sample will be at least 28 percent was determined. Was the probability a Left-tail, Right -tail or Interval Probability? Use Excel to find the probability that the mean bond percent for the sample will be at least 28 percent. Select the two correct answers that apply below.

Correct answer: Right-Tail P(x¯¯¯>28)=0.035 1. We can use the Central Limit Theorem for Means here as the sample size of 35 is greater than 30. 2. The following information is given: The mean of the population is μ=26.9, so we know μx¯=26.9 by the CLT; The standard deviation of the population σ=3.6, so we know that σx¯=3.635√=0.609; 3. Open Excel and click on any cell. 4. Because this is a greater than probability, we need to use the complement rule. First, type =NORM.DIST(28. 26.9, 0.609 TRUE). *Note, when using Excel, it's helpful to reference cells in the formula instead of typing in numbers by hand so as to utilize exact values in the calculation instead of rounded numbers. This tells us that P(x¯≤28)≈0.965. Then, the probability that x¯ is greater than 28 is 0.035. Therefore, the probability that the 35 sampled participants have a mean bond percent in their portfolio exceeding 28 percent is 0.035.

At a financial services company, the mean percent allocation for bonds in a retirement portfolio is 26.9 percent with a population standard deviation of 3.6 percent. A random sample of 35 retirement plan participants was taken and the probability that the mean bond percent for the sample will be at least 28 percent was determined. Was the probability a Left-tail, Right -tail or Interval Probability? What is the probability that the mean bond percent for the sample will be at least 28 percent? Select all that apply below. Select the two correct answers that apply below. You may use a calculator or the portion of the z-table given below. Round your answer to three decimal places if necessary

Correct answer: Right-Tail P(x¯¯¯>28)=0.035 The word exceeds means greater than, which are both key words meaning Right-tail probability. With a right tail probability you must subtract the probability from one. To find the probability a z-score must first be computed using the formula z=x¯−μx¯σx¯ σx¯=3.63-√5=0.609 z=28−26.90.609=1.81 By looking in the Normal Probability table for 1.81 we find the probability that the mean will be to the left of 28 is 0.965. In order to find the right tail value, that value must be subtracted from one, 1−0.965=0.035. Therefore, the probability that the 35 sampled participants have a mean bond percent in their portfolio exceeding 28 percent is 0.035. Your answer: Right-Tail P(x¯¯¯>28)=0.965 The probability that the mean will be to the left of 28 is 0.965. However, the question asked for the area to the right.

The average college major requires 34 credit hours to complete, with a standard deviation of 3 hours. A college's academic advisors conduct a study to see how many credit hours a sample size of 50 students will need to take to complete their majors. They calculate: μx¯=μ=34 and σx¯=μn−−√=3450−−√=4.81 What did they do wrong?

Correct answer: To get σx¯, the academic advisors divided μ by n−−√ instead of σ. The answer should be: μx¯=34 and σx¯=0.42 The Central Limit Theorem tells us that, if n≥30, μx¯=μ and σx¯=σn√. Here, n=50 so the Central Limit Theorem applies. So μx¯=μ=34. You must divide σ, not μ, by n−−√ to find σx¯. The correct answer is μx¯=34 and σx¯=0.42.

Assume that the heights of male baseball players with an unknown distribution has a mean of 55 inches and a standard deviation of 5 inches. A sample of size n=39 players is randomly selected from the population. Can we use the Central Limit Theorem here to find the probability that the sample mean is between 55.2 inches and 55.4 inches?

Correct answer: Yes, we can use the Central Limit Theorem. We can use the Central Limit Theorem to help us calculate the probability here! So long as n≥30 the Central Limit Theorem applies no matter what the sampling distribution is.

Question Assume that the standardized test scores of a certain group of high school students has an unknown distribution with a mean of 90 percent and a standard deviation of 15 percent. A sample size of n=31 is randomly drawn from a population. Can we use the Central Limit Theorem here to find the probability that the sample mean is between 85% and 92%?

Correct answer: Yes, we can use the Central Limit Theorem. We can use the Central Limit Theorem to help us calculate the probability here! So long as n≥30 the Central Limit Theorem applies no matter what the sampling distribution is.

A head librarian for a large city is looking at the overdue fees per user system-wide to determine if the library should extend its lending period. The average library user has $19.67 in fees, with a standard deviation of $7.02. The data is normally distributed and a sample of 72 library users is selected at random from the population. Select the expected mean and standard deviation of the sampling distribution from the options below

Correct answer: σx¯=$0.83 μx¯=$19.67 The standard deviation of the sampling distribution is σx¯=σn−−√=$7.0272−−√=$0.83 When the distribution is normal, the mean of the sampling distribution is equal to the mean of the population μx¯=μ=$19.67.

A family of statisticians is trying to decide if they can afford for their child to play youth baseball. The cost of joining a team is normally distributed with a mean of $750 and a standard deviation of $185. If a sample of 40 teams is selected at random from the population, select the expected mean and standard deviation of the sampling distribution below.

Correct answer: σx¯=$29.25 μx¯=$750 The standard deviation of the sampling distribution σx¯=σn−−√=$18540−−√=$29.25 When the distribution is normal the mean of the sampling distribution is equal to the mean of the population μx¯=μ=$750.

The heights of all basketball players are normally distributed with a mean of 72 inches and a population standard deviation of 1.5 inches. If a sample of 15 players are selected at random from the population, select the expected mean of the sampling distribution and the standard deviation of the sampling distribution below.

Correct answer: σx¯=0.387 inches μx¯=72 inches The standard deviation of the sampling distribution σx¯=σn√=1.51√5=0.387inches. Likewise, when the distribution is normal the mean of the sampling distribution is equal to the mean of the population μx¯=μ=72 inches

A high school track team has a grade point average of 3.36, with a standard deviation of 1.2. Their coach wants to identify the team members in the bottom 33% for extra tutoring help. What is the grade point average that will mean that a track team athlete will be recommended for extra tutoring help if the coach examines just 32 of her athletes? Use the z-table below:

Correct answers: 1$-0.44 2$0.21​ 3$3.27 Looking at the standard normal probabilities table the z-score value that corresponds to 33% would be z=−0.44 . σx¯ would be equal to the standard deviation divided by the square root of n. σx¯=1.232−−√=0.21 By plugging all the numbers into the formula z=x¯−μσx¯ we find that −0.44=x¯−3.360.21 −0.0924=x¯−3.36 3.2676=x¯ Therefore, any athlete with a grade point average of less than3.27will be recommended for extra tutoring.

Newborn baby boys weigh, on average, 7.5 pounds, with a standard deviation of 2 pounds. A group of pediatricians want to implement special treatments for newborn baby boys in the bottom 25% birth weight. How many babies will receive the special treatments if the pediatricians are only looking at a sample of 100 newborns? Use the z-table below: Round the z-score and σx¯ to two decimal places. Round x¯ to one decimal place

Correct answers: 1$-0.68$−0.68 ​2$0.2$0.2 ​3$7.4$7.4​ Looking at the standard normal probabilities table the z-score value that corresponds to 25% would be z=−0.68 . σx¯ would be equal to the standard deviation divided by the square root of n. σx¯=2100−−−√=0.2 By plugging all the numbers into the formula z=x¯−μσx¯ we find that −0.68=x¯−7.50.2 −0.136=x¯−7.5 7.364=x¯ So, the pediatricians should be administering the special treatments to newborn baby boys weighing less than7.4pounds.

The average home price in Massachusetts is $410,000, with a standard deviation of $65,000. A city planner is studying the distribution across the state of home prices in the bottom 10%. What price will indicate that a home is in the bottom 10% if the city planner examines 200 homes around the state? Use the z-table below:

Correct answers: 1$-1.28 2$4596.19 3$404116.88 Looking at the standard normal probabilities table the z-score value that corresponds to 10% would be z=−1.28 . σx¯ would be equal to the standard deviation divided by the square root of n. σx¯=$65,000200−−−√=$4,596.19 By plugging all the numbers into the formula z=x¯−μσx¯ we find that −1.28=x¯−$410,000$4,596.19 −$5,883.12=x¯−$410,000 $404,116.88=x¯ The city planner should examine homes that are priced less than $404,116.88 in order to consider the homes priced in the bottom 10% of all homes in Massachusetts.

An ICU nurse is recording the systolic blood pressure of incoming patients. Assume the systolic blood pressure has an unknown distribution with a mean of 110 mm Hg and a standard deviation of 12.5 mm Hg. The nurse believes incoming patients' systolic blood pressure readings are too high. If she randomly selects 40 patients and records their systolic blood pressure, what is the probability that the average systolic blood pressure reading from the sample is over 116 mm Hg? Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find μx¯ and σx¯; Calculate the z-score for x¯=116 and move the slider along the x-axis to the appropriate z-score; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve. Round to three decimal places.

Correct answers: 1$110 ​2$1.976​ 3$0.001 The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution, μx¯=μ. The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size, σx¯=σn√. So, the mean of this sampling distribution of the means with sample size n=40 is: μx¯=μ=110 The standard deviation of the sampling distribution is: σx¯=σn−−√=12.540−−√≈1.976 The z-score for 116 using the formula z=x¯¯¯−μx¯σx¯ is: 116−1101.976=3.04 Using the graph: Select the correct probability area graph: the one that has the blue area in the right tail represents a greater than probability; The Central Limit Theorem applies: μx¯=110 and σx¯=1.976; The z-score for x¯=116 is z=3.04, so we move the black slider along the x-axis to z=3.04; The purple area under the curve is: 0.0012. This represents the probability that the mean of the sample is greater than 116 mm Hg. Rounding to three decimal places we have 0.001. Therefore, the probability that the average systolic blood pressure of the sample will be larger than 116 mmHg is 0.001.

Assume that the amount of money professional gamers spend on tech devices per year has an unknown distribution with a mean of 1899 dollars and a standard deviation of 120 dollars. A gaming website claims most gamers spend more per year on tech devices. If a statistician, who works for the gaming website, randomly selects 36 gamers, what is the probability that the average spent per year on tech devices is greater than 1920 dollars? Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find μx¯ and σx¯; Calculate the z-score for x¯=1920 and move the slider along the x-axis to the appropriate z-score; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve. Round your answers to three decimal places.

Correct answers: 1$1899 ​2$20 3$0.147 The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution, μx¯=μ. The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size, σx¯=σn√. So, the mean of this sampling distribution of the means with sample size n=36 is: μx¯=μ=$1899 The standard deviation of the sampling distribution is: σx¯=σn−−√=12036−−√=20 The z-score for 1920 using the formula z=x¯¯¯−μx¯σx¯ is: z=1920−189920=1.05 Using the graph: Select the correct probability area graph: the one that has the blue area in the right tail represents a greater than probability; The Central Limit Theorem applies: μx¯=1899 and σx¯=20; The z-score for x¯=1920 is z=1.05, so we move the black slider along the x-axis to z=1.05; The purple area under the curve is: 0.1469. This represents the probability that the mean of the sample is greater than $1920. Rounding to three decimal places, we have 0.147. Therefore, the probability that the average of the sample will be larger than 1920 dollars is 0.147.

A football recruiting coordinator believes it is unlikely for quarterbacks to score over 22 touchdowns during a season. To test this, the coordinator randomly samples 33 quarterbacks from last season. Assume the number of touchdowns per season follows an unknown distribution with a mean of 19 touchdowns and a standard deviation of 10 touchdowns. Use Excel to calculate the probability that the average number of touchdowns for the sample is between 22 and 24 touchdowns. Round your answers to three decimals

Correct answers: 1$19 2$1.741 3$0.040 1. We can use the Central Limit Theorem for Means here as the sample size of 33 is greater than 30. 2. The following information is given: The mean of the population is μ=19, so we know μx¯=19 by the CLT; The standard deviation of the population σ=10, so we know that σx¯=1033√=1.741; 3. Open Excel and click on any cell. 4. Note that because this is a between probability we need to find two values and then the difference between them. First, type =NORM.DIST(22, 19, 1.741, TRUE). This tells us that P(x¯≤22)≈0.958. Then, type =NORM.DIST(24, 19, 1.741, TRUE). This tells us that P(x¯≤24)≈0.998. *Note, when using Excel, it's helpful to reference cells in the formula instead of typing in numbers by hand so as to utilize exact values in the calculation instead of rounded numbers. The difference between these two values gives us P(22≤x¯≤24)=0.040. Therefore, the probability that the sample mean will be between 22 touchdowns and 24 touchdowns is 0.040.

The average age of all students at a certain college is 22 years and the standard deviation is 2 years. What is the probability that the average age of a randomly selected sample of 100 students will be less than 21.8 years? Round the probability to three decimal places. Use the z-table below

Correct answers: 1$22 ​2$0.2 ​3$0.159 The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution. So, the mean of this sampling distribution of the means with sample size 100 is μx¯=μ=22 years The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size: σx¯=σn−−√=2100−−−√=0.2 years The z-score for 21.8 using the formula z=x−μσ is −1. Using the Standard Normal Table, the area to the left of −1 is approximately 0.159. Therefore, the probability that the sample mean will be less than 21.8 years is approximately 0.159.

A well known social media company is looking to expand their online presence by creating another platform. They know that they current average 2,500,000 users each day, with a standard deviation of 625,000 users. If they randomly sample 50 days to analyze the use of their existing technology, identify each of the following, rounding to the nearest whole number if necessary:

Correct answers: 1$2500000 2$625000 ​3$50 4$2500000 ​5$88388 We are given population mean μ=2,500,000 and population standard deviation σ=625,000, and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=50. By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=2,500,000 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=625,00050−−√≈88,388

Students at a university frequent the university cafeteria an average of 30 times each week, with a standard deviation of 5 visits, distributed normally. The dean of the university wants to know how many visits separate the bottom 35% from the top 65% visits in a sampling distribution of 100 students in order to make better decisions about the staffing of the university cafeteria. Use Excel to calculate this number. Round x¯ to two decimal places.

Correct answers: 1$29.81 1. We can use the Central Limit Theorem here since the sample size, n=100, is greater than 30. 2. We know that μ=30, so μx¯=30 by the Central Limit Theorem. We can calculate σx¯ as: σx¯=5100√=0.5. 3. Open Excel and click on any cell. We can either set up the spreadsheet with the information given, or we can continue with the =NORM.INV function on its own. 4. Type =NORM.INV(. Then enter the probability, 0.35, the mean 30, and the standard_deviation, 0.5. Hit enter. You get 29.807. If the area under the sampling distribution is 0.35 units, the value that separates this area from the other 0.65 units is approximately 29.81 visits.

At a physical therapy center, doctors work with patients to develop an exercise treatment plan to combat injuries. Assume that the time spent doing pool exercises per week has an unknown distribution with a mean of 4.5 hours and a standard deviation of 0.9 hours. A doctor believes most patients at the physical therapy center swim more per week. If the doctor randomly selects 32 physical therapy patients, what is the probability that the average time spent swimming per week is over 4.7 hours? Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find μx¯ and σx¯; Calculate the z-score for x¯=4.7 and move the slider along the x-axis to the appropriate z-score; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve. Round μx¯¯¯ to one decimal, σx¯¯¯ to three decimals, and P(x¯¯¯>4.7) to four decimal places.

Correct answers: 1$4.5 2$0.159 ​3$0.1038​ The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution, μx¯=μ. The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size, σx¯=σn√. So, the mean of this sampling distribution of the means with sample size n=32 is: μx¯=μ=4.5 The standard deviation of the sampling distribution is: σx¯=σn−−√=(0.9)(32−−√)≈0.159 The z-score for 4.7 using the formula z=x¯¯¯−μx¯σx¯ is: 4.7−4.50.159=1.26 Using the graph: Select the correct probability area graph: the one that has the purple area in the right tail represents a greater than probability; The Central Limit Theorem applies: μx¯=4.5 and σx¯=0.159; The z-score for x¯=4.5 is z=1.26, so we move the black slider along the x-axis to z=1.26; The purple area under the curve is: 0.1038. This represents the probability that the mean of the sample is greater than 4.7 hours. Therefore, the probability that the average of the sample will be more than 4.7 hours is 0.104.

Assume that the game playthrough times for a newly released puzzle game has a mean of 49.8 minutes and a standard deviation of 4.2 minutes. A sample of size n=39 playthrough times is randomly selected from a gaming population. What is the probability that the sample mean is in between 50 minutes and 51 minutes? You may use a calculator or the portion of the z-table given below. Round your answers to three decimal places where appropriate.

Correct answers: 1$49.8 2$0.673 ​3$0.344 The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution. The standard deviation is equal to the standard deviation of the population divided by the square root of the sample size. So, the mean of this sampling distribution of the means with sample size 39 is 49.8 minutes and the standard deviation is 4.239√≈0.673 minutes. Next two z-scores would need to be calculated: z1=x¯−μx¯σx¯=50−49.80.673=0.30 z2=x¯−μx¯σx¯=51−49.80.673=1.78 Using the Standard Normal Table, the area to the left of 0.30 is approximately 0.618, and the area to the left of 1.78 is approximately 0.962. Therefore, the probability that the sample mean will be between 50 minutes and 51 minutes is 0.962−0.618=0.344.

Since we've determined that the Central Limit Theorem applies, let's calculate the probability: Assume that the heights of the baseball players has an unknown distribution with a mean of 55 inches and a standard deviation of 5 inches. A sample of size n=39 players is randomly selected from the population. Use the graph below to calculate the probability that the sample mean is between 55.2 and 55.4 inches, and round to the nearest hundredth: Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find μx¯ and σx¯; Calculate the z-scores for x¯=55.2 and and x¯=55.4, then move the sliders along the x-axis to the appropriate z-scores; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve.

Correct answers: 1$55 ​ 2$0.800 ​3$0.09​ The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution, μx¯=μ. The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size, σx¯=σn√. So, the mean of this sampling distribution of the means with sample size n=39 is: μx¯=μ=55 inches The standard deviation of the sampling distribution is: σx¯=σn−−√=539−−√≈0.80 inches Next, the two z-scores would need to be calculated: z1=x¯−μx¯σx¯=55.2−550.8=0.25 z2=x¯−μx¯σx¯=55.4−550.8=0.5 Using the graph: Select the correct probability area graph: the one that has the blue area in the middle represents a between probability; The Central Limit Theorem applies: μx¯=55 and σx¯=0.8; The z-score for x¯=55.2 is z=0.25, so we move the green slider along the x-axis to z=0.25. Then, the z-score for x¯=55.4 is z=0.5, so we move the other green slider to 0.5 ; The purple area under the curve is: 0.0928. This represents the probability that this particular sample has a mean between 55.2 and 55.4 inches. Rounding to the nearest hundredth, we have P(55.2≤x¯≤55.4)=0.09.

A psychologist believes a majority of people have dreams that last less than 7 minutes. Assume that the length of a dream follows an unknown distribution with a mean of 6 minutes and a standard deviation of 2.4 minutes. The psychologist randomly selects 31 people for a sleep study and measures the length of each of their dreams. What is the probability that the sample mean dream length is less than 7 minutes? Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find μx¯ and σx¯; Calculate the z-score for x¯=7 and move the slider along the x-axis to the appropriate z-score; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve. Round to three decimal places

Correct answers: 1$6 2$0.431$ ​3$0.990​ The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution, μx¯=μ. The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size, σx¯=σn√. So, the mean of this sampling distribution of the means with sample size n=31 is: μx¯=μ=6 The standard deviation of the sampling distribution is: σx¯=σn−−√=2.431−−√=0.431 The z-score for 7 using the formula z=x¯¯¯−μx¯σx¯=7−60.431 is 2.32. Using the graph: Select the correct probability area graph: the one that has the purple area in the left tail represents a less than probability; The Central Limit Theorem applies: μx¯=6 and σx¯=0.431; The z-score for x¯=7 is z=2.32, so we move the black slider along the x-axis to z=2.32; The purple area under the curve is: 0.9898. This represents the probability that the mean of the sample is less than 7 minutes. Therefore, the probability that the sample mean will be less than 7 minutes is approximately 0.990.

The average credit card debt owed by Americans is $6375, with a standard deviation of $1200. Suppose a random sample of 36 Americans is selected. Identify each of the following:

Correct answers: 1$6375 ​2$1200 ​3$36$ ​4$6375 ​5$200​ We are given population mean μ=6375 and population standard deviation σ=1200, and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯ for samples of size n=36. By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=6375 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=120036−−√=200

Credit card holders carry an average debt of $3,027 per month, with a standard deviation of $1,060. A credit card company is looking to drop its accounts with their customers who are not using their cards as much, that is customers whose monthly debts are in the bottom 2.5% of all accounts. Use Excel to calculate how much debt must a customer have to be dropped by the credit card company if the company only looks at 30 accounts. Round the σx¯ and x¯ to two decimal places.

Correct answers: 1. 193.53 2. 2647.69​ 1. We can use the Central Limit Theorem here since the sample size is 30. 2. We know that μ=3,027, so μx¯=3,027 by the Central Limit Theorem. We can calculateσx¯ as: σx¯=106030√≈193.529. 3. Open Excel and click on any cell. We can either set up the spreadsheet with the information given, or we can continue with the =NORM.INV function on its own. 4. Type =NORM.INV(. Then enter the probability, 0.025, the mean 3,027, and the standard_deviation, 193.529. Hit enter. You get 2,647.691. The credit card company, in this case, will drop all customers whose cards have a balance less than or equal to $2,647.69 because it puts them in the bottom 2.5% of their customers.

Suppose lengths of text messages have an unknown distribution with mean 32 and standard deviation 3 characters. A sample of size n=47 is randomly taken from the population and the sum of the values is computed. Using the Central Limit Theorem for Sums, what is the mean for the sample sum distribution?

Correct answers: 1504 characters​ The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution multiplied by the sample size ((n)(μX)). So the mean of the sample sum distribution is (n)(μX) = 47 * 32 = 1504

Suppose weights, in pounds, of dogs in a city have an unknown distribution with mean 26 and standard deviation 3 pounds. A sample of size n=67 is randomly taken from the population and the sum of the values is computed. Using the Central Limit Theorem for Sums, what is the mean for the sample sum distribution?

Correct answers: 1742 pounds​ The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution multiplied by the sample size ((n)(μX)). So the mean of the sample sum distribution is (n)(μX)=(67)(26)=1742

Determine the area under the standard normal curve that lies to the right of the z-score of 1.73

Correct answers:$0.0418$0.0418​ We can look at the portion of the Standard Normal Table given in the problem. The ones and tenths digits of 1.73 corresponds with the row of the table labeled "1.7." The hundredths digit corresponds to the column of the table labeled "0.03." Finding where the row and column meet, we can see that the area to the left of the z-score, 1.73, is 0.9582. However, we want to know the area to the right of the z-score, which is P(Z>1.73). So, we need to use the Complement Rule. If 0.9582 is the area to the left, we can subtract this value from 1 to find the area to the right.P(Z>1.73)=1−0.9582=0.0418So the area to the right of the z-score of 1.73 is 0.0418.

Suppose pages per book in a library have an unknown distribution with mean 318 and standard deviation 21 pages. A sample of size n=44 is randomly taken from the population and the sum is taken. What is the probability that the resulting sum is between 13970 and 13999 pages?

Correct answers:$0.0828$0.0828​ The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution times the number of samples, so the mean is (318)(44)=13992. The standard deviation is equal to the original standard deviation multiplied by the square root of the sample size. So, the standard deviation is (21)(44−−√)≈139.298. To find the probability using the Standard Normal Table, we find that the z-scores for the two values, 13970 and 13999, are −0.16 and 0.05 respectively, using the formula z=x−μσ. Using the Standard Normal Table, the area to the left of z=−0.16 is 0.4364, and the area to the left of z=0.05 is 0.5199. 0.5199−0.4364=0.0835, so the probability is about 8%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(13970, 13999, 13992, 21×44−−√), which gives us a result of 0.0828.

The heights of dogs, in inches, in a city have an unknown distribution with mean 22 and standard deviation 3 inches. A sample, with size n=39, is randomly drawn from the population and the mean is taken. What is the probability that the mean is less than 21.7 inches?

Correct answers:$0.2661$0.2661​ The Central Limit Theorem for Means states that the mean of the normal distribution of means is equal to the mean of the original distribution. The standard deviation is equal to the original standard deviation divided by the square root of the sample size. So, the mean of this mean distribution is 22 and the standard deviation is 339√≈0.480. To find the probability using the Standard Normal Table, we find that the z-score for the value 21.7 is −0.624, using the formula z=x−μσ. Using the Standard Normal Table, the area to the left of z=−0.624 is 0.2676, so the probability is about 27%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(−1010, 21.7, 22, 3/39−−√), which gives us a result of 0.2661.

Find the area to the right of the z-score 0.03 under the standard normal curve.

Correct answers:$0.4880$0.4880​ We can look at the portion of the Standard Normal Table given in the problem. The ones and tenths digits of 0.03 corresponds with the row of the table labeled "−0.0." The hundredths digit corresponds to the column of the table labeled "0.03." Finding where the row and column meet, we can see that the area to the left of the z-score, 0.03, is 0.5120. However, we want to know the area to the right of the z-score, which is P(Z>0.03). So, we need to use the Complement Rule. If 0.5120 is the area to the left, we can subtract this value from 1 to find the area to the right.P(Z>0.03)=1−0.5120=0.4880So the area to the right of the z-score of 0.03 is 0.4880.

According to a recent study, the number of times a person visits a doctor per year follows an approximately normal distribution with a mean of 4.6 visits and a standard deviation of 0.8 visits. A tech company wishes to offer their employees health insurance next year. To find the best health insurance plan, the company randomly selects employees and notes how many times they visited the doctor last year. If the research resulted in a standard error of 0.025 visits, how many employees did they sample?

Correct answers:$1024\ \text{employees}$1024 employees​ The standard error, or standard deviation of the sampling distribution, is σx¯=σn−−√ . 0.025=0.8n−−√ n−−√=0.80.025=32 n=1024 Therefore, a sample size of 1024 employees results in a standard error of 0.025 doctor's office visits

Batting averages during a baseball season follow an approximately normal distribution with a mean of 0.260 and a standard deviation of 0.060. A statistician for a sports channel wishes to analyze the batting average from the last season. If the statistician is looking for a small standard error, what baseball player sample size is needed to get a standard error of 0.005?

Correct answers:$144\ \text{baseball players}$144 baseball players​ The standard error, or standard deviation of the sampling distribution, is σx¯=σn−−√ . 0.005=0.060n−−√ n−−√=0.0600.005=12 n=144 Therefore, a sample size of 144 baseball players results in a standard error of 0.005.

A local food delivery service wants to expand their outreach, but before investing in resources, the owner has been collecting data for the last couple of months. The data below represents weekly deliveries for the downtown area. An unknown distribution has a mean of 183 and a standard deviation of 7.5. A sample, with size n=80, was randomly drawn from a population. Using the Central Limit Theorem for Means, what is the mean for the sample mean distribution?

Correct answers:$183$183​ The central limit theorem for sample means states that the mean of the normal distribution of sample means is equal to the mean of the original distribution, 183.

Question A local food delivery service wants to expand their outreach, but before investing in resources, the owner has been collecting data for the last couple of months. The data below represents weekly deliveries for the downtown area. An unknown distribution has a mean of 183 and a standard deviation of 7.5. A sample, with size n=80, was randomly drawn from a population. Using the Central Limit Theorem for Means, what is the mean for the sample mean distribution?

Correct answers:$183$183​ The central limit theorem for sample means states that the mean of the normal distribution of sample means is equal to the mean of the original distribution, 183.

The test scores for a statewide real estate licensing test have an unknown distribution with mean 259 and standard deviation 38 points. A sample, with size n=55, was randomly drawn from the population. Using the Central Limit Theorem for Means, what is the mean for the sample mean distribution? Give just a number for your answer.

Correct answers:$259$259​ The Central Limit Theorem for Means states that the mean of the normal distribution of sample means is equal to the mean of the original distribution, 259.

What is the probability that the sample mean for a sample of size 32 will be more than 37? Use the results from above in your calculation and round your answer to the nearest percent.

Correct answers:$29$29​ From above, we know the sampling distribution is normally distributed with mean 36 and standard deviation 1.8. The problem is asking for the probability the sample mean will be more than 37. So, we need to compute P(x¯>37). To compute this, we will compute the z−score of 37. P(x¯>37)=P(Z>37−361.8)=P(Z>0.556)=1−P(Z<0.556)=1−0.711=0.289.

Suppose the birthweights of infants born in 2015 have a μ of 3056 grams and the σ is 514 grams. A sample of size n=81 is randomly taken from the population and the mean is taken. Using the Central Limit Theorem for Means, what is the mean for the sample mean distribution?

Correct answers:$3056$3056​ The mean of the sample mean distribution stays the same as the original mean.

What is the probability that the sample mean for a sample of size 150 will be more than 82? Use the results from above in your calculation and round your answer to the nearest percent.

Correct answers:$31$31​ From above, we know the sampling distribution is normally distributed with mean 81 and standard deviation 2.0. The problem is asking for the probability the sample mean will be more than 82. So, we need to compute P(x¯>82). To compute this, we will compute the z−score of 82. P(x¯>82)=P(Z>82−812.0)=P(Z>0.490)=1−P(Z<0.490)=1−0.688=0.312.

A hospital supervisor is looking for ways to increase efficiency of patients checking in to the emergency room. The check-in process follows a nearly normal distribution with a mean of 66 minutes and a standard deviation of 21 minutes. What sample size of patient check-ins does the supervisor need in order to achieve a standard error of 3 minutes?

Correct answers:$49\ \text{patient check-ins}$49 patient check-ins​ The standard error, or standard deviation of the sampling distribution, is σx¯=σn−−√ . 3=21n−−√ n−−√=213=7 n=49 Therefore, a sample size of 49 patient check-ins results in a standard error of 3 minutes.

What is the probability that the sample mean for a sample of size 37 will be less than 55? Use the results from above in your calculation and round your answer to the nearest percent.

Correct answers:$97$97​ From above, we know the sampling distribution is normally distributed with mean 50 and standard deviation 2.6. The problem is asking for the probability the sample mean will be less than 55. So, we need to compute P(x¯<55). To compute this, we will compute the z−score of 55. P(x¯<55)=P(Z<55−50/2.6)=P(Z<1.923)=0.972.

A video game company sells an average of 132 games a month, with a standard deviation of 9 games. The company is looking to reward stores that are selling in the top 7%. How many video games must a store sell in order to be eligible for a reward if the company is only looking at 36 of their stores. Use the z-table below: Round the z-score and σx¯ to two decimal places. Round x¯ up to the nearest whole number.

Correct answers:1$-0.53$−0.53​2$9.72$9.72​ Notice that even though our sample size is only n=15, we are given that the population is normally distributed so we know that the sampling distribution will also be normally distributed. Thus, we can use the standard normal table for probabilities. We know the separating percentage, so we can find the separating length by working backward (from percentage to the z-score in a standard normal distribution that separates the lowest 30% from the highest 70%). 30% in a standard normal distribution corresponds to a z-score of −0.53. The mean of the sampling distribution is also 10 cm. Knowing this, we can calculate the standard deviation of the sampling distribution of the means: σx¯¯¯=σn−−√=215−−√=0.52. Taking the z-score, μ , and σx¯ , and using the formula z=x¯¯¯−μx¯σx¯ we solve for the sample mean x¯¯¯ : z−0.53x¯¯¯=x¯¯¯−μx¯σx¯¯¯=x¯¯¯−100.52≈9.72 If the area under the sampling distribution is 0.3 units, the value that separates this area from the other 0.7 units is approximately 9.72 cm.

The state transportation department is conducting a study on the length of green lights in a certain city. The green lights' lengths are normally distributed with a mean of 45 seconds and a standard deviation of 15 seconds. How many seconds separate the lowest 24% of the means from the highest 76% in a sampling distribution of 75 traffic lights? Use the z-table below. Round the z-score and x¯ to two decimal places.

Correct answers:1$-0.71$−0.71​2$43.77$43.77​ We know the separating percentage, so we can find the separating length by working backward (from percentage to the z-score in a standard normal distribution that separates the lowest 24% from the highest 76%). 24% in a standard normal distribution corresponds to a z-score of −0.71. The mean of the sampling distribution is 45 seconds. The standard deviation of the sampling distribution of the means is σx¯¯¯=σn−−√=1575−−√≈1.7321 Using the formula z=x¯¯¯−μx¯σx¯ to find the sample mean x¯¯¯ , we get −0.71x¯¯¯=≈x¯¯¯−451.732143.77 If the area under the sampling distribution is 0.24 units, the value that separates this area from the other 0.76 units is approximately 43.77 seconds.

A hospital administrator is examining the amount of time its care teams spend with patients. He finds that the care teams are spending an average of 22 minutes per patient with a standard deviation of 17 minutes, distributed normally. What amount of time separates the lowest 60% of the means of time spent with patient from the highest 40% in a sample size of 40 care teams? Use the z-table below:

Correct answers:1$0.25$0.25​2$23$23​ We know the separating percentage, so we can find the separating length by working backward (from percentage to the z-score in a standard normal distribution that separates the lowest 60% from the highest 40%). 60% in a standard normal distribution corresponds to a z-score of 0.25. The mean of the sampling distribution is also 22 minutes, since we know the distribution is normal. The standard deviation of the sampling distribution of the means is σx¯¯¯=σn−−√=1740−−√=2.69. Using the formula z=x¯¯¯−μx¯σx¯ to find the sample mean x¯¯¯, we get 0.25x¯¯¯=≈x¯¯¯−222.6923 If the area under the sampling distribution is 0.6 units, the value that separates this area from the other 0.40 units is approximately 23 minutes.

The average library patron reads 42 books each year, with a standard deviation of 12 books. The library would like to reward its top 25% of patrons, as judged by the number of books read each year. How many books must a library patron read to be eligible for the reward if the library is only examining 100 patrons? Use the z-table below:

Correct answers:1$0.67$0.67​2$1.2$1.2​3$43$43​ The top 25% is the complement of the bottom 75%. Looking at the standard normal probabilities table the z-score value that corresponds to 75% would be z=0.67 σx¯ would be equal to the standard deviation divided by the square root of n. σx¯=12100−−−√=1.2 By plugging all the numbers into the formula z=x¯−μσx¯ we find that 0.67=x¯−421.2 0.804=x¯−42 42.804=x¯ In this case you'd have to round up to 43 books, since 42 wouldn't be enough to put the patron into the top25%.

A financial advisor is analyzing a family's estate plan. The amount of money that the family has invested in different real estate properties is normally distributed with a mean of $225,000 and a standard deviation of $50,000. How much money separates the lowest 80% of the amount invested from the highest 20% in a sampling distribution of 10 of the family's real estate holdings? Use the z-table below

Correct answers:1$0.84$0.84​2$238,281.57$238,281.57​ We know the separating percentage, so we can find the separating length by working backward (from percentage to the z-score in a standard normal distribution that separates the lowest 80% from the highest 20%). 80% in a standard normal distribution corresponds to a z-score of 0.84. The mean of the sampling distribution is 225,000. The standard deviation of the sampling distribution of the means is σx¯¯¯=σn−−√=50,00010−−√=15,811.39. Using the formula z=x¯¯¯−μx¯σx¯ to find the sample mean x¯¯¯, we get 0.84x¯¯¯=≈x¯¯¯−225,00015,811.39238,281.57 If the area under the sampling distribution is 0.8 units, the value that separates this area from the other 0.2 units is approximately $238,281.57

The average library patron reads 42 books each year, with a standard deviation of 10 books. The library would like to reward its top 25% of patrons, as judged by the number of books read each year. Use Excel to find how many books must a library patron read to be eligible for the reward if the library is only examining 100 patrons. Round x¯ and σx¯ up to the nearest whole number.

Correct answers:1$1$1​2$43$43​ 1. We can use the Central Limit Theorem here since the number of patrons, n=100, is greater than or equal to 30. 2. We know that μ=42, so μx¯=42 by the Central Limit Theorem. We can calculateσx¯ as: σx¯=10100√=1. We are looking for the number of books that defines the lowest 75% of books read, which corresponds to the highest 25%. 3. Open Excel and click on any cell. We can either set up the spreadsheet with the information given, or we can continue with the =NORM.INV function on its own. 4. Type =NORM.INV(. Then enter the probability, 0.75, the mean 42, and the standard_deviation, 1. Hit enter. You get 42.674. In this case you'd have to round up to 43 books, since 42 wouldn't be enough to put the patron into the top 25%.

A video game company sells an average of 132 games a month, with a standard deviation of 9 games. The company is looking to reward stores that are selling in the top 7%. Use Excel to find the number of video games must a store sell in order to be eligible for a reward if the company is only looking at 36 of their stores. Round σx¯ to one decimal place. Round x¯ up to the nearest whole number.

Correct answers:1$1.5$1.5​2$135$135​ 1. We can use the Central Limit Theorem here since the sample size, 36, is greater than 30. 2. We know that μ=132, so μx¯=132 by the Central Limit Theorem. We can calculateσx¯ as: σx¯=936√=1.5. We are looking for the average number of video games sold to put a store in top 7% of the company. Remember that the =NORM.INV shows us left-tail values. So, the top 7% corresponds to a left-tail value of 93%. 3. Open Excel and click on any cell. We can either set up the spreadsheet with the information given, or we can continue with the =NORM.INV function on its own. 4. Type =NORM.INV. Then enter the probability, 0.93, the mean 132, and the standard_deviation, 1.5. Hit enter. We see that the sample mean given is 134.2136. In this case you'd have to round up to 135 video games, since 134 wouldn't be enough games sold to put the store into the top 7%.

A video game company sells an average of 132 games a month, with a standard deviation of 9 games. The company is looking to reward stores that are selling in the top 7%. Use Excel to find the number of video games must a store sell in order to be eligible for a reward if the company is only looking at 36 of their stores. Round σx¯ to one decimal place. Round x¯ up to the nearest whole number

Correct answers:1$1.5$1.5​2$135$135​ 1. We can use the Central Limit Theorem here since the sample size, 36, is greater than 30. 2. We know that μ=132, so μx¯=132 by the Central Limit Theorem. We can calculateσx¯ as: σx¯=936√=1.5. We are looking for the average number of video games sold to put a store in top 7% of the company. Remember that the =NORM.INV shows us left-tail values. So, the top 7% corresponds to a left-tail value of 93%. 3. Open Excel and click on any cell. We can either set up the spreadsheet with the information given, or we can continue with the =NORM.INV function on its own. 4. Type =NORM.INV. Then enter the probability, 0.93, the mean 132, and the standard_deviation, 1.5. Hit enter. We see that the sample mean given is 134.2136. In this case you'd have to round up to 135 video games, since 134 wouldn't be enough games sold to put the store into the top 7%.

A computer technician believes the upload speed offered by a new internet company in town is not as fast as advertised. Assume the upload speed follows an unknown distribution with a mean of 38 Mbps and a standard deviation of 11 Mbps. If the technician randomly selects n=32 internet connections, use the graph below to calculate the probability that the sample mean is between 32 Mbps and 34 Mbps. Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find μx¯ and σx¯; Calculate the z-score for x¯=32 and x¯=34 and move the slider along the x-axis to the appropriate z-scores; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve. Round your answers to two decimal places.

Correct answers:1$1.94$1.94​2$0.02$0.02​ The Central Limit Theorem for Means states that the mean of any sampling distribution is equal to the mean of the population, μx¯=μ. The standard deviation is equal to the standard deviation of the population divided by the square root of the sample size, σx¯=σn√. So, the mean of this sampling distribution with sample size n=32 is μx¯=μ=38 and the standard deviation is 1132−−√≈1.94 . Then the z-score for x¯=34 using the formula z=x¯−μx¯σx¯ is: z1=x¯−μx¯σx¯=34−381.94=−2.06 The z-score for x¯=32 is: z2=x¯−μx¯σx¯=32−381.94=−3.09 Using the graph: Select the correct probability area graph: the one that has the purple area in the middle represents a between probability; The Central Limit Theorem applies: μx¯=38 and σx¯=1.94; The z-score for x¯=34 is z=−2.06, so we move the black slider along the x-axis to z=−2.06. The z-score for x¯=32 is z=−3.09 so we move the other black slider to z=−3.09 on the x-axis; The purple area under the curve is: 0.0187. This represents the probability that the mean of the sample is between 32 and 34 Mbps. Rounding to two decimal places, we have P(32≤x¯≤34)=0.02. That is, the probability that the average Mbps for the sample will be between 32 and 34 is 0.02.

A computer technician believes the upload speed offered by a new internet company in town is not as fast as advertised. Assume the upload speed follows an unknown distribution with a mean of 38 Mbps and a standard deviation of 11 Mbps. If the technician randomly selects n=32 internet connections, use Excel to calculate the probability that the sample mean is between 32 Mbps and 34 Mbps. Round your answers to three decimal places.

Correct answers:1$1.945$1.945​2$0.0187$0.0187​ 1. We can use the Central Limit Theorem for Means here as the sample size of 32 is greater than 30. 2. The following information is given: The mean of the population is μ=38, so we know μx¯=38 by the CLT; The standard deviation of the population σ=11, so we know that σx¯=1132√=1.945; 3. Open Excel and click on any cell. 4. Note that because this is a between probability we need to find two values and then the difference between them. First, type =NORM.DIST(32, 38, 1.945, TRUE). This tells us that P(x¯≤32)=0.001. Then, type =NORM.DIST(34, 38, 1.945, TRUE). *Note, when using Excel, it's helpful to reference cells in the formula instead of typing in numbers by hand so as to utilize exact values in the calculation instead of rounded numbers. This tells us that P(x¯≤34)=0.020. The difference between these two values gives us P(32≤x¯≤34)=0.019. Rounding to three decimal places, we have P(32≤x¯≤34)=0.019. That is, the probability that the average Mbps for the sample will be between 32 and 34 is 0.019.

The average adult gets 7.45 hours of sleep each night, with a standard deviation of 0.65 hours. A pharmaceutical company developing a sleep aid is researching how much sleep the top 1% of adults get each night, on average. How many hours of sleep must an adult get each night to be in the top 1% if the company is only basing their initial research on the sleep habits of 30 adults? Be sure to round to the nearest hundredth. Use the z-table below:

Correct answers:1$2.31$2.31​2$0.12$0.12​3$7.73$7.73​ The top 1% is the complement of the bottom 99%. Looking at the standard normal probabilities table the z-score value that corresponds to 99% would be z=2.31 . σx¯ would be equal to the standard deviation divided by the square root of n. σx¯=0.653-√0=0.12 By plugging all the numbers into the formula z=x¯−μσx¯ we find that 2.31=x¯−7.450.12 0.28=x¯−7.45 7.73=x¯ In this case, an adult would have to get at least 7.73 hours of sleep to be in the top 1%.

The average age of all students at a certain college is 22 years and the standard deviation is 2 years. Use Excel to find the probability that the average age of a randomly selected sample of 100 students will be less than 21.8 years? Round the probability to three decimal places.

Correct answers:1$22$22​2$0.2$0.2​3$0.159$0.159​ 1. Open Excel and click on any cell. We can either set up our spreadsheet with the information given, or we can continue with the =NORM.DIST function on its own. The given information is: The population mean, μ=22. Then we known that μx¯=22 years. The population standard deviation, σ=2. Then we know that σx¯=σn√=0.2. 2. Type =NORM.DIST. Then enter the value 21.8 as x, the mean of the sampling distribution μx¯=22 as the mean, the standard deviation of the sampling distribution σx¯=0.2 as the standard_deviation, and TRUE for cumulative. Therefore, the probability that the sample mean will be less than 21.8 years is approximately 0.159, when rounded to three decimal places.

The average age of all students at a certain college is 22 years and the standard deviation is 2 years. What is the probability that the average age of a randomly selected sample of 100 students will be less than 21.8 years? Round the probability to three decimal places. Use the z-table below

Correct answers:1$22$22​2$0.2$0.2​3$0.159$0.159​ The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution. So, the mean of this sampling distribution of the means with sample size 100 is μx¯=μ=22 years The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size: σx¯=σn−−√=2100−−−√=0.2 years The z-score for 21.8 using the formula z=x−μσ is −1. Using the Standard Normal Table, the area to the left of −1 is approximately 0.159. Therefore, the probability that the sample mean will be less than 21.8 years is approximately 0.159

An airline is studying the average length of time that each plane spends on the ground at an airport. The plane's ground stays have an unknown distribution with a mean of 48 minutes and a standard deviation of 22 minutes. A sample of 50 planes is randomly taken from the population. Use Excel to calculate the probability that the average plane's ground stay for the sample will be greater than 55 minutes. Round to three decimal places.

Correct answers:1$48$48​2$3.111$3.111​3$0.012$0.012​ 1. We can use the Central Limit Theorem for Means here as the sample size of 50 is greater than 30. 2. The following information is given: The mean of the population is μ=48, so we know μx¯=48 by the CLT; The standard deviation of the population σ=22, so we know that σx¯=2250√=3.111; 3. Open Excel and click on any cell. 4. This is a greater than probability, so we will need to use the complement rule. First, type =NORM.DIST(55, 48, 3.111, TRUE). *Note, when using Excel, it's helpful to reference cells in the formula instead of typing in numbers by hand so as to utilize exact values in the calculation instead of rounded numbers. This tells us that P(x¯≤55)≈0.988. Then, using the complement rule, P(x¯≥55)=0.012. Therefore, the probability that the average ground stay for the sample will be greater than 55 minutes is 0.012.

For employees at a large company, the mean number of overtime hours worked each week is 9.2 hours with a population standard deviation of 1.6 hours. A random sample of 49 employees was taken and the probability that the mean number of overtime hours will exceed 9.3 hours was determined. Was the probability a Left-tail, Right -tail or Interval Probability? What is the probability that the mean overtime hours for the sample will exceed 9.3 hours? Select all that apply below. Select the two correct answers that apply below. You may use a calculator or the portion of the z-table given below. Round your answer to three decimal places if necessary.

Correct answers:1$55$55​2$0.800$0.800​3$0.09$0.09​ The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution, μx¯=μ. The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size, σx¯=σn√. So, the mean of this sampling distribution of the means with sample size n=39 is: μx¯=μ=55 inches The standard deviation of the sampling distribution is: σx¯=σn−−√=539−−√≈0.80 inches Next, the two z-scores would need to be calculated: z1=x¯−μx¯σx¯=55.2−550.8=0.25 z2=x¯−μx¯σx¯=55.4−550.8=0.5 Using the graph: Select the correct probability area graph: the one that has the blue area in the middle represents a between probability; The Central Limit Theorem applies: μx¯=55 and σx¯=0.8; The z-score for x¯=55.2 is z=0.25, so we move the green slider along the x-axis to z=0.25. Then, the z-score for x¯=55.4 is z=0.5, so we move the other green slider to 0.5 ; The purple area under the curve is: 0.0928. This represents the probability that this particular sample has a mean between 55.2 and 55.4 inches. Rounding to the nearest hundredth, we have P(55.2≤x¯≤55.4)=0.09.

A basketball coach is looking over the possessions per game during last season. Assume that the possessions per game follows an unknown distribution with a mean of 56 points and a standard deviation of 12 points. The basketball coach believes it is unusual to score less than 50 points per game. To test this, she randomly selects 36 games. Use Excel to calculate the probability that the sample mean is less than 50 points. Round your answer to three decimal places if necessary.

Correct answers:1$56$56​2$2$2​3$0.001$0.001​ 1. We can use the Central Limit Theorem for Means here as the sample size of 36 is greater than 30. 2. The following information is given: The mean of the population is μ=56, so we know μx¯=56 by the CLT; The standard deviation of the population σ=12, so we know that σx¯=1236√=2; 3. Open Excel and click on any cell. 4. Type =NORM.DIST(50, 56, 2, TRUE). *Note, when using Excel, it's helpful to reference cells in the formula instead of typing in numbers by hand so as to utilize exact values in the calculation instead of rounded numbers. This tells us that P(x¯≤50)≈0.001. Therefore, the probability that the sample mean will be less than 50 points is approximately 0.001.

Question The height of seaweed of all plants in a body of water are normally distributed with a mean of 10 cm and a standard deviation of 2 cm. Use Excel to find which length separates the lowest 30% of the means of the plant heights in a sampling distribution of sample size 15 from the highest 70%? Round your answer to the nearest hundredth. Provide your answer below:

Correct answers:1$9.73$9.73​ 1. We can use the Central Limit Theorem here since the population is normally distributed and the population standard deviation is known. 2. We know that μ=10, so μx¯=10 by the Central Limit Theorem. We can calculateσx¯ as: σx¯=215√=0.52. We are looking for the length separates the lowest 30% of the means of the plant heights in a sampling distribution of sample size 15 from the highest 70%. Remember that the =NORM.INV shows us left-tail values. So, the we will use the lowest 30% as it is a left-tail value. 3. Open Excel and click on any cell. We can either set up the spreadsheet with the information given, or we can continue with the =NORM.INV function on its own. 4. Type =NORM.INV. Then enter the probability, 0.30, the mean 10, and the standard_deviation, 0.52. Hit enter. If the area under the sampling distribution is 0.3 units, the value that separates this area from the other 0.7 units is approximately 9.73 cm.

The height of seaweed of all plants in a body of water are normally distributed with a mean of 10 cm and a standard deviation of 2 cm. Use Excel to find which length separates the lowest 30% of the means of the plant heights in a sampling distribution of sample size 15 from the highest 70%? Round your answer to the nearest hundredth

Correct answers:1$9.73$9.73​ 1. We can use the Central Limit Theorem here since the population is normally distributed and the population standard deviation is known. 2. We know that μ=10, so μx¯=10 by the Central Limit Theorem. We can calculateσx¯ as: σx¯=215√=0.52. We are looking for the length separates the lowest 30% of the means of the plant heights in a sampling distribution of sample size 15 from the highest 70%. Remember that the =NORM.INV shows us left-tail values. So, the we will use the lowest 30% as it is a left-tail value. 3. Open Excel and click on any cell. We can either set up the spreadsheet with the information given, or we can continue with the =NORM.INV function on its own. 4. Type =NORM.INV. Then enter the probability, 0.30, the mean 10, and the standard_deviation, 0.52. Hit enter. If the area under the sampling distribution is 0.3 units, the value that separates this area from the other 0.7 units is approximately 9.73 cm

Since we've determined that the Central Limit Theorem applies, let's calculate the probability: Assume that the standardized test scores of a certain group of high school students has an unknown distribution with a mean of 90 percent and a standard deviation of 15 percent. A sample size of n=31 is randomly drawn from a population. Use Excel to calculate the probability that the sample mean is between 85 and 92 percent, and round to the nearest hundredth.

Correct answers:1$90$90​2$2.69$2.69​3$0.74$0.74​ Note that because this is a probability between two values, we will need to find two probabilities in Excel, P(x¯≤85) and P(x¯≤92), and the difference between them. This will tell us P(85≤x¯≤92). 1. Open Excel and click on any cell. We can either set up our spreadsheet with the information given, or we can continue with the =NORM.DIST function on its own. The given information is: The population mean, μ=90. Then we known that μx¯=90. The population standard deviation, σ=15. Then we know that σx¯=σn√=2.69. 2. Type =NORM.DIST. Then enter the value 92 as x, the mean of the sampling distribution μx¯=90 as the mean, the standard deviation of the sampling distribution σx¯=2.69 as the standard_deviation, and TRUE for cumulative. 3. Click on another cell. Type =NORM.DIST. Then enter the value 85 as x, the mean of the sampling distributionμx¯=90 as the mean, the standard deviation of the sampling distribution σx¯=2.69 as the standard_deviation, and TRUE for cumulative. 6. To find the difference between the two probabilities, click on another empty cell. Then type =A1-A2. Notice here that A1 is the cell with the first probability we calculated, and we are subtracting A2, the cell with the second probability. So, the area between these two values is: 0.77107−0.03172=0.73934 Thus, the probability that the sample mean is in between 85 and 92 is about 0.74, when we round to the nearest hundredth.

Suppose life expectancy for gas ranges has an unknown distribution with mean 19 years and standard deviation 2 years. A sample of size n=100 is randomly taken from the population and the mean is taken. Using the Central Limit Theorem for Means, what is the mean for the sample mean distribution?

Correct answers:19​ The mean of the sample mean distribution stays the same as the original mean.

An unknown distribution is created by the test scores of a nationwide standardized test. This distribution has a mean of 90 and a standard deviation of 15. A sample, with size n=35, was randomly drawn from a population. Using the Central Limit Theorem for Means, what is the mean and the standard deviation for the sample mean distribution?

Finding the Mean: The Central Limit Theorem for Means states that the mean of the normal distribution of sample means is equal to the mean of the original distribution. So the mean of the sample mean distribution is also 90. Finding the Standard Deviation: The Central Limit Theorem for Means states the standard deviation of the normal distribution of sample means is equal to the original distribution's standard deviation divided by the square root of the sample size, σXn√. The original standard deviation is 15, and the sample size is 35. So, the standard deviation of the distribution of sample means is: σXn−−√=1535−−√≈2.535

Central Limit Theorem for Sums

Given the mean of a population (μ) and standard deviation (σ), the parameters for sampling distribution of the sums of sample means are: Mean (μ)= (n)(μX) Standard Deviation (σ)= (σX)(n−−√) It is safe to assume that the distribution is normal if the sample size is at least 30 The z-score is given by x¯−μσ=x¯−(n)(μX)(σX)(n−−√)

Central Limit Theorem for Means

Given the mean of a population (μX) and the standard deviation (σX), the parameters for sampling distribution of the means are: Mean (μ)= μX Standard Deviation (σ)= σXn√ It is safe to assume that the distribution is normal if the sample size is at least 30 The z-score is given by x¯−μσ=x¯−μXσXn√

Key Terms Central limit theorem: the fact that the distribution of the sample mean of a population becomes closer to the normal distribution as the sample size increasesThe Central limit theorem can be abbreviated as CLT Central limit theorem for means: says that if you keep drawing larger and larger samples and calculate the means of those samples, the sample means will form their own normal distribution or "bell curve" Central limit theorem for sums: says that if you keep drawing larger and larger samples and continue to take their sums, the sums will form their own normal distribution or "bell curve"

Key Terms Central limit theorem: the fact that the distribution of the sample mean of a population becomes closer to the normal distribution as the sample size increasesThe Central limit theorem can be abbreviated as CLT Central limit theorem for means: says that if you keep drawing larger and larger samples and calculate the means of those samples, the sample means will form their own normal distribution or "bell curve" Central limit theorem for sums: says that if you keep drawing larger and larger samples and continue to take their sums, the sums will form their own normal distribution or "bell curve"

For runners competing in a marathon, the mean time to run a mile is 4.2 minutes with a population standard deviation of 0.6 minutes. A news reporter is looking to include statistics for his article about fast runners. At a marathon, a random sample of 36 runners was taken and the probability that the mean time to run a mile for the sample will be less than 3.9 minutes was found. Was the probability a Left-tail, Right -tail or Interval Probability? What is the probability that the mean time to run a mile for the sample will be less than 3.9 minutes? Select the two correct answers that apply below. Use Excel to calculate the probability.

Left- Tail P(x¯¯¯<3.9)=0.0013 1. We can use the Central Limit Theorem for Means here as the sample size of 36 is greater than 30. 2. The following information is given: The mean of the population is μ=4.2, so we know μx¯=4.2 by the CLT; The standard deviation of the population σ=0.6, so we know that σx¯=0.636√=0.1; 3. Open Excel and click on any cell. 4. Type =NORM.DIST(3.9, 4.2, 0.1, TRUE). *Note, when using Excel, it's helpful to reference cells in the formula instead of typing in numbers by hand so as to utilize exact values in the calculation instead of rounded numbers. This tells us that P(x¯≤3.9)≈0.0013. Therefore, the probability that a sample of 31 runners will have a mean mile time of less than 3.9 minutes is 0.003.

At a financial services company, the mean percent allocation for bonds in a retirement portfolio is 26.9 percent with a population standard deviation of 3.6 percent. A random sample of 35 retirement plan participants was taken and the probability that the mean bond percent for the sample will be at least 28 percent was determined. Was the probability a Left-tail, Right -tail or Interval Probability? What is the probability that the mean bond percent for the sample will be at least 28 percent? Select all that apply below. Select the two correct answers that apply below. You may use a calculator or the portion of the z-table given below. Round your answer to three decimal places if necessary.

Right-Tail P(x¯¯¯>28)=0.035 The word exceeds means greater than, which are both key words meaning Right-tail probability. With a right tail probability you must subtract the probability from one. To find the probability a z-score must first be computed using the formula z=x¯−μx¯σx¯ σx¯=3.63-√5=0.609 z=28−26.90.609=1.81 By looking in the Normal Probability table for 1.81 we find the probability that the mean will be to the left of 28 is 0.965. In order to find the right tail value, that value must be subtracted from one, 1−0.965=0.035. Therefore, the probability that the 35 sampled participants have a mean bond percent in their portfolio exceeding 28 percent is 0.035.

The mean number of touchdowns per football game is 3.7 with a population standard deviation of 0.9. A random sample of 31 football games was taken and the probability that the mean number of touchdowns for the sample will be more than 3.9 was found. Was the probability a Left-tail, Right -tail or Interval Probability? What is the probability that the mean number of touchdowns for the sample will be more than 3.9? Select the two correct answers that apply below. You may use a calculator or the portion of the z-table given below. Round your answer to three decimal places if necessary.

Right-Tail P(x¯¯¯>3.9)=0.106 The word exceeds means greater than, which are both key words meaning Right-tail probability. With a right tail probability you must subtract the probability from one. To find the probability a z-score must first be computed using the formula z=x¯−μx¯σx¯ σx¯=0.93-√1=0.16 z=3.9−3.70.16=1.25 By looking in the Normal Probability table for 1.25 we find the probability that the mean will be to the left of 3.9 is 0.894. In order to find the right tail value, that value must be subtracted from one, 1−0.894=0.106. Therefore, the probability that the mean for the sample of 31 football games that have more than 3.9 touchdowns is 0.106.

The mean number of touchdowns per football game is 3.7 with a population standard deviation of 0.9. A random sample of 31 football games was taken and the probability that the mean number of touchdowns for the sample will be more than 3.9 was found. Was the probability a Left-tail, Right -tail or Interval Probability? Use Excel to find the probability that the mean number of touchdowns for the sample will be more than 3.9. Select the two correct answers that apply below.

Right-Tail P(x¯¯¯>3.9)=0.108 1. We can use the Central Limit Theorem for Means here as the sample size of 31 is greater than 30. 2. The following information is given: The mean of the population is μ=3.7, so we know μx¯=3.7 by the CLT; The standard deviation of the population σ=0.9, so we know that σx¯=0.931√=0.162; 3. Open Excel and click on any cell. 4. Because this is a greater than probability, we need to use the complement rule. First, type =NORM.DIST(3.9, 3.7, 0.162, TRUE). *Note, when using Excel, it's helpful to reference cells in the formula instead of typing in numbers by hand so as to utilize exact values in the calculation instead of rounded numbers. This tells us that P(x¯≤3.9)≈0.892. Then, the probability that x¯ is greater than 3.9 is 0.108. Therefore, the probability that the mean for the sample of 31 football games that have more than 3.9 touchdowns is 0.108.

For employees at a large company, the mean number of overtime hours worked each week is 9.2 hours with a population standard deviation of 1.6 hours. A random sample of 49 employees was taken and the probability that the mean number of overtime hours will be at least 9.3 hours was determined. Was the probability a Left-tail, Right -tail or Interval Probability? Use Excel to find the probability that the mean overtime hours for the sample will be at least 9.3 hours. Select the two correct answers that apply below.

Right-Tail P(x¯¯¯>9.3)=0.331 1. We can use the Central Limit Theorem for Means here as the sample size of 49 is greater than 30. 2. The following information is given: The mean of the population is μ=9.2, so we know μx¯=9.2 by the CLT; The standard deviation of the population σ=49, so we know that σx¯=1.649√=0.229; 3. Open Excel and click on any cell. 4. Because this is a greater than probability, we need to use the complement rule. First, type =NORM.DIST(9.3, 9.2, 0.229, TRUE). *Note, when using Excel, it's helpful to reference cells in the formula instead of typing in numbers by hand so as to utilize exact values in the calculation instead of rounded numbers. This tells us that P(x¯≤9.3)≈0.669. Then, the probability that x¯ is greater than 9.3 is 0.331. Therefore, the probability that the mean for the sample of 49 employees who have weekly overtime hours exceeding 9.3 hours is 0.331.

An airline reports that, if its flights are delayed, they are delayed for an average of 22 minutes with a standard deviation of 18 minutes. The airline's flight analyst wants to know the average delay time at 40 of the airline's West Coast airports. She calculates: μx¯=μn−−√=2240−−√=3.48 and σx¯=σ=18 What did she do wrong?

She divided μ by n−−√ instead of σ. She should have had: μx¯=22 and σx¯=2.85 The flight analyst did not need to divide μ by n−−√, though she did forget to divide σ by n−−√. Remember that, so long as n≥30, the Central Limit Theorem tells us that μx¯=μ. The sample size is 40, so here μx¯=μ=22. The Central Limit Theorem also tells us that the standard deviation of the sampling distribution is σx¯=σn√, so we must remember to divide the population's standard deviation, σ, by the square root of the sample size, n, to get σx¯.

Question An unknown distribution is created by the test scores of a nationwide standardized test. This distribution has a mean of 90 and a standard deviation of 15. A sample, with size n=35, was randomly drawn from a population. Using the Central Limit Theorem for Means, what is the mean and the standard deviation for the sample mean distribution?

Solution Finding the Mean: The Central Limit Theorem for Means states that the mean of the normal distribution of sample means is equal to the mean of the original distribution. So the mean of the sample mean distribution is also 90. Finding the Standard Deviation: The Central Limit Theorem for Means states the standard deviation of the normal distribution of sample means is equal to the original distribution's standard deviation divided by the square root of the sample size, σXn√. The original standard deviation is 15, and the sample size is 35. So, the standard deviation of the distribution of sample means is: σXn−−√=1535−−√≈2.535

Example - CLT for Means Question An unknown population distribution has a mean of 90 and a standard deviation of 15. A sample of size n=31 is drawn randomly from the population. What is the probability that the sample mean is in between 85 and 92? (Find P(85<x¯<92).)

Solution We can use the Central Limit Theorem for Means here as the sample size of 31 is greater than 30. The following information is given: mean of the population (μX) = 90 standard deviation of the population (σX) = 15 sample size (n) = 31 We can also write this as X¯∼N(90,1531√) as the sampling distribution is normal. We want to compute P(85<x¯<92), the probability that the sample mean is between 85 and 92. This is represented by the shaded area between 85 and 92 in the graph below. We can determine the probability either by using a Standard Normal Table or a calculator. If using a Standard Normal Table: Find the z-scores of 92 and 85, and determine the areas to the left of both the z-scores in a standard normal distribution. The formula for the z-score is z=x¯−μXσn√. z1=92−90(1531√)≈0.74 We can see that the area to the left of z=0.74 is 0.7704. z2=85−901531√≈−1.86 From the above table, the area to the left of z=−1.86 is 0.0314. So, the area between these two z-scores is 0.7704−0.0314=0.7390. So, the probability that the sample mean is in between 85 and 92 is about 74%.

Question An unknown population distribution has a mean of 90 and a standard deviation of 15. A sample of size n=80 is drawn randomly from the population. What is the probability that the total sum of the 80 values is greater than 7500 (P(x¯>7500))?

Solution We can use the Central Limit Theorem for Sums here as the sample size of 80 is greater than 30. The following information is given: mean (μX) = 90 standard deviation (σX) = 15 sample size (n) = 80 We can write this as ΣX∼N((80)(90),(80−−√)(15)) as the sampling distribution is normal. We want to compute P(x¯>7500), the probability that the sample sum is greater than 7500. This is represented by the shaded region to the right of 7500 in the graph below. We can determine the probability either using a by using a Standard Normal Table or a calculator: If using a Standard Normal Table: Find the z-score of 7500, and determine the area to the right of the z-score in a standard normal distribution. The formula for the z-score is z=x¯−μXσn√. z=7500−(80)(90)(80−−√)(15)≈7500−7200(8.944)(15)≈300134.16≈2.24 From the above table, the area to the left of z=2.24 is 0.9875. As we are calculating the probability that the sample sum is greater than 7500, we need to find the area to the right of z=2.24. For this, we can use the complement rule and subtract 0.9875 from 1 which gives 1−0.9875=0.0125. So, the probability that the total sum of the 80 values is greater than 7500 is about 0.0125 or 1.3%. If using a Calculator: Input the parameters into the normalcdf( ) function as follows: normalcdf(7500,1099,(80)(90),(80−−√)(15)) as the format for normalcdf function is normalcdf(lower value, upper value, μ, σn√). After pressing enter, the calculator displays a value closer to 0.0125.Note: We entered 1099 for the upper value as we want to find the probability that the total sum of the 80 values is greater than 7500. As calculator mandates in using an upper value, we choose that large value.

Question: According to researchers the mean number of calories that an adult eats each day is 3500 calories, with a population standard deviation of 320 calories. A nutritionist believes that number is too low and has decided to take a random sample of 50 people to challenge the findings. Use the graph below to calculate the probability that the sample will have a mean number of calories that exceeds 3600: Drag and move the blue dot to select the appropriate probability graph area from the four options on the left. (Note - there are four graphs available to choose from. Only select between less than, greater than, and area between graphs.); Use the Central Limit Theorem to find μx¯ and σx¯; Calculate the z-score for x¯=3600 and move the slider along the x-axis to the appropriate z-score; The purple area under the curve represents the probability of the event occurring. Interpret the purple area under the curve.

Solution: The Central Limit Theorem for Means states that the mean of any sampling distribution of the means is equal to the mean of the population distribution. So, the mean of this sampling distribution of the means with sample size 50 is: μx¯=μ=3500 calories The standard deviation of the sampling distribution is equal to the standard deviation of the population divided by the square root of the sample size:: σx¯=σn−−√=32050−−√≈45.25 calories The z-score for 3600 using the formula z=x¯−μx¯σx¯ is 2.21. We can find this area given the graph above, or given the Standard Normal table. On the graph: Select the correct probability area graph: the one that has the blue area in the right tail represents a greater than probability; The Central Limit Theorem applies: μx¯=3500 and σx¯=45.25; Calculate the z-score for x¯=3600 and move the slider along the x-axis to z=2.21; The purple area under the curve is: 0.0136. This represents the probability that the mean of the sample will be greater than 3600 calories. Using the Standard Normal Table, the area to the left of 2.21 is 0.986. Therefore, the probability that the sample mean will be greater than 3600 calories is 1−0.986=0.014 Remember that since it is a Right Tail probability you have to subtract the value from 1 to get the right tail value.

Question: The body temperatures of all mosquitoes in a county are normally distributed with a mean of 57∘F and a standard deviation of 12∘F. Which temperature separates the highest 10% of the means of the mosquito temperatures in a sampling distribution of sample size 36 from the lowest 90%?

Solution: The question is asking us to find the TOP 10%, which means the probability we want to use is to 90%, or 0.90. To use Excel to help us solve this problem, we need to know the mean of the sampling distribution, the standard deviation of the sampling distribution, and the sample size, n. 1. Open Excel and write in the following cells: A1: Population Mean. A2: Population Standard Deviation A3: Sample Size A5: Sampling Distribution Mean A6: Sampling Distribution Standard Deviation A7: Probability 2. Now, we will set up the following equations in the cells to help us calculate the probability. The sampling distribution mean is equal to the population mean when n≥30, so in B5 type =B1. The sampling distributions standard deviation is equal to the population standard deviation divided by the square root of n when n≥30, so in B6 type =B2/SQRT(B3) To find the sample mean that corresponds to the given probability, =NORM.INV(B7, B5, B6). Now that the spreadsheet is set up, we can calculate our probability by typing in our known values. 4. In cell B1, we have our population mean =57. Notice that B5, μx¯, will also automatically populate. It is 57. 5. In cell B2, type the population standard deviation =12. Then type the sample size, n, in cell B3, =36. Notice here that B6, our σx¯, automatically populates as well. It is 2. 6. Type the probability 0.90 into cell B7. In cell B8, we see that x¯=59.5631. This tells us that the top 10% of mosquitoes in the given county will have a body temperature above 59.56∘F

Question: The body temperatures of all mosquitoes in a county are normally distributed with a mean of 57∘F and a standard deviation of 12∘F. Which temperature separates the highest 10% of the means of the mosquito temperatures in a sampling distribution of sample size 36 from the lowest 90%?

Solution: The question is asking us to find the TOP 10%, which since we are going to need to use the table backwards that means we need to look for the BOTTOM 90% in the table. Thus we need to go to the table and look for the z-score that corresponds to 0.9000. If we look at the table below we can see that: A z-score the corresponds to 0.9000 is z=1.28. Notice that the probability for z=1.28 is not quite 0.9000 however, the next number, 1.29, is above 0.9000. You always want to select the value that is closest to, but does not exceed, the desired probability. We are given that μ=57∘F, so the mean of the sampling distribution is μx¯=57∘F. We can then calculate the standard deviation of the sampling distribution: σx¯=σn−−√=1236−−√=126=2 Now that we know μ, σx¯ , and z , we can substitute these values into the formula z=x¯¯¯−μσx¯ . Then we will find the mean of the sampling distribution corresponding to the top 10% by solving the formula for x¯: z1.28x¯¯¯=x¯−μσx¯=x¯−572=59.56 This tells us that the top 10% of mosquitoes in the given county will have a body temperature above 59.56∘F

The mean salary for all employees in a company is $50,000 with population standard deviation of $4,000. Each bar on the histogram represents the mean salary for a sample of n employees. Move the circle on the slider to change the sample size, n. If repeated samples of 100 employees are taken and the sample mean average salary is calculated for each sample, what will be the mean and standard deviation of the distribution of sample means?

Solution: We are given population mean μ=$50,000 and population standard deviation σ=$4,000, and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯. By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=$50,000 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=4000100−−−√=400010=400

Question: A sample of size 31 is drawn randomly from a population with an unknown distribution. The population has a mean of 90 and a standard deviation of 15. What is the probability that the sample mean is between 85 and 92? (Find P(85<x¯<92).)

Solution: We can use the Central Limit Theorem for Means here as the sample size of 31 is greater than 30. The following information is given: mean of the population (μ) = 90 standard deviation of the population (σ) = 15 sample size (n) = 31 We can also write this as X¯∼N(90,1531√) as the sampling distribution is normal. We want to compute P(85<x¯<92), the probability that the sample mean is between 85 and 92. This is represented by the shaded area between 85 and 92 in the graph below. We can determine the probability either by using a Standard Normal Table: Find the z-scores of 92 and 85, and determine the areas to the left of both the z-scores in a standard normal distribution. The formula for the z-score is z=x¯−μXσn√. z1=92−90(1531√)≈0.74 z2=85−90(1531√)≈−1.86 We can see that the area to the left of z=0.74 is 0.7704. From the above table, the area to the left of z=−1.86 is 0.0314. So, the area between these two z-scores is: 0.7704−0.0314=0.7390 Thus, the probability that the sample mean is in between85and92is about74%.

How To Use Excel to Find a Mean Given a Probability

Step 1: Determine if the conditions of the Central Limit Theorem are met. If yes, proceed with finding the sample mean that yields a specific probability. Step 2: Calculate μx¯ and σx¯ using the formulas μx¯=μ and σx¯=σn√. Identify the probability given so that you can find the sample mean that corresponds to it. Step 3: Open Excel and click on any cell. Step 4: Type =NORM.INV. Notice the order of the probability, mean, and standard deviation. =NORM.INV(probability, mean, standard_dev). Step 4: Enter the probability you are given probability, the mean of the sampling distribution μx¯ as the mean and the standard deviation of the sampling distribution σx¯ as the standard_deviation. Step 5: Hit ENTER and analyze your result. Remember, =NORM.INV will always give you the sample mean corresponding to a left-tail probability.

Sampling distribution of a sample mean

Suppose simple random samples of size n are drawn from a population with mean μ and standard deviation σ. Then the sampling distribution of the sample means, x¯, has: Mean: μx¯=μ. Some μx¯ will overestimate μ and some will underestimate, , but the distribution of such estimates (the sampling distribution) will share the population's mean. and Standard deviation, also known as standard error: σx¯=σn−−√. It depends on, but is not equal to σ, the population standard deviation. Since larger samples generally produce better estimates, σx¯ decreases as the sample size increases. Just a note: to distinguish the sampling distribution's mean and standard deviation from the population's mean μ and standard deviation σ, we use a subscript: μx¯ and σx¯. We noticed that the sampling distribution of the sample means is approximately normal in shape as n increases, as shown in the above histograms. The following theorem asserts that this is generally true for all sampling distributions, regardless of the shape of the underlying population distribution.

The lengths, in inches, of adult corn snakes have an unknown distribution with mean 51 and standard deviation 9 inches. A sample, with size n=43, is randomly drawn from the population and the mean is taken. What is the probability that the mean is less than 50.2 inches?

The Central Limit Theorem for Means states that the mean of the normal distribution of means is equal to the mean of the original distribution. The standard deviation is equal to the original standard deviation divided by the square root of the sample size. So, the mean of this mean distribution is 51 and the standard deviation is 943√≈1.372. To find the probability using the Standard Normal Table, we find that the z-score for the value 50.2 is −0.583, using the formula z=x−μσ. Using the Standard Normal Table, the area to the left of z=−0.583 is 0.2810, so the probability is about 28%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(−1010, 50.2, 51, 9/43−−√), which gives us a result of 0.2800.

Use the Central Limit Theorem for Means to Find the Sample Mean and the Sample Standard Deviation in Business Examples

The Central Limit Theorem for Means: There are many challenges in the use of statistics for business purposes. One of these obstacles is the size of sample data and the accuracy for representing the entire population. The Central Limit Theorem is one of the methods to approach this problem. In the case of investors, this theorem is of assistance for financial portfolios by measuring the return of an investment in given stocks. Managers in delivery services use this strategy to determine the probability for on-time delivery.

The heights, in inches, of male orangutans have an unknown distribution with mean 51 and standard deviation 4 inches. A sample, with size n=65, is randomly drawn from the population and the values are added together. Using the Central Limit Theorem for Sums, what is the mean for the sample sum distribution?

The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution multiplied by the sample size ((n)(μX)). So the mean of the sample sum distribution is (n)(μX)=(65)(51)=3315

Suppose heights of seasonal pine saplings have an unknown distribution with mean 280 and standard deviation 12 millimeters. A sample of size n=42 is randomly taken from the population and the sum is taken. What is the probability that the resulting sum is between 11815 and 11840 millimeters?

The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution times the number of samples, so the mean is (280)(42)=11760. The standard deviation is equal to the original standard deviation multiplied by the square root of the sample size. So, the standard deviation is (12)(42−−√)≈77.769. To find the probability using the Standard Normal Table, we find that the z-scores for the two values, 11815 and 11840, are 0.71 and 1.03 respectively, using the formula z=x−μσ. Using the Standard Normal Table, the area to the left of z=0.71 is 0.7611, and the area to the left of z=1.03 is 0.8485. 0.8485−0.7611=0.0874, so the probability is about 9%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(11815, 11840, 11760, 1242−−√), which gives us a result of 0.0879.

The lengths, in inches, of adult corn snakes have an unknown distribution with mean 56 and standard deviation 10 inches. A sample, with size n=46, is randomly drawn from the population and the sum is taken. What is the probability that the sum is less than 2516 inches?

The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution times the number of samples, so the mean is (56)(46)=2576. The standard deviation is equal to the original standard deviation multiplied by the square root of the sample size. So, the standard deviation is (10)(46−−√)≈67.823. To find the probability using the Standard Normal Table, we find that the z-score for the value 2516 is −0.8847, using the formula z=x−μσ. Using the Standard Normal Table, the area to the left of z=−0.8847 is 0.1894, so the probability is about 19%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(−1010, 2516, 2576, (10)(46−−√)), which gives us a result of 0.1882.

The population of per capita personal income for a certain year in the United States has a mean of μ=$50,000 and a standard deviation of σ=$8,000. An economist is interested in how the spread of incomes is affected by sampling a different number of working adults in the United States. He calculates that, for 200 adults, the mean and standard deviation are: μx¯=μn=50,000200=$250 and σx¯=σn−−√=8,000200−−−√=$565.69 What did he do wrong?

The economist divided μ by n, but didn't need to do anything to μ. The answer should be: μx¯=$50,000 and σx¯=$565.69. The Central Limit Theorem tells us that, if n≥30, then μx¯=μ and σx¯=σn√. Here, n=200, so the Central Limit Theorem applies. Then μx¯=μ=$50,000 and σx¯=σn√=8,000200√=$565.69.

The average car dealership for a particular brand of car sells 15 vehicles each week, with a standard deviation of 5. The car brand's distribution center wants to analyze how many cars to ship to its 50 dealerships in Florida. An employee in charge of distribution wants to find the sample mean and standard deviation to determine how many cars to send to the 50 dealerships. He calculates that: μx¯=μ=15 and σx¯=σn=0.10 What did the employee do wrong?

The employee divided σ by n instead of n−−√. The answer should be: μx¯=15 and σx¯=0.71 The Central Limit Theorem tells us that, for n≥30, μx¯=μ and σx¯=σn√. Here, n=50 so the Central Limit Theorem applies. The employee remembered that μx¯=μ, but forgot to divide σ by the n−−√. The correct answer is: μx¯=15 and σx¯=0.71.

Finding the Mean and Standard Deviation of the CLT for Means

The first alternative is the Central Limit Theorem for Means. This theorem says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten dice) and calculate their means, the sample means will form their own normal distribution (the "bell curve"). There are two important parameters to this formed normal distribution, for the CLT for Means: The normal distribution has the same mean as the original distribution. The normal distribution has a standard deviation that is equal to the original standard deviation (σ) divided by the square root of the sample size (n−−√). NOTE: The variable n is the number of values that are averaged together, not the number of times the experiment is done. Suppose X is a random variable with some distribution (that may be known or unknown). This is the original distribution. Using a subscript that matches the random variable, suppose the: mean of X=μX standard deviation of X=σX Call the sampling distribution of the mean, the random variable X¯. If you draw random samples of size n, then as n increases (or as the samples get larger), the distribution of the random variable, X¯ which consists of sample means, tends to be normally distributed. We can write this as: X¯∼N(μX,σXn√), where X¯ can be simulated as a normal distribution with a mean, μX, and a standard deviation of σXn√.

The histogram above shows the sampling distribution for samples of size n=2. But would happen to the histogram's shape if we took increasingly larger samples? Move the circle on the slider to see what happens to the sampling distribution as the sample size, n, increases. What do you notice about the sampling distribution's histogram?

The histogram's bars cluster more towards the middle of the graph: As n increases, the histogram's bars get closer together in the middle of the graph. This is because as n increases, the sample means are getting closer and closer together. Another way of saying this is that the standard deviation of the sampling distribution is decreasing. The histogram's bars increase in height: Remember that the histogram's bars measure the frequency with which sample means fall in that specific interval. So, as the bars cluster towards the middle and also increase in height, it shows that the sample means are getting closer and closer together. The histogram's shape also follows more closely with the orange line, the normal curve: This means that as n increases, the sampling distribution is more closely approximating the normal distribution. This is a very important consequence!

A manufacturing company wants to study the rate of production for lumber siding that is produced in one of their factories. Past data reveals that the production of ten boxes of lumber siding has a mean of 29 days with standard deviation 9 days. The normal distribution for the population is shown by the dotted black line. The company plans to take a random sample of 34 such batches of lumber siding and will calculate the mean production period of the sample to compare to the known production periods. Compute the the mean and standard deviation of the sampling distribution of sample means for a sample of size 34. Round your answers to the nearest tenth. Show your answer by moving the two draggable points to build the sampling distribution.

The population has mean μ=29 and standard deviation σ=9. This distribution is shown with the black dotted line. We are asked for the mean and standard deviation of the sampling distribution for a sample of size 34. Central Limit Theorem states that the sample mean of a sample of size n is normally distributed with mean μx¯=μ and σx¯=σn√. In our case, we have μ=29, σ=9, and n=34. So, μx¯=29 and σx¯=934−−√=1.5. This distribution is shown with the red solid line. Notice the sampling distribution, which represents the sample mean of random values of the population, has the same mean as the population distribution. However, the sample mean will vary less than a random value from the population, and therefore has a smaller standard deviation.

A textile manufacturing company wants to study the rate of production for a merino wool blanket that is produced in one of their textile facilities. Past data reveals that the production of one batch of blankets has a mean of 36 days with standard deviation 10 days. The normal distribution for the population is shown by the dotted black line. The company plans to take a random sample of 32 such batches of blankets and will calculate the mean production period of the sample to compare to the known production periods. Compute the the mean and standard deviation of the sampling distribution of sample means for a sample of size 32. Round your answers to the nearest tenth. Show your answer by moving the two draggable points to build the sampling distribution.

The population has mean μ=36 and standard deviation σ=10. This distribution is shown with the black dotted line. We are asked for the mean and standard deviation of the sampling distribution for a sample of size 32. Central Limit Theorem states that the sample mean of a sample of size n is normally distributed with mean μx¯=μ and σx¯=σn√. In our case, we have μ=36, σ=10, and n=32. So, μx¯=36 andσx¯=1032−−√=1.8. This distribution is shown with the red solid line. Notice the sampling distribution, which represents the sample mean of random values of the population, has the same mean as the population distribution. However, the sample mean will vary less than a random value from the population, and therefore has a smaller standard deviation. FEEDBACK

An accounting manager wants to study how many accountants work at least 45 hours per week. Past data reveals that the hours worked for an accountant have a mean of 40 with standard deviation 10 hours. The normal distribution for the population is shown by the dotted black line. The manager plans to take a random sample of 30 such accountants and will calculate the mean hours worked of the sample to compare to the known hours worked. Compute the the mean and standard deviation of the sampling distribution of sample means for a sample of size 30. Round your answers to the nearest tenth. Show your answer by moving the two draggable points to build the sampling distribution.

The population has mean μ=40 and standard deviation σ=10. This distribution is shown with the black dotted line. We are asked for the mean and standard deviation of the sampling distribution for a sample of size 30. Central Limit Theorem states that the sample mean of a sample of size n is normally distributed with mean μx¯=μ and σx¯=σn√. In our case, we have μ=40, σ=10, and n=30. So, μx¯=40 and σx¯=1030−−√=1.8. This distribution is shown with the red solid line. Notice the sampling distribution, which represents the sample mean of random values of the population, has the same mean as the population distribution. However, the sample mean will vary less than a random value from the population, and therefore has a smaller standard deviation.

An online marketing agency wants to study the rates that other marketing agencies charge for services. Past data reveals that the hourly charge for a social media marketing service has a mean of $50 with standard deviation $16. The normal distribution for the population is shown by the dotted black line. The agency plans to take a random sample of 37 such services and will calculate the mean hourly rate of the sample to compare to the known hourly rate. Compute the the mean and standard deviation of the sampling distribution of sample means for a sample of size 37. Round your answers to the nearest tenth. Show your answer by moving the two draggable points to build the sampling distribution.

The population has mean μ=50 and standard deviation σ=16. This distribution is shown with the black dotted line. We are asked for the mean and standard deviation of the sampling distribution for a sample of size 37. Central Limit Theorem states that the sample mean of a sample of size n is normally distributed with mean μx¯=μ and σx¯=σn√. In our case, we have μ=50, σ=16, and n=37. So, μx¯=50 and σx¯=1637−−√=2.6. This distribution is shown with the red solid line. Notice the sampling distribution, which represents the sample mean of random values of the population, has the same mean as the population distribution. However, the sample mean will vary less than a random value from the population, and therefore has a smaller standard deviation.

Mary, an entrepreneur, started a shoe store online and wants to compare her online startup costs to other similar businesses. Past data reveals that similar business have a mean startup cost of $81 with standard deviation $25 (in hundreds of dollars). The normal distribution for the population is shown by the dotted black line. Mary plans to take a random sample of 150 similar businesses and will calculate the mean startup costs of the sample to compare to the known startup costs. Compute the the mean and standard deviation of the sampling distribution of sample means for a sample of size 150. Round your answers to the nearest tenth. Show your answer by moving the two draggable points to build the sampling distribution.

The population has mean μ=81 and standard deviation σ=25. This distribution is shown with the black dotted line. We are asked for the mean and standard deviation of the sampling distribution for a sample of size 150. Central Limit Theorem states that the sample mean of a sample of size n is normally distributed with mean μx¯=μ and σx¯=σn√. In our case, we have μ=81, σ=25, and n=150. So, μx¯=81 and σx¯=25150−−−√=2.0. This distribution is shown with the red solid line. Notice the sampling distribution, which represents the sample mean of random values of the population, has the same mean as the population distribution. However, the sample mean will vary less than a random value from the population, and therefore has a smaller standard deviation.

Casino regulators in Las Vegas find that the game of craps has the best odds of winning. In one day, 300 craps players on average per casino reported winning some money, with a standard deviation of 45. The regulators want to compare how many players win, on average, over 31 days. They calculate: μx¯=μn−−√=30031−−√=53.88 and σx¯=σn−−√=4531−−√=8.08 What did they do wrong?

The regulators incorrectly divided μ by n−−√. The correct answer should be: μx¯=300 and σx¯=8.08 The Central Limit Theorem tells us, for n≥30, μx¯=μ and σx¯=σn√. Here, the regulators have incorrectly divided μ by n−−√. You should have μx¯=μ=300. The regulators have σx¯ correct. So the correct answer is: μx¯=300 and σx¯=8.08

The student test scores for a nationwide standardized test have an unknown distribution with mean 233 and standard deviation 40 points. A sample, with size n=43, is randomly drawn from the population and the values are added together. Using the Central Limit Theorem for Sums, what is the mean for the sample sum distribution? Give just a number for your answer, and do not include the units.

The student test scores for a nationwide standardized test have an unknown distribution with mean 233 and standard deviation 40 points. A sample, with size n=43, is randomly drawn from the population and the values are added together. Using the Central Limit Theorem for Sums, what is the mean for the sample sum distribution? Give just a number for your answer, and do not include the units.

The average one-bedroom apartment rents for $2,000 a month in Washington, D.C., with a standard deviation of $400. A real estate agent is looking to find a one-bedroom apartment for a client that is in the top 15% of all rentals in the city. How much should a client expect to pay for a one-bedroom apartment in the top 15% of rentals if the agent only surveys 50 apartments in the city? Use the z-table below

The top 15% is the complement of the bottom 85%. Looking at the standard normal probabilities table, the z-score value that corresponds to 85% would be z=1.04 σx¯ would be equal to the standard deviation divided by the square root of n. σx¯=4005-√0=$56.57 By plugging all the numbers into the formula z=x¯−μσx¯ we find that 1.04=x¯−$2,000$56.57 $58.83=x¯−$2,000 $2,058.83=x¯ The client should expect to pay at least $2,058.83 for an apartment in the top 15% of all rentals in the city.

The body temperature of men is normally distributed with an average of 97.49 degrees Fahrenheit and a standard deviation of 0.43 degrees Fahrenheit. If a sample of 9 men is selected, what will the average body temperature be for the sample? What will be the standard deviation?

We are given population mean μ=97.49 and population standard deviation σ=0.43, and want to find the mean and standard error of the sampling distribution, μx¯ and σx¯. By the Central Limit Theorem, the means of the two distributions are the same: μx¯=μ=97.49 To find the Standard Deviation of the sampling distribution, we divide the population standard deviation by the square root of the sample size: σx¯=σn−−√=0.439-√=0.14

A company is offering 401k matching for its employees who stay with the company for more than 10 years. The company's CFO finds that the average retirement account holds $490,000, with a standard deviation of $55,000, distributed normally. What amount of money separates the lowest 20% of the means of retirement accounts from the highest 80% in a sampling of 80 employees? Use the z-table below:

We know the separating percentage, so we can find the separating length by working backward (from percentage to the z-score in a standard normal distribution that separates the lowest 20% from the highest 80%) 20% in a standard normal distribution corresponds to a z-score of −0.85. The mean of the sampling distribution is also $490,000 since the distribution is normal. The standard deviation of the sampling distribution of the means is σx¯¯¯=σn−−√=$55,00080−−√=$6,149.19 Using the formula z=x¯¯¯−μx¯σx¯ to find the sample mean x¯¯¯, we get −0.85x¯¯¯=≈x¯¯¯−$490,000$6,149.19$484,773.19 If the area under the sampling distribution is 0.20 units, the value that separates this area from the other 0.80 units is approximately $484,773.19.

A tire manufacturer is measuring the margin of error in the thickness of their tires to make sure it is within safety limits. Overall, the tires' thickness is normally distributed with a mean of 0.45 inches and a standard deviation of 0.05 inches. What thickness separates the lowest 5% of the means from the highest 95% in a sample size of 65 tires? Use the z-table below: Round the z-score and x¯ to two decimal places.

We know the separating percentage, so we can find the separating thickness by working backward (from percentage to the z-score in a standard normal distribution that separates the lowest 5% from the highest 95%). 5% in a standard normal distribution corresponds to a z-score of −1.65. The mean of the sampling distribution is also 0.45 cm. The standard deviation of the sampling distribution of the means is σx¯¯¯=σn−−√=0.0565−−√=0.01. Using the formula z=x¯¯¯−μx¯σx¯ to find the sample mean x¯¯¯, we get −1.65x¯¯¯=≈x¯¯¯−0.450.010.43 If the area under the sampling distribution is 0.05 units, the value that separates this area from the other 0.95 units is approximately 0.43 inches.

Define and Apply Sampling Distribution for the Sample Mean

We often want to be able to use statistics to support studies or experiments in our chosen fields. For example, a business owner may want to present evidence of the growth of his business to his shareholders, or a research scientist may want to show his colleagues proof of the success of new patient treatments. It is often difficult to analyze an entire population for a statistical study, so we use "large enough" random samples and statistical inference to analyze a characteristic (mean, standard deviation) and make conclusions to support or refute our claims.

The Central Limit Theorem for Means

When the population is normally distributed, it is not surprising that the sampling distribution of the sample means will also be normally distributed. The Central Limit Theorem makes an even stronger assertion: for large enough sample sizes, the sampling distribution will be approximately normal for any population, regardless of the shape of population distribution. This holds true, for large enough n, even when the population distribution may be skewed left or right, or is otherwise non-symmetric. Suppose a simple random samples of size n is drawn from a population with mean μ and standard deviation σ. Then the sampling distribution of the sample mean, x¯, is approximately a normal distribution, with Mean: μx¯=μ Standard deviation: σx¯=σn−−√ For any population distribution, provided n≥30, the sampling distribution becomes closer to normal as n increases. When the population itself has a normal distribution, the sampling distribution is normal for any sample size n. What The Central Limit Theorem tells us is that if you have any distribution (i.e. Skewed right, Skewed left) with a mean μ and standard deviation σ, then the sample mean of x¯¯¯ with a sample size of n will have an approximately normal distribution. How large is "large enough" depends on the distribution, but typically n≥30 will assure an approximately normal sampling distribution for any population distribution. This approximation will improve as the sample size increases: the larger the sample size, the closer to normal the sampling distribution of the sample mean will become. As you will see, the near-normality of all sampling distributions asserted by the Central Limit Theorem is what will allow us to make inferences and test hypotheses later in the course.

Introduction to the Central Limit Theorem

Why are we so concerned with means? Two reasons are: they give us a middle ground for comparison, and they are easy to calculate. The central limit theorem (CLT) is one of the most powerful and useful ideas in all of statistics. There are two alternative forms of the theorem - the Central Limit Theorem for Means, and the Central Limit Theorem for Sums. Both are concerned with drawing finite samples (with size n) from a population with a known mean (μ) and standard deviation (σ). The important fact about the CLT is that the distribution of sample means tend to follow the normal distribution, no matter what the distribution of the original population is. The sample size, n, that is required in order to be "large enough," depends on the original population from which the samples are drawn. But as a rule of thumb, the sample size should be at least 30, or the data should have originally come from a normal distribution. If the original population is far from normal, then more observations are needed for the sample means or sums to be normal. Sampling is done with replacement.

The average time it takes a certain brand of ibuprofen to start working is 25 minutes, with a standard deviation of 13 minutes, distributed normally. A pharmacist randomly samples 20 pills from this brand, because she is researching different brands in order to find the quickest acting ibuprofen to recommend to her customers. Identify the following to help her make her recommendations, rounding to the nearest hundredth if necessary:

μ=​ $$25 minutes σ=​ $$13 minutes n=​ $$20 μx​=​ $$25 minutes σx​=​ $$2.91 minutes

A sociologist is studying the number of years of education of students whose mothers have bachelor's degrees or higher. The data is normally distributed with a population mean of 14.5 years and a population standard deviation of 2.5 years. If a sample of 55 students is selected at random from the population, select the mean and standard deviation of the sampling distribution below.

σx¯=0.34 years μx¯=14.5 years The standard deviation of the sampling distribution σx¯=σn√=2.555√=0.34 years. When the distribution is normal, the mean of the sampling distribution is equal to the mean of the population μx¯=μ=14.5 years.