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Knewton Alta - Chapter 5 - Continuous Random Variables

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An organization has members who possess IQs in the top 4% of the population. If IQs are normally distributed, with a mean of 100 and a standard deviation of 15, what is the minimum IQ required for admission into the organization? Use Excel, and round your answer to the nearest integer.

126 =NORM.INV(0.96,100,15)

Question: The weight of a 500-gram box of cereal is normally distributed with a mean of 503 grams and A standard deviation of 5 grams. Find the highest weight of a box such that all the weights less than this unknown weight represent the lower 75% of the distribution.

Solution: Here, the probability to the left of x is 75%=0.75, the mean is 503, and the standard deviation is 5. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.75,503,5), and press ENTER. Thus, the answer rounded to one decimal place is 506.4. Thus, the weights of boxes under 506.4 grams represent the lower 75% of the distribution.

Uniform distribution

happens when each of the values within an interval are equally likely to occur, so each value has the exact same probability as the others over the entire interval givenA Uniform distribution may also be referred to as a Rectangular distribution

An appliance dealer's website features the mean life expectancy values for major household appliances. A gas range's mean life expectancy is 19 years with a standard deviation of 2 years. Using the Empircal Rule, for what range of years are 68% of gas ranges expected to last?

Answer Explanation Correct answers:$\text{68% of the gas ranges are expected to last }\ 17\ \text{ to }21.$68% of the gas ranges are expected to last 17 to 21. The Empiral Rule states that 68% of data falls within one standard deviation from the mean. Calculate the value of one standard deviation below the mean, and one standard deviation above the mean to find the range. 1⋅σ 1⋅2=2 19+2=21 19−2=17 68% of data falls in the range of 17 to 21 years.

After collecting the data, Maria finds that the employee turnover for a certain company is normally distributed with mean 14 employees and standard deviation 2 employees. What is the probability that in a randomly selected company, the employee turnover is less than 8 employees? Enter your answer as a percent rounded to 2 decimal places if necessary. Include the percent symbol % in your answer.

Answer Explanation Correct answers:$0.15\%$0.15% Notice that 8 employees is three standard deviations less than the mean. Based on the Empirical Rule, 99.7% of the company's employee turnover is within three standard deviations of the mean. Since the normal distribution is symmetric, this implies that 0.15% of the company's employee turnover is less than three standard deviation less than the mean.

The United Nations stores statistics about all the live births in the United States. The latest data concerns the birthweights of infants born in 2015. The μ is 3056 grams and the σ is 514 grams. Your neighbor's baby had a birthweight of 1000 grams. How many standard deviations is that away from the mean?

Correct answers: This means that x=1,000 grams is 4 standard deviation(s) below or to the left of the mean, μ=3056 grams. z=x−μσ z=1000−3056514 z=2056/514 z=−4 So, x=1,000 grams is 4 standard deviations below or to the left of the mean.

A worn, poorly set-up machine is observed to produce components whose length X follows a normal distribution with mean 14 centimeters and standard deviation 3 centimeters. Calculate the probability that the length of a component lies between 19 and 21 centimeters. Round your answer to four decimal places.

Correct answers:$0.0380$0.0380 The mean is μ=14, and the standard deviation is σ=3. As the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 21 for X, 14 for Mean, 3 for Standard_dev, and TRUE for Cumulative, all for the higher value of X. Thus, the answer, rounded to four decimal places, is 0.9902. 5. Click on any other empty cell. Click Insert function, fx. 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of NORMDIST function, enter 19 for X, 14 for Mean, 3 for Standard_dev, and TRUE for Cumulative, all for the lower value of X. Thus, the answer, rounded to four decimal places, is 0.9522. Now subtract, 0.9902−0.9522=0.0380. Thus, the probability that the length of a component lies between 19 and 21 centimeters is 0.0380.

A multiple-choice test consists of 160 questions with possible answers of A, B, C, and D. Estimate the probability that with random guessing, the number of correct answers is between 50 and 55, inclusive. Use Excel to find the probability. Round your answer to four decimal places.

Correct answers:$0.0391$0.0391 Notice that n=160 and p=14=25%=0.25. So, np=160×0.25=40 and n(1−p)=160×(1−0.25)=160×0.75=120 are both greater than 5, which means that the normal distribution can be used to approximate the binomial. Because the probabilities of fewer than 50 and at most 55 have to be calculated, the values of interest are 50−0.5=49.5 and 55+0.5=55.5, as the probability includes 50 and 55. Use Excel to calculate the probability. Because the probability between two values is to be calculated, subtract the cumulative probability of 49.5 from the cumulative probability of 55.5. In this case, use the NORM.DIST function twice. 1. Open Excel and click on an empty cell. Enter =NORM.DIST(55.5,160∗0.25,SQRT(160∗0.25∗0.75),1)−NORM.DIST(49.5,160∗0.25,SQRT(160∗0.25∗0.75),1) and press ENTER. The probability of correctly answering between 50 and 55 questions, inclusive, is approximately 0.0391.

The probability of winning on a certain slot machine is 7%. If the slot machine is played 500 times, find the probability of winning exactly 30 times. Use Excel to find the probability. Round your answer to four decimal places.

Correct answers:$0.0476$0.0476 Here, n=500 and p=7%=0.07. So, np=500×0.07=35 and n(1−p)=500×(1−0.07)=500×0.93=465 are both greater than 5, which means that the normal distribution can be used to approximate the binomial. Because the probability of exactly 30 has to be calculated, the values of interest are 30+0.5=30.5 and 30−0.5=29.5. Because the probability between two values is to be calculated, subtract the cumulative probability of 29.5 from the cumulative probability of 30.5. In this case, use the NORM.DIST function twice. 1. Open Excel and click on an empty cell. Enter =NORM.DIST(30.5,500∗0.07,SQRT(500∗0.07∗0.93),1)−NORM.DIST(29.5,500∗0.07,SQRT(500∗0.07∗0.93),1) and press ENTER. The probability, rounded to four decimal places, is 0.0476. Thus, the probability of winning exactly 30 times is approximately 0.0476.

On average, 28 percent of 18 to 34 year olds check their social media profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a random variable X, which has a standard deviation of five percent. Find the probability that the percent of 18 to 34 year olds who check social media before getting out of bed in the morning is, at most, 32. Round your answer to four decimal places.

Correct answers:$0.7881$0.7881 The mean is μ=28, and the standard deviation is σ=5. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of the NORMDIST function, enter 32 for X, 28 for Mean, 5 for Standard_dev, and TRUE for Cumulative. Thus, the answer, rounded to four decimal places, is P(X<32)≈0.7881

The average speed of a car on the highway is 85 kmph with a standard deviation of 5 kmph. Assume the speed of the car, X, is normally distributed. Find the probability that the speed is less than 80 kmph. Round your answer to four decimal places.

Correct answers:$0.1587$0.1587 The mean is μ=85, and the standard deviation is σ=5. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 80 for X, 85 for Mean, 5 for Standard_dev, and TRUE for Cumulative. Thus, the answer, rounded to four decimal places, is P(X<80)≈0.1587.

According to a census, 2% of all births in a country are twins. If there are 2200 births in one month, calculate the probability that more than 50 births in one month would result in twins. Use Excel to find the probability. Round your answer to four decimal places.

Correct answers:$0.1611$0.1611 Here, n=2200 and p=2%=0.02. So, np=2200×0.02=44 and n(1−p)=2200×(1−0.02)=2200×0.98=2156 are both greater than 5, which means that the normal distribution can be used to approximate the binomial. Because the probability of more than 50 has to be calculated, the value of interest is 50+0.5=50.5, as the probability does not include 50. Use Excel to calculate the probability. 1. Open Excel and click on an empty cell. Enter =NORM.DIST(50.5,2200∗0.02,SQRT(2200∗0.02∗0.98),1) and press ENTER. The answer, rounded to four decimal places, is 0.8389. Because the probability to the right of 50.5 has to be calculated, subtract, 1−0.8389=0.1611. Thus, the probability that more than 50 births in one month would result in twins is approximately 0.1611.

A tire company finds the lifespan for one brand of its tires is normally distributed with a mean of 47,500 miles and a standard deviation of 3,000 miles. What mileage would correspond to the the highest 3% of the tires? Use Excel, and round your answer to the nearest integer.

Correct answers:$53142$53142 Here, the mean, μ, is 47,500 and the standard deviation, σ, is 3,000. Let x be the minimum number of miles for a tire to be in the top 3%. The area to the right of x is 3%=0.03. So, the area to the left of x is 1−0.03=0.97. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.97,47500,3000) and press ENTER. The answer, rounded to the nearest integer, is x≈53,142. Thus, the approximate number of miles for the highest 3% of the tires is 53,142 miles.

Dakota is a quality control manager for a manufacturing plant that produces batteries for tablet computers. Based on prior history, Dakota knows that 0.6% of all tablet computer batteries produced are defective. He selects a random sample of n=1,300 tablet computer batteries from a recently produced lot and tests each battery to determine whether it is defective. What is the probability that Dakota finds 6 or fewer defective batteries? Use Excel to find the probability. Round your answer to four decimal places.

Correct answers:$0.3203$0.3203 Here, n=1,300 and p=0.6%=0.006. So, np=1,300×0.006=7.8 and n(1−p)=1,300×(1−0.006)=1,300×0.994=1,292.2 are both greater than 5, which means that the normal distribution can be used to approximate the binomial. Because the probability of at most 6 has to be calculated, the value of interest is 6+0.5=6.5, as the probability includes 6. Use Excel to calculate the probability. 1. Open Excel and click on an empty cell. Enter =NORM.DIST(6.5,1300∗0.006,SQRT(1300∗0.006∗0.994),1) and press ENTER. The answer, rounded to four decimal places, is 0.3203. Thus, the probability that Dakota finds 6 or fewer defective batteries in the sample is approximately 0.3203.

The weight of bags of green landscaping gravel, X, is modeled by normal distribution with a mean 26.7 kilograms and standard deviation 0.3 kilogram. Determine the probability that a randomly selected bag of green gravel will weigh between 26.5 and 27.5 kilograms. Round your answer to four decimal places.

Correct answers:$0.7437$0.7437 The mean is μ=26.7, and the standard deviation is σ=0.3. As the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 27.5 for X, 26.7 for Mean, 0.3 for Standard_dev, and TRUE for Cumulative, all for the higher value of X. Thus, the answer, rounded to four decimal places, is 0.9962. 5. Click on any other empty cell. Click Insert function, fx. 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of NORMDIST function, enter 26.5 for X, 26.7 for Mean, 0.3 for Standard_dev, and TRUE for Cumulative, all for the lower value of X. Thus, the answer, rounded to four decimal places, is 0.2525. Now subtract, 0.9962−0.2525=0.7437. Thus, the probability that a randomly selected bag of green gravel will weigh between 26.5 and 27.5 kilograms is 0.7437.

A worn, poorly set-up machine is observed to produce components whose length X follows a normal distribution with mean 14 centimeters and variance 9. Calculate the probability that a component is at least 12 centimeters long. Round your answer to four decimal places.

Correct answers:$0.7475$0.7475 The mean is μ=14, and the standard deviation is σ=9-√=3. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 12 for X, 14 for Mean, 3 for Standard_dev, and TRUE for Cumulative. The probability, rounded to four decimal places, is P(X<12)≈0.2525. The desired probability is P(X≥12), so subtract from 1 to get P(X≥12)=1−0.2525=0.7475.

A professional basketball player made 51.6% of field goals in one season. If 110 field goals are randomly selected from the season, what is the probability that more than 52 were made? Use Excel to find the probability. Round your answer to four decimal places.

Correct answers:$0.7918$0.7918 Here, n=110 and p=51.6%=0.516. So, np=110×0.0516=56.76 and n(1−p)=110×(1−0.516)=200×0.484=53.24 are both greater than 5, which means that the normal distribution can be used to approximate the binomial. Because the probability of more than 52 has to be calculated, the value of interest is 52+0.5=52.5, as the probability does not include 52. Use Excel to calculate the probability. 1. Open Excel and click on an empty cell. Enter =NORM.DIST(52.5,110∗0.516,SQRT(110∗0.516∗0.484),1) and press ENTER. The output, rounded to four decimal places, is 0.2082. Because the probability to the right of 52.5 has to be calculated, subtract 0.2082 from 1. 1−0.2082=0.7918 Thus, the probability that more than 52 field goals in the sample were made is approximately 0.7918.

The time, X minutes, taken by Tim to install a satellite dish is assumed to be a normal random variable with mean 127 and standard deviation 20. Determine the probability that Tim will takes less than 150 minutes to install a satellite dish. Round your answer to four decimal places.

Correct answers:$0.8749$0.8749 The mean is μ=127, and the standard deviation is σ=20. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 150 for X, 127 for Mean, 20 for standard_dev, and TRUE for Cumulative. Thus, the answer, rounded to four decimal places, is P(X<150)≈0.8749.

At a store, 40% of all the customers pay for their purchases with a credit card. If a random sample of 300 customers is selected, what is the approximate probability that at most 140 customers pay with a credit card? Use the normal distribution to approximate the binomial distribution. Round your answer to four decimal places.

Correct answers:$0.9922$0.9922 Here, n=300 and p=40%=0.4. So, np=300×0.4=120 and n(1−p)=300×(1−0.4)=300×0.6=180 are both greater than 5, which means that the normal distribution can be used to approximate the binomial. As the probability of at most 140 has to be calculated, the value of interest is 140+0.5=140.5, as the probability does include 140. Use Excel to calculate the probability. 1. Open Excel and click on an empty cell. Enter =NORM.DIST(140.5,300∗0.4,SQRT(300∗0.4∗0.6),1) and press ENTER. The output, rounded to four decimal places, is 0.9922. Thus, the probability that at most 140 customers paid by credit card is 0.9922.

The weights of oranges are normally distributed with a mean of 12.4 pounds and a standard deviation of 3 pounds. Find the minimum value that would be included in the top 5% of orange weights. Use Excel, and round your answer to one decimal place.

Correct answers:$17.3$17.3 Here, the mean, μ, is 12.4 and the standard deviation, σ, is 3. Let x be the minimum value that would be included in the top 5% of orange weights. The area to the right of x is 5%=0.05. So, the area to the left of x is 1−0.05=0.95. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.95,12.4,3) and press ENTER. The answer, rounded to one decimal place, is x≈17.3. Thus, the minimum value that would be included in the top 5% of orange weights is 17.3 pounds.

A credit card company receives numerous phone calls throughout the day from customers reporting fraud and billing disputes. Most of these callers are put "on hold" until a company operator is free to help them. The company has determined that the length of time a caller is on hold is normally distributed with a mean of 2.5 minutes and a standard deviation 0.5 minutes. If 1.5% of the callers are put on hold for longer than x minutes, what is the value of x? Use Excel, and round your answer to two decimal places.

Correct answers:$3.59$3.59 Here, the mean, μ, is 2.5 and the standard deviation, σ, is 0.5. The area to the right of x is 1.5%=0.015. The area to the left of x is 1−0.015=0.985. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.985,2.5,0.5) and press ENTER. The answer, rounded to two decimal places, is x≈3.59. Thus, 1.5% of the callers are put on hold for longer than 3.59 minutes.

The number of walnuts in a mass-produced bag is modeled by a normal distribution with a mean of 44 and a standard deviation of 5. Find the number of walnuts in a bag that has more walnuts than 80% of the other bags. Use Excel, and round your answer to the nearest integer.

Correct answers:$48$48 Here, the mean, μ, is 44 and the standard deviation, σ, is 5. Let x be the number of walnuts in the bag. The area to the left of x is 80%=0.80. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.80,44,5) and press ENTER. The answer, rounded to the nearest integer, is x≈48. Thus, there are 48 walnuts in a bag that has more walnuts than 80% of the other bags.

Suppose that the weight of navel oranges is normally distributed with a mean of μ=6 ounces and a standard deviation of σ=0.8 ounces. Find the weight below that one can find the lightest 90% of all navel oranges. Use Excel, and round your answer to two decimal places.

Correct answers:$7.03$7.03 Here, the mean, μ, is 6 and the standard deviation, σ, is 0.8. The area to the left of x is 90%=0.90. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.90,6,0.8) and press ENTER. The answer, rounded to two decimal places, is x≈7.03. Thus, navel oranges that weigh less than 7.03 ounces compose the lightest 90% of all navel oranges.

The top 5% of applicants on a test will receive a scholarship. If the test scores are normally distributed with a mean of 600 and a standard distribution of 85, how low can an applicant score to still qualify for a scholarship? Use Excel, and round your answer to the nearest integer.

Correct answers:$740$740 Here, the mean, μ, is 600 and the standard deviation, σ, is 85. Let x be the score on the test. As the top 5% of the applicants will receive a scholarship, the area to the right of x is 5%=0.05. So the area to the left of x is 1−0.05=0.95. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.95,600,85) and press ENTER. The answer rounded to the nearest integer, is x≈740. Thus, an applicant can score a 740 and still be in the top 5% of applicants on a test in order to receive a scholarship.

In a survey of men aged 20-29 in a country, the mean height was 73.4 inches with a standard deviation of 2.7 inches. Find the minimum height in the top 10% of heights. Use Excel, and round your answer to one decimal place.

Correct answers:$76.9$76.9 Here, the mean, μ, is 73.4 and the standard deviation, σ, is 2.7. Let x be the minimum height in the top 10% of heights. The area to the right of x is 10%=0.10. So, the area to the left of x is 1−0.10=0.90. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.90,73.4,2.7) and press ENTER. The answer, rounded to one decimal place, is x≈76.9. Therefore, the minimum height in the top 10% of heights is 76.9 inches.

Suppose that the weight, X, in pounds, of a 40-year-old man is a normal random variable with mean 147 and standard deviation 16. Calculate P(120≤X≤153). Round your answer to four decimal places.

Correct answers:$P\left(120\le X\le153\right)=0.6004$P(120≤X≤153)=0.6004 The mean is μ=147, and the standard deviation is σ=16. As the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 153 for X, 147 for Mean, 16 for Standard_dev, and TRUE for Cumulative, all for the higher value of X. Thus, the answer, rounded to four decimal places, is 0.6462. 5. Click on any other empty cell. Click Insert function, fx. 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of NORMDIST function, enter 120 for X, 147 for Mean, 16 for Standard_dev, and TRUE for Cumulative, all for the lower value of X. Thus, the answer, rounded to four decimal places, is 0.0458. Now subtract, 0.6462−0.0458=0.6004. Thus, P(120≤X≤153)=0.6004.

The average number of acres burned by forest and range fires in a county is 4,500 acres per year, with a standard deviation of 780 acres. The distribution of the number of acres burned is normal. What is the probability that between 3,000 and 4,800 acres will be burned in any given year? Round your answer to four decimal places.

Correct answers:$P\left(3,000<X<4,800\right)=0.6225$P(3,000<X<4,800)=0.6225 The mean is μ=4500 and the standard deviation is σ=780. Because the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of the NORMDIST function, enter 4800 for X, 4500 for mean, 780 for Standard_dev, and TRUE for Cumulative, all for the higher value of X. Thus, the answer, rounded to four decimal places, is 0.6497. 5. Click on any other empty cell. Click Insert function, fx. 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of the NORMDIST function, enter 3000 for X, 4500 for mean, 780 for Standard_dev, and TRUE for Cumulative, all for the lower value of X. Thus, the answer, rounded to four decimal places, is 0.0272. Now subtract: 0.6497−0.0272=0.6225. Thus, the probability that between 3,000 and 4,800 acres will be burned in any given year is 0.6225.

A firm's marketing manager believes that total sales for next year will follow the normal distribution, with a mean of $3.2 million and a standard deviation of $250,000. Determine the sales level that has only a 3% chance of being exceeded next year. Use Excel, and round your answer to the nearest dollar.

Correct answers:$\$3670198$$3670198 Here, the mean, μ, is 3.2 million =3,200,000 and the standard deviation, σ, is 250,000. Let x be sales for next year. To determine the sales level that has only a 3% chance of being exceeded next year, the area to the right of x is 0.03. So the area to the left of x is 1−0.03=0.97. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.97,3200000,250000) and press ENTER. The answer, rounded to the nearest dollar, is x≈3,670,198. Thus, the sales level that has only a 3% chance of being exceeded next year is $3,670,198.

The average price of a laptop is $965. Assume laptop prices are approximately normally distributed with a standard deviation of $100. The least expensive 5% of laptops cost less than what amount? Use Excel, and round your answer to two decimal places.

Correct answers:$\$800.51$$800.51 Here, the mean, μ, is 965, and the standard deviation, σ, is 100. Let x be the cost under which the least expensive 5% of laptops cost. The area to the left of x is 5%=0.05. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.05,965,100) and press ENTER. The answer, rounded to two decimal places, is x≈800.51. Thus, the least expensive 5% of laptops cost less than $800.51.

Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean of μ=81 points and a standard deviation of σ=4 points. The middle 50% of the exam scores are between what two values? Use Excel, and round your answers to the nearest integer.

Correct answers:$\left(78,\ 84\right)$(78, 84) Here, the mean, μ, is 81, the standard deviation, σ, is 4, and the probability between two x-values is 50%=0.50. The following figure illustrates the points x1 and x2, where the probability between x1 and x2 is 0.50. As the x1 and x2 are symmetric about the mean, the tail probabilities are calculated using 1−0.502=0.25. The probability to the left of x1 is 0.25. Use Excel to find x1. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.25,81,4) and press ENTER. Rounding to the nearest integer, x1≈78. The probability to the left of x2 is 0.25+0.50=0.75. Use Excel to find x2. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.75,81,4) and press ENTER. Rounding to the nearest integer, x2≈84. Thus, the middle 50% of the exam scores are between 78 and 84.

A group of students with normally distributed salaries earn an average of $6,800 with a standard deviation of $2,500. Find the cutoff for the salary that corresponds to the lower 25% of all salaries. Use Excel, and round your answer to the nearest integer.

Correct answers:$\text{$}5114$$5114 Here, the probability to the left of x is 25%=0.25, the mean is 6,800, and the standard deviation is 2,500. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.25,6800,2500), and press ENTER. Thus, the answer rounded to the nearest integer is 5,114. Therefore, the lowest 25% of all salaries corresponds to salaries of $5,114 or less.

The weights of bags of raisins are normally distributed with a mean of 175 grams and a standard deviation of 11 grams. Bags in the upper 4.5% are too heavy and must be repackaged. Also, bags in the lower 5% do not meet the minimum weight requirement and must be repackaged. What are the ranges of weights for raisin bags that need to be repackaged? Use Excel, and round your answers to the nearest integer.

Correct answers:1$157$1572$194$194 Here, the mean, μ, is 175 and the standard deviation, σ, is 11. Let x1 be the highest weight such that the bags are in the lower 5% and x2 be the lowest weight such that the bags are in the upper 4.5%. The probability below x1 is 5%, and the probability above x2 is 4.5%. If the raisin bags have to be repackaged, then they should fall below x1and above x2. The figure below illustrates the points x1 and x2. The probability between x1 and x2 is 1−0.045−0.05=0.905. The normal distribution curve for the weights of bags of raisins is shown below. As x1 and x2 are to be calculated, first find x1. The probability to the left of x1 is 0.05. Use Excel to find x1. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.05,175,11) and press ENTER. The answer, rounded to the nearest integer, is 157. Thus, x1≈157. To calculate x2, the probability to the left of x2 is 0.05+0.905=0.955. Use Excel to find x2. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.955,175,11) and press ENTER. The answer, rounded to the nearest integer, is 194. Thus, x2≈194. Therefore, the ranges of weights for raisin bags that need to be repackaged are less than 157 grams and more than 194 grams.

The number of miles a motorcycle, X, will travel on one gallon of gasoline is modeled by a normal distribution with mean 44 and standard deviation 5. If Mike starts a journey with one gallon of gasoline in the motorcycle, find the probability that, without refueling, he can travel more than 50 miles. Round your answer to four decimal places.

First find the probability to the left of 50 and subtract from 1. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 50 for X, 44 for Mean, 5 for Standard_dev, and TRUE for Cumulative. This probability, rounded to four decimal places, is 0.8849. Now subtract, 1−0.8849=0.1151. Thus, the desired probability is P(X>50)=0.1151.

Here, the mean, μ, is 100 and the standard deviation, σ, is 15. Let x be the minimum IQ needed to get into the organization. As the top 4% of the population are members of the organization, the area to the right of x is 4%=0.04. So the area to the left of x is 1−0.04=0.96. Use Excel to find x. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.96,100,15) and press ENTER. The answer, rounded to the nearest integer, is x≈126. Thus, the minimum IQ required for admission into the organization is 126.

Here, the mean, μ, is 81, the standard deviation, σ, is 4, and the probability between two x-values is 50%=0.50. The following figure illustrates the points x1 and x2, where the probability between x1 and x2 is 0.50. As the x1 and x2 are symmetric about the mean, the tail probabilities are calculated using 1−0.502=0.25. The probability to the left of x1 is 0.25. Use Excel to find x1. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.25,81,4) and press ENTER. Rounding to the nearest integer, x1≈78. The probability to the left of x2 is 0.25+0.50=0.75. Use Excel to find x2. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.75,81,4) and press ENTER. Rounding to the nearest integer, x2≈84. Thus, the middle 50% of the exam scores are between 78 and 84.

Recall that the binomial distribution with n trials and probability of success p has mean μ=np and standard deviation σ=npq−−−√, where q=1−p. We can denote this distribution with B(n,p). When approximating the binomial distribution with the normal distribution, we use the same mean and standard deviation. That is to say, B(n,p)≈N(np,npq−−−√).

The mean is μ=85, and the standard deviation is σ=5. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 80 for X, 85 for Mean, 5 for Standard_dev, and TRUE for Cumulative. Thus, the answer, rounded to four decimal places, is P(X<80)≈0.1587

At a university, 30% of the incoming freshmen elect to enroll in a statistics course offered by the university. Approximating the binomial distribution with a normal distribution, find the probability that of 600 randomly selected incoming freshmen, at least 200 have elected to enroll in the course.

Solution: Here, n=600 and p=30%=0.3. So, np=600×0.3=180 and n(1−p)=600×(1−0.3)=600×0.7=420 are both greater than 5, which means that the normal distribution can be used to approximate the binomial. As the probability of at least 200 has to be calculated, the value of interest is 200−0.5=199.5, as the probability includes 200. The following image represents the probability that at least 200 incoming freshmen have elected to enroll in the statistics course. Use Excel to calculate the probability. 1. Open Excel and click on an empty cell. Enter =NORM.DIST(199.5,600∗0.3,SQRT(600∗0.3∗0.7),1) and press ENTER. The answer, rounded to four decimal places, is 0.9588. As the probability to the right of 199.5 has to be calculated, subtract, 1−0.9588=0.0412. Thus, the probability that at least 200 incoming freshmen have elected to enroll in the statistics course is 0.0412.

Question: At a university, 30% of the incoming freshmen elect to enroll in a statistics course offered by the university. Approximating the binomial distribution with a normal distribution, find the probability that of 600 randomly selected incoming freshmen, at most 200 have elected to enroll in the course.

Solution: Here, n=600 and p=30%=0.3. So, np=600×0.3=180 and n(1−p)=600×(1−0.3)=600×0.7=420 are both greater than 5, which means that the normal distribution can be used to approximate the binomial. As the probability of at most 200 has to be calculated, the value of interest is 200+0.5=200.5, as the probability does include 200. The following image represents the probability that at most 200 incoming freshmen have elected to enroll in the statistics course. Use Excel to calculate the probability. 1. Open Excel and click on an empty cell. Enter =NORM.DIST(200.5,600∗0.3,SQRT(600∗0.3∗0.7),1) and press ENTER. The answer, rounded to four decimal places, is 0.9661. Thus, the probability that at most 200 incoming freshmen have elected to enroll in the statistics course is 0.9661.

Solution: Here, n=600 and p=30%=0.3. So, np=600×0.3=180 and n(1−p)=600×(1−0.3)=600×0.7=420 are both greater than 5, which means that the normal distribution can be used to approximate the binomial. As the probability of exactly 200 has to be calculated, the value of interests are 200+0.5=200.5 and 200−0.5=199.5. The following image represents the probability that exactly 200 incoming freshmen have elected to enroll in the statistics course. Because the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORM.DIST function twice. 1. Open Excel and click on an empty cell. Enter =NORM.DIST(200.5,600∗0.3,SQRT(600∗0.3∗0.7),1)−NORM.DIST(199.5,600∗0.3,SQRT(600∗0.3∗0.7),1) and press ENTER. The output, rounded to four decimal places, is 0.0073. Thus, the probability that exactly 200 incoming freshmen have elected to enroll in the statistics course is 0.0073.

Solution: Here, n=600 and p=30%=0.3. So, np=600×0.3=180 and n(1−p)=600×(1−0.3)=600×0.7=420 are both greater than 5, which means that the normal distribution can be used to approximate the binomial. As the probability of fewer than 200 has to be calculated, the value of interest is 200−0.5=199.5, as the probability does not include 200. The following image represents the probability that fewer than 200 incoming freshmen have elected to enroll in the statistics course Use Excel to calculate the probability. 1. Open Excel and click on an empty cell. Enter =NORM.DIST(199.5,600∗0.3,SQRT(600∗0.3∗0.7),1) and press ENTER. The answer, rounded to four decimal places, is 0.9588. Thus, the probability that fewer than 200 incoming freshmen have elected to enroll in the statistics course is 0.9588.

Solution: Here, n=600 and p=30%=0.3. So, np=600×0.3=180 and n(1−p)=600×(1−0.3)=600×0.7=420 are both greater than 5, which means that the normal distribution can be used to approximate the binomial. As the probability of more than 200 has to be calculated, the value of interest is 200+0.5=200.5, as the probability does not include 200. The following image represents the probability that more than 200 incoming freshmen have elected to enroll in the statistics course. Use Excel to calculate the probability. 1. Open Excel and click on an empty cell. Enter =NORM.DIST(200.5,600∗0.3,SQRT(600∗0.3∗0.7),1) and press ENTER. The answer, rounded to four decimal places, is 0.9661. As the probability to the right of 200.5 has to be calculated, subtract, 1−0.9661=0.0339. Thus, the probability that more than 200 incoming freshmen have elected to enroll in the statistics course is 0.0339.

Question: The weight of a 500-gram box of cereal is normally distributed with a mean of 503 grams and a standard deviation of 5 grams. Find the weights of boxes if the probability between the two weights corresponds to the middle 66% of the distribution. That is, assume that the two weights are symmetric about the mean of 503 grams.

Solution: Here, the mean, μ, is 503, the standard deviation, σ, is 5, and the probability between two x-values is 66%=0.66. The following figure illustrates x1 and x2, where the probability between x1 and x2 is 0.66. Because the entire graph must add up to 1, the unshaded portion must be equal to 1−0.66, which is 0.34. Each side must account for one half of that amount, so each unshaded region is 0.34/2=0.17. As x1 and x2 are to be calculated, first find x1. The probability to the left of x1 is 0.17. Use Excel to find x1. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.17,503,5), and press ENTER. The answer rounded to one decimal place is 498.2. Thus, x1≈498.2. To calculate x2, the probability to the left of x2 is 0.17+0.66=0.83. Use Excel to find x2. 1. Open Excel. Click on an empty cell. Type =NORM.INV(0.83,503,5), and press ENTER. The answer rounded to one decimal place is 507.8. Thus, x2≈507.8. Therefore, weights of boxes between 498.2 and 507.8 grams correspond to the middle 66% of the distribution.

An appliance dealer's website features the mean life expectancy values for major household appliances. A standard refrigerator's mean life expectancy is 18 years with a standard deviation of 2 years. What is the probability that your standard refrigerator will last between 12 and 20 years? Express the probability as a percentage rounded to the nearest hundredth.

The Empirical Rule states that you can calculate a probability by summing up the % areas under the normal curve by using the 68−95−99 rule. We know that σ=2, and 18+2=20. Therefore, 20 is one standard deviation more than the mean. Based on the Empiral Rule, 68% of data falls within one standard deviation from the mean. Since we are only looking for the upper half, we get 682=34%. Now for 12, we know that σ=2, and 18−2−2−2=12. Therefore, 12 is three standard deviations less than the mean. Based on the Empiral Rule, 99.7% of data falls within three standard deviations from the mean. Since we are only looking for the lower half of that, we get 99.72=49.85%. We now add the two percentages: 34%+49.85%=83.85% That means you have an 83.85% chance of having a standard refrigerator last between 12 and 20 years.

Suppose x has a normal distribution with μ=50 and σ=6. Use the Empirical Rule to find the data that is within 1, 2, and 3 standard deviations of the mean.

The mean is 50, and the standard deviation is 6. So, the data that lies within one standard deviation of 50 (between −1σ and 1σ) will be the data that lies in the range that is 6 units less than 50 and 6 units more than 50. So, The values 50-6=44 and 50+6=56 are within one standard deviation of the mean. About 68% of the x-values lie between 44 and 56. The data that lies within two standard deviations of 50 (between −2σ and 2σ) will be the data that lies in the range that is (6)(2)=12 units less than 50 and (6)(2)=12 units more than 50. So, The values 50-12=38 and 50+12=62 are within two standard deviations of the mean. About 95% of the x-values lie between 38 and 62. The data that lies within three standard deviations of 50 (between −3σ and 3σ) will be the data that lies in the range that is (6)(3)=18 units less than 50 and (6)(3)=18 units more than 50. So, The values 50-18=32 and 50+18=68 are within three standard deviations of the mean. About 99.7% of the x-values lie between 32 and 68.

The mean is μ=14, and the standard deviation is σ=9-√=3. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 12 for X, 14 for Mean, 3 for Standard_dev, and TRUE for Cumulative. The probability, rounded to four decimal places, is P(X<12)≈0.2525. The desired probability is P(X≥12), so subtract from 1 to get P(X≥12)=1−0.2525=0.7475.

Suppose that the weight, X, in pounds, of a 40-year-old man is a normal random variable with mean 147 and standard deviation 16. Calculate P(120≤X≤153). Round your answer to four decimal places.

The mean is μ=147, and the standard deviation is σ=16. As the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 153 for X, 147 for Mean, 16 for Standard_dev, and TRUE for Cumulative, all for the higher value of X. Thus, the answer, rounded to four decimal places, is 0.6462. 5. Click on any other empty cell. Click Insert function, fx. 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of NORMDIST function, enter 120 for X, 147 for Mean, 16 for Standard_dev, and TRUE for Cumulative, all for the lower value of X. Thus, the answer, rounded to four decimal places, is 0.0458. Now subtract, 0.6462−0.0458=0.6004. Thus, P(120≤X≤153)=0.6004.

Question The weight of bags of green landscaping gravel, X, is modeled by normal distribution with a mean 26.7 kilograms and standard deviation 0.3 kilogram. Determine the probability that a randomly selected bag of green gravel will weigh less than 26 kilograms. Round your answer to four decimal places.

The mean is μ=26.7, and the standard deviation is σ=0.3. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 26 for X, 26.7 for Mean, 0.3 for Standard_dev, and TRUE for Cumulative. Thus, the answer, rounded to four decimal places, is P(X<26)≈0.0098.

The mean is μ=26.7, and the standard deviation is σ=0.3. As the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 27.5 for X, 26.7 for Mean, 0.3 for Standard_dev, and TRUE for Cumulative, all for the higher value of X. Thus, the answer, rounded to four decimal places, is 0.9962. 5. Click on any other empty cell. Click Insert function, fx. 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of NORMDIST function, enter 26.5 for X, 26.7 for Mean, 0.3 for Standard_dev, and TRUE for Cumulative, all for the lower value of X. Thus, the answer, rounded to four decimal places, is 0.2525. Now subtract, 0.9962−0.2525=0.7437. Thus, the probability that a randomly selected bag of green gravel will weigh between 26.5 and 27.5 kilograms is 0.7437.

The mean is μ=28, and the standard deviation is σ=5. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of the NORMDIST function, enter 32 for X, 28 for Mean, 5 for Standard_dev, and TRUE for Cumulative. Thus, the answer, rounded to four decimal places, is P(X<32)≈0.7881. FEEDBACK

Sugar canes have lengths, X, that are normally distributed with mean 365.45 centimeters and standard deviation 4.9 centimeters. What is the probability of the length of a randomly selected cane being between 360 and 370 centimeters? Round your answer to four decimal places.

The mean is μ=365.45, and the standard deviation is σ=4.9. As the probability between two values is to be calculated, subtract the probability of the lower value from the higher value. In this case, you have to use the NORMDIST function twice. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 370 for X, 365.45 for Mean, 4.9 for Standard_dev, and TRUE for Cumulative, all for the higher value of X. Thus, the answer, rounded to four decimal places, is 0.8234. 5. Click on any other empty cell. Click Insert function, fx. 6. Search for NORMDIST in the search for a function dialog box and click GO. 7. Make sure NORMDIST is on top in select a function. Then click OK. 8. In the function arguments of NORMDIST function, enter 360 for X, 365.45 for Mean, 4.9 for Standard_dev, and TRUE for Cumulative, all for the lower value of X. Thus, the answer, rounded to four decimal places, is 0.1330. Now subtract, 0.8234−0.1330=0.6904. Thus, the probability of the length of a randomly selected cane being between 360 and 370 centimeters is 0.6904.

A certain type of mango that weighs over 450 grams is considered to be large in size. The weights for this certain type of mango, X grams, are normally distributed with a mean of 400 and a standard deviation of 20. Find the probability that a mango of this type, selected at random, will not be considered as large. Round your answer to four decimal places.

The mean is μ=400, and the standard deviation is σ=20. 1. Open Excel and click on any empty cell. Click Insert function, fx. 2. Search for NORMDIST in the search for a function dialog box and click GO. 3. Make sure NORMDIST is on top in select a function. Then click OK. 4. In the function arguments of NORMDIST function, enter 450 for X, 400 for Mean, 20 for standard_dev, and TRUE for Cumulative. Thus, the answer, rounded to four decimal places, is P(X<450)≈0.9938.

An appliance dealer's website features the mean life expectancy values for major household appliances. A freezer's mean life expectancy is 16 years with a standard deviation of 2 years. Using the Empirical Rule, what is the probability that your freezer will function less than 10 years

We know that σ=2, and 16−2−2−2=10. Therefore, 10 is three standard deviations less than the mean. Based on the Empiral Rule, 99.7% of data falls within three standard deviations from the mean. However, we want to know the probability of the data values being less than 10 years, not falling within three standard deviations of the mean. So, we know that 0.3% is divided into two tails, one on the far left and one on the far right. Only 0.15% of the data lies below 10. That means you only have a 0.15% chance of having a freezer last less than 10 years.

How To Approximate the Binomial Distribution Using the Normal Distribution in Excel Every binomial distribution is completely described by its two parameters number of trials n, and probability of success in each trial p. Each unique normal distribution is completely described by its two parameters, mean μ, and standard deviation σ. For the appropriate ranges of n and p, the normal distribution approximates the binomial distribution by substituting np for μ and np(1−p)−−−−−−−−√ for σ. First compute np and n(1−p) and check the conditions for the random variable. If the conditions are satisfied, compute the mean μ=np, the standard deviation σ=np(1−p)−−−−−−−−√, and identify the value(s) of interest using continuity correction(s). Use the Excel formula NORM.DIST to calculate the desired probability.

1. Open Excel and click on any empty cell. Enter =NORM.DIST( followed by value of interest, mean, standard deviation, cumulative and press ENTER. The syntax is NORM.DIST(X,mean,standard deviation,cumulative). Enter 0 for the cumulative entry to get a value from the probability density function, and enter 1 to get a value from the cumulative distribution function, which is the area under the normal curve to the left of the value of interest.

"Use the empirical rule for normal distributions to estimate probability in business contexts" The number of pages per book on a bookshelf is normally distributed with mean 248 pages and standard deviation 21 pages. What is the probability that a randomly selected book has less than 206 pages?

Answer Explanation Correct answers: probability=2.5% Notice that 206 pages is two standard deviations less than the mean. Based on the Empirical Rule, 95% of the number of pages in the books are within two standard deviations of the mean. Since the normal distribution is symmetric, this implies that 2.5% of the number of pages in the books are less than two standard deviation less than the mean.

You flip a fair coin 16 times. This binomial distribution can be approximated with a normal distribution, N(8,2), where the mean (μ) is 8, and the standard deviation (σ) is 2. Approximating the binomial distribution with a normal curve, what is the probability of flipping 9 or 10 heads in the 16 tosses? You may use this Standard Normal Table below or a calculator.

Correct answer: 0.30 We approximate the binomial with a normal distribution to determine the probability of flipping 9 or 10 heads. We must choose values just to the left of 9 and to the right of 10. So, we can choose 8.5 and 10.5, and compute the area between these two values. First, find the z-scores that correspond with 8.5 and 10.5. Using the formula, z=x−μσ, we compute z-scores of 0.25 and 1.25 that correspond with 8.5 and 10.5 respectively. Using the Standard Normal Table, the area to the left of z=1.25 is 0.8944, and the area to the left of z=0.25 is 0.5987. The area between these two scores is the difference between 0.8944−0.5987, which is 0.2957. So, 0.2957 or about 30% is the probability of flipping 9 or 10 heads in 16 tosses. (Using a calculator, Find the area in between 10.5 and 8.5, using normalcdf (8.5,10.5,8,2), which is approximately 0.2956.)

The amount of time it takes Isabella to solve a Rubik's cube is continuous and uniformly distributed between 5 minutes and 11 minutes. What is the probability that it takes Isabella more than 9 minutes to solve a Rubik's cube?

Correct answer: 0.333 Let X be the time it takes to solve a Rubik's cube. The range of possible values is 5 to 11, so X∼U(5,11). We are interested in values greater than 9. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a). So we find P(X>9)=(11−9)(111−5)=(2)(16)=0.333

The lifespan of a halogen light bulb is exponentially distributed with a mean of 1000 hours. Given that a halogen bulb has already worked for 400 hours, what is the probability that the bulb will continue to work for more than an additional 900 hours?

Correct answer: 0.407 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units.Mathematically, this can be written as P(X>k+t∣X>k)=P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the light bulb will work for more than an additional 900 hours given that it has already worked for 400 hours is P(X>1300|X>400)=P(X>900)=1−(1−e−1/1000(900))≈0.407

"Use the memoryless property of the exponential distribution to compute conditional probability" The time it takes for a person to choose a birthday gift is exponentially distributed with an average value of 30 minutes. Given that it has already taken 12 minutes for a person to choose a birthday gift, what is the probability that it will take more than an additional 22 minutes?

Correct answer: 0.480 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units. Mathematically, this can be written as P(X>k+t∣∣X>k)=P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the time it takes is more than an additional 22 minutes given that it has already taken 12 minutes isP(X>34 | X>12) = P(X>22) = e−1/30(22) ≈0.480

The amount of time it takes Evelyn to solve a Rubik's cube is continuous and uniformly distributed between 3 minutes and 12 minutes. What is the probability that it takes Evelyn more than 8 minutes given that it takes more than 4 minutes for her to solve a Rubik's cube?

Correct answer: 0.500 Let X be the time it takes to solve a Rubik's cube. The range of possible values is 3 to 12, so X∼U(3,12). We are given that it takes more than 4 minutes, so this means that the possible values are restricted to the range 4 to 12. This interval will help us write our probability density function (height of the rectangle). f(x)=1//12−4=18 We are interested in values greater than 8. So the range of desired outcomes is 8 to 12. This will be the length of our base, 12−8=4. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we find P(X>8∣∣X>4)=(12−8)(112−4)=(4)(18)=48=0.500

Josh's performance score in the last employee evaluation was 87. If the performance scores were approximately normally distributed with μ=80 and z=7, what is the standard deviation of the data?

Correct answer: 1 We can work backwards using the z-score formula to find the standard deviation. The problem gives us the values for z, x and μ. Substituting these numbers back into the formula, we get: z=(x−μ)σ 7=(87−80)σ 7=7/σ σ=1

If X is a continuous random variable with the uniform distribution U(5.5,20.5), what is P(X<8)?

Correct answer: 1/6 We are interested in values less than 8. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we findP(X<8)=(8−5.5)(120.5−5.5)=(2.5)(1/15)=1/6

In 2014, the CDC estimated that the mean height for adult women in the U.S. was 64 inches with a standard deviation of 4 inches. Suppose X, height in inches of adult women, follows a normal distribution. Which of the following gives the probability that a randomly selected woman has a height of greater than 68 inches?

Correct answer: 16% We know that σ=4, and 64+4=68. Therefore 68 is one standard deviation more than the mean. Based on the Empirical Rule, 68% of the data is within one standard deviation of the mean. However, we want to know the probability of the data values being greater than 68 inches, not the probability of the data falling within one standard deviation of the mean. We know that the upper half of the data (above 64) represents 50% of the data. We also know that, by the Empirical Rule, one standard deviation more than the mean represents 34% of the data (half of 68). So, to find the probability of the heights of students being greater than 68 centimeters, we need to subtract this section of one standard deviation more than the mean (34%) from the upper half (50%). 50%−34%=16%, so 16% of the women's heights are greater than 68 inches.

The probability density function for the random variable X is shown below. Find P(X>11).

Correct answer: 19 Remember that the probability of a range of values is given by the area under the graph over that range of values. So we want the area under the curve for the values of x greater than 11. A coordinate system has a horizontal x-axis labeled from 0 to 15 in increments of 1 and a vertical y-axis labeled at StartFraction 1 Over 18 EndFraction and StartFraction 1 Over 12 EndFraction. A random probability function f left-parenthesis x right-parenthesis is depicted by two solid line segments. The first solid line segment connects the points left-parenthesis 0 comma StartFraction 1 Over 12 EndFraction right-parenthesis and left-parenthesis 9 comma StartFraction 1 Over 12 EndFraction right-parenthesis. The second solid line segment decreases from left to right connecting the points left-parenthesis 9 comma StartFraction 1 Over 12 EndFraction right-parenthesis to the point left-parenthesis 15 comma 0 right-parenthesis. A dashed horizontal line segment goes from the point left-parenthesis 0 comma StartFraction 1 Over 18 EndFraction right-parenthesis to the point left-parenthesis 11 comma StartFraction 1 Over 18 EndFraction right-parenthesis. A vertical dashed line segment connects the points left-parenthesis 9 comma 0 right-parenthesis and left-parenthesis 9 comma StartFraction 1 Over 12 EndFraction right-parenthesis. Another vertical dashed line segment connects the points left-parenthesis 11 comma 0 right-parenthesis and left-parenthesis 11 comma StartFraction 1 Over 18 EndFraction right-parenthesis. The triangular region enclosed by the points left-parenthesis 11 comma StartFraction 1 Over 18 EndFraction right-parenthesis, left-parenthesis 11 comma 0 right-parenthesis, and left-parenthesis 15 comma 0 right-parenthesis is shaded. Note that this region is a triangle, with base 15−11=4 and height 1/18, so the area (and hence the probability) is P(X>11)=1/2(4)(1/18)=1/9

The probability density function for the random variable X is shown below. Find P(X<2).

Correct answer: 2/9 Remember that the probability for a range of values is equal to the area under the curve. So we want the area under the curve for the values of x less than 2. Note that this region is a rectangle, with base 2 and height 1/9, so the area (and hence the probability) is P(X<2)=(base)(height)=(2)(1/9)=2/9 So we find that P(X<2)=2/9.

The amount of time it takes Lexie to go grocery shopping is continuous and uniformly distributed between 14.5 minutes and 43.5 minutes. What is the probability that it takes Lexie between 19 and 41 minutes to go grocery shopping?

Correct answer: 22/29 Let X be the time it takes to go grocery shopping. The range of possible values is 14.5 to 43.5, so X∼U(14.5,43.5). We are interested in values greater than 19 and less than 41. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a). So we find P(19<X<41)=(41−19)(1/43.5−14.5)=(22)(1/29)=22/29

If X∼U(4.5,18.5) is a continuous uniform random variable, what is P(X>6)?

Correct answer: 25/28 We are interested in values greater than 6. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we find P(X>6)=(18.5−6)(1/18.5−4.5)=(12.5)(1/14)=25/28

The probability density function for the random variable X is shown below. Find P(X>4).

Correct answer: 25/81 Remember that the probability of a range of values is given by the area under the graph over that range of values. So we want the area under the curve for the values of x greater than 4. Note that this region is a triangle, with base 9−4=5 and height 1081, so the area (and hence the probability) is P(X>4)=1/2(5)(10/81)=25/81

The amount of time it takes William to wait for the bus is continuous and uniformly distributed between 4.5 minutes and 14.5 minutes. What is the probability that it takes William more than 7 minutes to wait for the bus?

Correct answer: 3/4 Let X be the time it takes to wait for the bus. The range of possible values is 4.5 to 14.5, so X∼U(4.5,14.5).We are interested in values greater than 7. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we findP(X>7)=(14.5−7)(1/14.5−4.5)=(7.5)(1/10)=3/4

The GMAC conducted research studies to estimate the mean GMAT score between 2013−2015 and its standard deviation. The estimated mean was 551.94 points out of 800 possible points, and the estimated standard deviation was 120.88 points. Assume SAT scores follow a normal distribution. Using the Empirical Rule, about 95% of the scores lie between which two values?

Correct answer: 310.18 to 793.7 The Empirical Rule says that 95% of the data lies within two standard deviations of the mean. The standard deviation is 120.88. So, the data that lie within two standard deviations of 551.94 (between −2σ and 2σ) will be the data that lie in the range that is (120.88)(2)=241.76 units less than the mean (551.94) and more than the mean (551.94). So, the values 551.94−241.76=310.18 and 551.94+241.76=793.7 are within two standard deviations of the mean. About 95% of the x-values lie between 310.18 and 793.7.

The probability density function for the random variable X is shown below. Find P(X<6).

Correct answer: 67 Remember that the probability for a range of values is equal to the area under the curve. So we want the area under the curve for the values of x less than 6. Note that this region is a rectangle, with base 6 and height 17, so the area (and hence the probability) isP(X<6)=(base)(height)=(6)(17)=67So we find that P(X<6)=67.

If X∼U(5.5,18.5) is a continuous uniform random variable, what is P(X<9)?

Correct answer: 7/26 We are interested in values less than 9. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we findP(X<9)=(9−5.5)(118.5−5.5)=(3.5)(113)=7/26

Mr. Karly's business statistic test scores are normally distributed with a mean score of 87 (μ) and a standard deviation of 4 (σ). Using the Empirical Rule, about 99.7% of the data values lie between which two values?

Correct answer: 75−99 The Empirical Rule says that 99.7% of the data lies within three standard deviations of the mean. The standard deviation is 4. So, the data that lies within three standard deviations of 80 (between −3σ and 3σ) will be the data that lies in the range that is (4)(3)=12 units less than the mean (87) and more than the mean (87). So, the values 87−12=75 and 87+12=99 are within three standard deviations of the mean. About 99.7% of the x-values lie between 75 and 99.

Question Mr. Karly's business statistic test scores are normally distributed with a mean score of 87 (μ) and a standard deviation of 4 (σ). Using the Empirical Rule, about 99.7% of the data values lie between which two values?

The amount of time it takes Isabella to do a math problem is continuous and uniformly distributed between 38.5 seconds and 79.5 seconds. What is the probability that it takes Isabella between 51 and 60 seconds to do a math problem?

Correct answer: 9/41 Let X be the time it takes to do a math problem. The range of possible values is 38.5 to 79.5, so X∼U(38.5,79.5).We are interested in values greater than 51 and less than 60. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we findP(51<X<60)=(60−51)(1/79.5−38.5)=(9)(141)=9/41

The probability density function for the random variable X is shown below. Find P(X>5).

Correct answer: 9/64 Remember that the probability of a range of values is given by the area under the graph over that range of values. So we want the area under the curve for the values of x greater than 5. Note that this region is a triangle, with base 8−5=3 and height 3/32, so the area (and hence the probability) is P(X>5)=1/2(3)(3/32)=9/64

The graphs below represent the performance evaluations of 3 managers, labeled A, B, and C, on the same axis. Determine which normal distribution has the largest mean.

Correct answer: A Remember that the mean of a normal distribution is the x-value of its central point (the top of the "hill"). Therefore, a distribution with a larger mean will be centered farther to the right than a distribution with a smaller mean.The distribution that is farthest to the right is A, so that has the largest mean. The manager A has the largest mean of performance ratings, meaning they rate higher on average.

"Understand the notation and interpret the parameters of a normal distribution in business examples" The following data set provides information from the World Meterological Organization that defines climatological standard normals as averages of climatological data computed for the following consecutive periods of 30 years: (WMO, 1984). The data is from random countires around the world from January 1961 to December 1990. Looking at the data set, for which month is the mean monthly value the highest for Dar Es Salaam in the United Republic of Tanzania?

Correct answer: April Dar Es Salaam reaches its highest mean monthly value at 269 mm in August.

The following data set provides information from the World Meterological Organization that defines climatological standard normals as averages of climatological data computed for the following consecutive periods of 30 years: (WMO, 1984). The data is from random countries around the world from January 1961 to December 1990. Looking at the data set, which station in the United Republic of Tanzania reported the largest standard deviation of mean value for the month of January?

Correct answer: Mtwara The standard deviation of mean value in January is 123 mm for Mtwara, which is larger than the standard deviation of mean values for Songea (101 mm), Mbeya (61 mm), and Bukoba (61 mm).

A fair coin is flipped 60 times. Let X be the number of heads. Which of the following distributions is a good estimate of X?

Correct answer: N(30,3.9) Remember that when n is fairly large, a binomial distribution with parameters n (the number of trials), p (the probability of success), and q=1−p (the probability of failure), can be approximated by the normal distribution with mean μ=np and standard deviation σ=npq−−−√. In symbols, the binomial distribution is approximated by N(np,npq−−−√). In this case, because this is a fair coin, p=q=12, and n=60. So we find that a good approximation of the distribution of X is μ=12(60)=30. And σ is σ=npq−−−√=(60)(12)(12)−−−−−−−−−−−−√=15−−√≈3.9 So the best approximation is N(30,3.9).

"Understand the notation and interpret the parameters of a normal distribution in business examples" The following data set provides information from the World Meterological Organization that defines climatological standard normals as averages of climatological data computed for the following consecutive periods of 30 years: (WMO, 1984). The data is from random countires around the world from January 1961 to December 1990. Looking at the data set, which station in the United Republic of Tanzania reported the largest mean monthly value for the month of January?

Correct answer: Songea The highest mean monthly value of January rainfall is 279 mm for Songea, which is larger than the amounts for Bukoba (154 mm), Mbeya (194 mm), and Mtwara (219 mm).

The waiting time for a patient in a doctor's office is exponentially distributed with an average wait time of 18 minutes. What is the probability that the wait time is greater than 20 minutes? Round the final answer to three decimal places.

Correct answers: 0.329 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 18, so the decay parameter is λ=1/18. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=1/18, the probability that X is greater than 20 is P(X>20)=1−P(X<20)=1−(1−e−1/18(20))≈0.329

Let X be the waiting time for a train to pass by, where X has an average value of 25 minutes. If the random variable X is known to be exponentially distributed, what is the probability that the wait time is less than 26 minutes? Round the final answer to three decimal places.

Correct answers: 0.647 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 25, so the decay parameter is λ=1/25. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=1/25, the probability that X is less than 26 is P(X<26)=1−e−1/25(26) ≈0.647

A survey of 500 motorists, in Los Angeles, found a mean commute time of 55 minutes. A TV entertainer drove for an hour and ten minutes, which meant a z-score of 1. What is the standard deviation?

Correct answers: 15min z=x−μ/σ 1=70−55/σ σ=70−55 σ=15 A mean commute time of 55 minutes, z-score of 1, and a x value of one hour and ten minutes or 70 minutes gives a standard deviation of 15 minutes.

Marc's ads per website are normally distributed with a standard deviation of 13 ads. If Marc has 231 ads, and the z-score of this value is 4, then what is his mean ads per website? Do not include the units in your answer. For example, if you found that the mean is 150 ads, you would enter 150.

Correct answers: 179 We can work backwards using the z-score formula to find the mean. The problem gives us the values for z, x and σ. So, let's substitute these numbers back into the formula: z=x−μ/σ 4= 231−μ/13 52=231−μ -179=−μ 179=μ We can think of this conceptually as well. We know that the z-score is 4, which tells us that x=231 is four standard deviations to the right of the mean, and each standard deviation is 13. So four standard deviations is (4)(13)=52 points. So, now we know that 231 is 52 units to the right of the mean. (In other words, the mean is 52 units to the left of x=231.) So the mean is 231−52=179.

A survey of 500 motorists, in Chicago, had a mean commute of 64 minutes. The standard deviation was 16 minutes. The z-score for a banking executive is −1.5. How long does she drive to work?

Correct answers: 40 min z=x−μ/σ −1.5=x−64/16 −24=x−64 x=40 A banking executive takes 40 minutes to commute to work in Chicago.

A survey of 500 motorists, in Miami, calculated a standard deviation of 11 minutes for their commute times. The z-score for a commute time of an hour is 1. What is the mean commute time?

Correct answers: 49min It is given that for a commute of 1 hour, or 60 minuets, the z-score is equal to 1. Therefore, we have that x=60, z=1, and σ=11. We can use the equation for z-score to calculate the mean, μ as follows. z=1 1=x−μ/σ 1=60−μ/11 11=60−μ=49 The mean commute time, in Miami, is 49 minutes.

Floretta is waiting for a tram. If X is the amount of time before the next tram arrives, and X is uniform with values between 1 and 7 minutes, then what is the average (mean) time for how long she will wait?

Correct answers: mean=4min The mean of a random variable with a uniform distribution U(a,b) is μ=a+b/2 (this is the midpoint of the interval [a,b]). So in this case, a=1 and b=7, so the mean is a+b/2=1+72=8/2=4 So the average waiting time is 4 minutes.

Floretta is waiting for a tram. If X is the amount of time before the next tram arrives, and X is uniform with values between 1 and 7 minutes, then what is the approximate standard deviation for how long she will wait, rounded to one decimal place?

Correct answers: std=1.7min The standard deviation for the waiting time is σσ=(b−a)^2/12−−−−−−−√=(b−a)12−−√ In this case we find that the standard deviation isσ=7−1/12−−√=6/12−−√=3-√≈1.7 So the standard deviation for waiting is approximately 1.7 minutes.

As in the previous problem, a fair coin is flipped 28 times. If X is the number of heads, then the distribution of X can be approximated with a normal distribution, N(14,2.6), where the mean (μ) is 14 and standard deviation (σ) is 2.6.Using this approximation, find the probability of flipping 18 or 19 heads. You may use the portion of the Standard Normal Table below.

Correct answers:$0.07$0.07 We approximate the binomial with a normal distribution to determine the probability of flipping 18 or 19 heads. We must choose values just to the left of 18 and to the right of 19. So, we can choose 17.5 and 19.5, and compute the area between these two values.First, find the z-scores that correspond with the values 17.5 and 19.5. Using the formula z=x−μσ, we find that the z-scores are approximately 1.35 and 2.12, which correspond with 17.5 and 19.5 respectively.Using the Standard Normal Table, the area to the left of z=2.12 is 0.9830, and the area to the left of z=1.35 is 0.9115. The area between these two scores is the difference 0.9830−0.9115, which is 0.0715. So, 0.0715 or about 7% is the probability of flipping 18 or 19 heads in 28 tosses. (Using a calculator, we find the area in between 17.5 and 19.5 using normalcdf (17.5,19.5,14,2.6), which is 0.0719.)

As in the previous problem, a fair coin is flipped 32 times. If X is the number of heads, then the distribution of X can be approximated with a normal distribution, N(16,2.8), where the mean (μ) is 16 and standard deviation (σ) is 2.8.Using this approximation, find the probability of flipping 19 or 20 heads. You may use the portion of the Standard Normal Table below.

Correct answers:$0.13$0.13 We approximate the binomial with a normal distribution to determine the probability of flipping 19 or 20 heads. We must choose values just to the left of 19 and to the right of 20. So, we can choose 18.5 and 20.5, and compute the area between these two values.First, find the z-scores that correspond with the values 18.5 and 20.5. Using the formula z=x−μσ, we find that the z-scores are approximately 0.89 and 1.61, which correspond with 18.5 and 20.5 respectively.Using the Standard Normal Table, the area to the left of z=1.61 is 0.9463, and the area to the left of z=0.89 is 0.8133. The area between these two scores is the difference 0.9463−0.8133, which is 0.1330. So, 0.1330 or about 13% is the probability of flipping 19 or 20 heads in 32 tosses. (Using a calculator, we find the area in between 18.5 and 20.5 using normalcdf (18.5,20.5,16,2.8), which is 0.1320.)

Company A is reviewing each site's net loss. They collected one site's net loss at a z-score of −0.89. Determine the area under the standard normal curve that lies to the left of the z-score of −0.89.

Correct answers:$0.1867$0.1867 We can look at the portion of the Standard Normal Table given in the problem. The ones and tenths digits of −0.89 corresponds with the row of the table labeled "−0.8." The hundredths digit corresponds to the column of the table labeled "0.09." Finding where the row and column meet, we can see that the area to the left of the z-score, −0.89, is 0.1867. This also is the probability, P(Z<−0.89). So the probability that each site's net loss is to the left or less than, z-score, −0.89, is 0.1867.

As in the previous problem, 20 people are randomly selected on a Saturday night at a bowling alley. If X is the number of people who get a strike, then the distribution of X can be approximated with a normal distribution, N(10,2.2), where the mean (μ) is 10 and standard deviation (σ) is 2.2.Using this approximation, find the probability of 12 or 13 people getting a strike. You may use the portion of the Standard Normal Table below.

Correct answers:$0.19$0.19 We approximate the binomial with a normal distribution to determine the probability of 12 or 13 people getting a strike. We must choose values just to the left of 12 and to the right of 13. So, we can choose 11.5 and 13.5, and compute the area between these two values.First, find the z-scores that correspond with the values 11.5 and 13.5. Using the formula z=x−μσ, we find that the z-scores are approximately 0.68 and 1.59, which correspond with 11.5 and 13.5 respectively.Using the Standard Normal Table, the area to the left of z=1.59 is 0.9441, and the area to the left of z=0.68 is 0.7517. The area between these two scores is the difference 0.9441−0.7517, which is 0.1924. So, 0.1924 or about 19% is the probability of 12 or 13 people getting a strike from 20 randomly selected people at a bowling alley on a Saturday night. (Using a calculator, we find the area in between 11.5 and 13.5 using normalcdf (11.5,13.5,10,2.2), which is 0.1919.)

The waiting time to receive a custom ordered package from a company is exponentially distributed with an average wait time of 39 days. Given that it has already taken 28 days, what is the probability that the wait time will be more than an additional 37 days? Round your answer to three decimal places.

Correct answers:$0.387$0.387 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units. Mathematically, this can be written as P(X>k+t∣∣X>k)=P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the waiting time is more than an additional 37 days given that it has already taken 28 days isP(X>65∣∣X>28)=P(X>37)= e−1/39(37)≈0.387

An exponential distribution is formed by the time it takes for a person to choose a birthday gift. The average time it takes for a person to choose a birthday gift is 41 minutes. Given that it has already taken 24 minutes for a person to choose a birthday gift,what is the probability that it will take more than an additional 34 minutes? Round your answer to three decimal places.

Correct answers:$0.436$0.436 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units. Mathematically, this can be written as P(X>k+t∣∣X>k)=P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the time it takes is more than an additional 34 minutes given that it has already taken 24 minutes isP(X>58∣∣X>24)=P(X>34)= e−1/ 41(34)≈0.436

" Use the memoryless property of the exponential distribution to compute conditional probability" An exponential distribution is formed by the waiting times for a train to pass by. The average waiting time for this distribution is 33 minutes. Given that it has already taken 19 minutes, what is the probability that the wait time will be more than an additional 27 minutes? Round your answer to three decimal places.

Correct answers:$0.441$0.441 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units. Mathematically, this can be written as P(X>k+t | X>k)= P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the waiting time is more than an additional 27 minutes given that it has already taken 19 minutes isP(X>46 | X>19)= P(X>27)= e−1/33(27)≈0.441

The random variable X is exponentially distributed, where X represents the waiting time for a patient in the emergency room. If X has an average value of 32 minutes, what is the probability that the wait time is less than 23 minutes? Round the final answer to three decimal places.

Correct answers:$0.513$0.513 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 32, so the decay parameter is λ=132. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=1/32, the probability that X is less than 23 is P(X<23)=1−e−1/32(23)≈0.513

Gail manages 112 rentals per year and her rentals per year are normally distributed. Suppose Gail has 180 rentals in 2009, and this value has a z-score of 4. What is the standard deviation? Do not include the units in your answer. For example, if you found that the standard deviation was 10 rentals, you would enter 10.

Correct answers:$17$17 We can work backwards using the z-score formula to find the standard deviation. The problem gives us the values for z, x and μ. So, let's substitute these numbers back into the formula: z = x−μσ =180−112/σ 4 =68/σ = 68=17 We can think of this conceptually as well. We know that the z-score is 4, which tells us that x=180 is four standard deviations to the right of the mean, 112. So we can think of the distance between 180 and 112, which is 68. This distance is broken up into four standard deviations, so 68/4=17. So 17 is the standard deviation.

Floretta works at a call center for a healthcare company and the supervisor is preparing her performance evaluation. Her calls per hour are normally distributed with a standard deviation of 4 calls. If Floretta scores 10 calls, and the z-score of this value is −4, then what is her mean calls in an hour? Do not include the units in your answer. For example, if you found that the mean is 33 calls, you would enter 33.

Correct answers:$26$26 We can work backwards using the z-score formula to find the mean. The problem gives us the values for z, x and σ. So, let's substitute these numbers back into the formula: z=(x−μ)σ −4=(10−μ)/4 −16=10−μ −26=−μ 26=μ We can think of this conceptually as well. We know that the z-score is −4, which tells us that x=10 is four standard deviations to the left of the mean, and each standard deviation is 4. So four standard deviations is (−4)(4)=−16 calls. So, now we know that 10 is 16 units to the left of the mean. (In other words, the mean is 16 units to the right of x=10.) So the mean is 10+16=26.

"Calculate the mean and standard deviation of a standard normal distribution in business examples" Company A was told that their market share was 1 standard deviation below the mean. If the market shares were approximately normal with μ=86 and σ=4, what was Company A's market share? Do not include units in your answer. For example, if you found that the answer was a 86 market share, you would enter 86.

Correct answers:$82$82 We can work backwards using the z-score formula to find the x-value. The problem gives us the values for z, μ and σ. So, let's substitute these numbers back into the formula: z−1−482=x−μσ=x−86/4=x−86=x We can think of this conceptually as well. We know that the z-score is −1, which tells us that x is one standard deviations to the left of the mean, 86. So we can think of the distance between 86 and the x-value as (1)(4)=4. So Company A's market share is 86−4=82.

Correct answers:$83.85\%$83.85% The Empirical Rule states that you can calculate a probability by summing up the % areas under the normal curve by using the 68−95−99 rule. We know that σ=2, and 18+2=20. Therefore, 20 is one standard deviation more than the mean. Based on the Empiral Rule, 68% of data falls within one standard deviation from the mean. Since we are only looking for the upper half, we get 682=34%. Now for 12, we know that σ=2, and 18−2−2−2=12. Therefore, 12 is three standard deviations less than the mean. Based on the Empiral Rule, 99.7% of data falls within three standard deviations from the mean. Since we are only looking for the lower half of that, we get 99.72=49.85%. We now add the two percentages: 34%+49.85%=83.85% That means you have an 83.85% chance of having a standard refrigerator last between 12 and 20 years.

An appliance dealer's website features the mean life expectancy values for major household appliances. A washing machine's mean life expectancy is 12 years with a standard deviation of 2 years. Using the Empirical Rule, what is the probability that your washing machine will last less than 14 years? Do not include the percent sign and round to the nearest whole number.

Correct answers:$84\%$84% The Empirical Rule states that you can calculate a probability by summing up the % areas under the normal curve by using the 68−95−99 rule. We know that σ=2, and 12+2=14. Therefore, 14 is one standard deviation more than the mean. Based on the Empiral Rule, 68% of data falls within one standard deviation from the mean. Since we are only looking for the upper half, we get 682=34%. We also know that 50% of the data falls below the mean. Therefore, we take the one standard devation more than the mean (34%) plus the lower half of the mean (50%). 34%+50%=84% That means you have an 84% chance of having a washing machine last less than 14 years.

After collecting the data, Raymond finds that the standardized test scores of the students in a school is normally distributed with mean 79 points and standard deviation 3 points. What is the probability that a randomly selected student score is less than 85 points? Enter your answer as a percent rounded to 2 decimal places if necessary. Include the percent symbol % in your answer.

Correct answers:$97.5\%$97.5% Notice that 85 points is two standard deviations greater than the mean. Based on the Empirical Rule, 95% of the scores are within two standard deviations of the mean. Since the normal distribution is symmetric, this implies that 2.5% of the scores are greater than two standard deviations above the mean. Alternatively, 97.5% of the scores are less than two standard deviations above the mean.

An appliance dealer's website features the mean life expectancy values for major household appliances. A compact refrigerator's mean life expectancy is 12 years with a standard deviation of 2 years. Using the Empirical Rule, what is the probability that your compact refrigerator will last more than 6 years? Do not include the percent sign and round to the nearest hundredth

Correct answers:$99.85$99.85 We know that σ=2, and 12−2−2−2=6. Therefore, 6 is three standard deviations less than the mean. Based on the Empiral Rule, 99.7% of data falls within three standard deviations from the mean. However, we want to know the probability of the data values being more than 6 years, not falling within three standard deviations of the mean. So, we know that 0.3% is divided into two tails, one on the far left and one on the far right. Since only 0.15% of the data lies below 6, we subject 0.15% from 100%. 100%−0.15%=99.85 That means you have a 99.85% chance of having a compact refrigerator last more than 6 years.

The amount of time it takes Victoria to solve crossword puzzles can be modeled by a continuous and uniformly distributed random time between 4 minutes and 12 minutes. What is the probability that it will take Victoria greater than 8 minutes (total) for a puzzle that she has already been working on for 7 minutes? Provide the final answer as a fraction.

Correct answers:$\frac{4}{5}$45 Let X be the time it takes to solve a crossword puzzle. We are told that the probability density function for X is uniform, so we can conclude that the conditional probability density function for X given that X>7 is also a uniform distribution. The possible values are restricted to the range 7 to 12. This interval will help us write the conditional probability density function (height of the rectangle). f(x)=112−7=15 We are interested in values greater than 8. So the range of desired outcomes is 8 to 12. This will be the length of the base, 12−8=4. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(X>8∣∣X>7)=(12−8)(112−7)=(4)(15)= 4/5

If X∼U(7,20) follows a uniform distribution, what is the mean of X?

Correct answers:$\text{mean=}13.5$mean=13.5 The mean of a random variable with a uniform distribution U(a,b) is μ=a+b2 (this is the midpoint of the interval [a,b]). So in this case, a=7 and b=20, so the mean is a+b/2=7+20/2=2/72=13.5

Lexie is waiting for a train. If X is the amount of time before the next train arrives, and X is uniform with values between 4 and 11 minutes, then what is the average (mean) time for how long she will wait?

Correct answers:$\text{mean=}7.5\text{min}$mean=7.5min The mean of a random variable with a uniform distribution U(a,b) is μ=a+b2 (this is the midpoint of the interval [a,b]). So in this case, a=4 and b=11, so the mean isa+b2=4+112=152=7.5So the average waiting time is 7.5 minutes.

Correct answers:$\text{probability=}0.15\text{%}$probability=0.15% We know that σ=2, and 16−2−2−2=10. Therefore, 10 is three standard deviations less than the mean. Based on the Empiral Rule, 99.7% of data falls within three standard deviations from the mean. However, we want to know the probability of the data values being less than 10 years, not falling within three standard deviations of the mean. So, we know that 0.3% is divided into two tails, one on the far left and one on the far right. Only 0.15% of the data lies below 10. That means you only have a 0.15% chance of having a freezer last less than 10 years.

The number of pages per book on a bookshelf is normally distributed with mean 248 pages and standard deviation 21 pages. What is the probability that a randomly selected book has less than 206 pages?

Correct answers:$\text{probability=}2.5\text{%}$probability=2.5% Notice that 206 pages is two standard deviations less than the mean. Based on the Empirical Rule, 95% of the number of pages in the books are within two standard deviations of the mean. Since the normal distribution is symmetric, this implies that 2.5% of the number of pages in the books are less than two standard deviation less than the mean.

An appliance dealer's website features the mean life expectancy values for major household appliances. A range hood's mean life expectancy is 14 years with a standard deviation of 2.5 years. Using the Empirical Rule, what is the probability that the range hood you buy would last more than 19 years?

Correct answers:$\text{probability=}2.5\text{%}$probability=2.5% We know that σ=2.5, and 14+2.5+2.5=19. Therefore, 19 is two standard deviations more than the mean. Based on the Empiral Rule, 95% of data falls within two standard deviations of the the mean. However, we want to know the probability of the data values being more than 19 years, not falling within two standard deviations of the mean. So, we know that 5% is divided into the two tails, one on the far left and one on the far right. Only 2.5% of the data lies above 19. That means you have only a 2.5% chance of having a range hood last more than 19 years.

After collecting the data, a market research finds that the ads on Facebook for each page is normally distributed with mean 149 ads and standard deviation 16 ads. Based on the empirical rule, what is the probability that a randomly selected Facebook page has less than 165 ads?

Correct answers:$\text{probability=}84\text{%}$probability=84% Notice that 165 ads is one standard deviation greater than the mean. Based on the Empirical Rule, 68% of the ads are within one standard deviation of the mean. Since the normal distribution is symmetric, this implies that 16% of the ads are greater than one standard deviation above the mean. Alternatively, 84% of the ads are less than one standard deviation above the mean.

A normal distribution is observed from the number of insurance claims per month for a certain insurance company. If the mean is 20 claims and the standard deviation is 3 claims, what is the probability that in a randomly selected month, the insurance company's claims are less than 26 claims?

Correct answers:$\text{probability=}97.5\text{%}$probability=97.5% Notice that 26 claims is two standard deviations greater than the mean. Based on the Empirical Rule, 95% of the insurance company's monthly claims are within two standard deviations of the mean. Since the normal distribution is symmetric, this implies that 2.5% of the insurance company's weekly claims are greater than two standard deviations above the mean. Alternatively, 97.5% of the insurance company's weekly claims are less than two standard deviations above the mean.

The times to complete an obstacle course is normally distributed with mean 87 seconds and standard deviation 7 seconds. Use the empirical rule for normal distributions to estimate the probability that a randomly selected finishing time is greater than 80 seconds?

Correct answers:$\text{probablity=}84\text{%}$probablity=84% Notice that 80 seconds is one standard deviation less than the mean. Based on the Empirical Rules for normal distributions, 68% of the finishing times are within one standard deviation of the mean. Since the normal distribution is symmetric, this implies that 16% of the finishing times are less than one standard deviation below the mean. Alternatively, 84% of the finishing times are greater than one standard deviation below the mean.

Lexie is waiting for a train. If X is the amount of time before the next train arrives, and X is uniform with values between 4 and 11 minutes, then what is the approximate standard deviation for how long she will wait, rounded to one decimal place?

Correct answers:$\text{std=}2.0\text{min}$std=2.0min The standard deviation for the waiting time is σσ=(b−a)212−−−−−−−√=(b−a)12−−√ In this case we find that the standard deviation isσ=11−412−−√=712−−√=73-√6≈2.0So the standard deviation for waiting is approximately 2.0 minutes.

If X∼U(7,20) follows a uniform distribution, what is the standard deviation of X, rounded to three decimal places?

Correct answers:$\text{std=}3.753$std=3.753 The standard deviation is σσ=(b−a)212−−−−−−−√=(b−a)12−−√ In this case we find that the standard deviation isσ=20−7/12−−√=1312−−√=133-√6≈3.753

If X follows a uniform distribution between 19 and 33, what is the standard deviation of X, rounded to three decimal places?

Correct answers:$\text{std=}4.041$std=4.041 The standard deviation is σσ=(b−a)212−−−−−−−√=(b−a)12−−√ In this case we find that the standard deviation isσ=33−19/12−−√= 14/12−−√=73/√3≈4.041

On a Saturday night, 50% of people bowling at a bowling alley get a strike. 20 people are randomly selected. Let X be the number of people who get a strike. What normal distribution best approximates X? Round to one decimal place if entering a decimal answer below.

Correct answers:1$10$102$2.2$2.2 Remember that when n is fairly large, a binomial distribution with parameters n (the number of trials), p (the probability of success), and q=1−p (the probability of failure), can be approximated by the normal distribution with mean μ=np and standard deviation σ=npq−−−√. In symbols, the binomial distribution is approximated by N(np,npq−−−√). In this case, because this situation can be modeled using a fair coin, p=q=1/2, and n=20. So we find that a good approximation of the distribution of X is μ=1/2(20) And σ is σ=npq−−−√=(20)(12)(12)−−−−−−−−−−−−√=5-√≈2.2 So the best approximation is N(10,2.2).

A fair coin is flipped 28 times. Let X be the number of heads. What normal distribution best approximates X? Round to one decimal place if entering a decimal answer below.

Correct answers:1$14$142$2.6$2.6 Remember that when n is fairly large, a binomial distribution with parameters n (the number of trials), p (the probability of success), and q=1−p (the probability of failure), can be approximated by the normal distribution with mean μ=np and standard deviation σ=npq−−−√. In symbols, the binomial distribution is approximated by N(np,npq−−−√). In this case, because this is a fair coin, p=q=12, and n=28. So we find that a good approximation of the distribution of X is μ=12(28)=14. And σ is σ=npq−−−√=(28)(12)(12)−−−−−−−−−−−−√=7-√≈2.6 So the best approximation is N(14,2.6)

Suppose X is a binomial random variable with n trials and probability of success p. If n is large and both np≥5 and n(1−p)≥5, then the random variable X is approximately normal with mean μ=np and standard deviation σ=np(1−p)−−−−−−−−√. As the binomial distribution is discrete and the normal distribution is continuous, we must use a continuity correction. A continuity correction is made to a discrete whole number x in the binomial distribution by representing the single value x by the interval from x−0.5 to x+0.5.

Determine whether the value of x itself should be included in the probability or not, and then find the value of interest using x−0.5 or x+0.5 appropriately. The probability can be determined using Excel.

How To The Empirical Rule states:

For a normal distribution, nearly all of the data will fall within 3 standard deviations of the mean. The empirical rule can be broken down into three parts: 68% of data falls within one standard deviation from the mean (between −1σ and 1σ). 95% fall within two standard deviations from the mean (between −2σ and 2σ). 99.7% fall within three standard deviations from the mean (between −3σ and 3σ). The empirical rule is also known as the 68−95−99.7 Rule.

Question IQ scores are normally distributed with a mean score of 100 (μ) and a standard deviation of 15 (σ). Use the Empirical Rule to find the data that is within 1, 2, and 3 standard deviations of the mean.

Here is a graph depicting the Empirical Rule, with the mean, 100, in the middle of the graph. Each of the 68−95−99.7 percentages are labeled on the distribution. Because we know the standard deviation is 15, each standard deviation from the mean is either 15 above or below that value. Data within one standard deviation of the mean: all scores between 85 and 115 (blue section on the graph) Data within two standard deviations of the mean: all scores between 70 and 130 (yellow section on the graph) Data within three standard deviations of the mean: all scores between 55 and 145 (green section on the graph)

If z follows the standard normal distribution, find P(z>1). Remember that σ from −1 to 1 is 0.68.

Key Terms Binomial distribution: a common discrete probability function that includes a a fixed number of identical trials in an experiment where the only possible outcomes are success and failure Binomial distribution is sometimes known as Bernoulli distribution Normal distribution: often described as a bell curve with parameters defined by mean and standard deviationNormal distribution is also known as Gaussian distribution

Memoryless property: knowing what happened in the past has no effect on future probabilities Memoryless property is sometimes referred to as the forgetfulness property Cumulative distribution function: the value of the area under the continuous distribution function to the left of a given value

The Empirical Rule

Organizations need to anticipate events to stay competitive. For an accurate estimation decision makers apply the empirical rule. In the areas of quality control, managers utilize this rule to regulate standards in production. Budget analysts rely on this rule to forecast acquisitions in equipment. Data that is normally distributed can be standardized, which allows us to compare data that are scaled differently, using z-scores. These z-scores describe where data falls relative to their means, and because of this standardization, we can apply a rule to estimate the probability of the distance of a set of data from its mean. This rule is called the Empirical Rule. The Empirical Rule applies generally to a random variable, X, following the shape of a normal distribution, or bell-curve, with a mean (μ) and a standard deviation (σ). In statistics, the Empirical Rule is often used for forecasting, especially because correct data is normally difficult to obtain, and sometimes even impossible. This rule gives a rough estimate of what the data would look like if we could survey the entire population.

The additional time spent waiting for the next customer does not depend on how much time has already elapsed since the last customer. This is referred to as the memoryless property. Specifically, the memoryless property says:

P(X>k+t∣X>k)=P(X>t) for all k≥0 and t≥0

A fair coin is flipped 32 times. Let X be the number of heads. What normal distribution best approximates X? Round to one decimal place if entering a decimal answer below.

Remember that when n is fairly large, a binomial distribution with parameters n (the number of trials), p (the probability of success), and q=1−p (the probability of failure), can be approximated by the normal distribution with mean μ=np and standard deviation σ=npq−−−√. In symbols, the binomial distribution is approximated by N(np,npq−−−√). In this case, because this is a fair coin, p=q=12, and n=32. So we find that a good approximation of the distribution of X is μ=12(32)=16. And σ is σ=npq−−−√=(32)(12)(12)−−−−−−−−−−−−√=8-√≈2.8 So the best approximation is N(16,2.8).

In a small town, 50% of single family homes have a front porch. 48 single family homes are randomly selected. Let X represent the number of single family homes with a porch. What normal distribution best approximates X? Round to one decimal place if entering a decimal answer below.

Remember that when n is fairly large, a binomial distribution with parameters n (the number of trials), p (the probability of success), and q=1−p (the probability of failure), can be approximated by the normal distribution with mean μ=np and standard deviation σ=npq−−−√. In symbols, the binomial distribution is approximated by N(np,npq−−−√). In this case, p=q=12, and n=48. So we find that a good approximation of the distribution of X is μ=12(48)=24. And σ is σ=npq−−−√=(48)(12)(12)−−−−−−−−−−−−√=12−−√≈3.5 So the best approximation is N(24,3.5).

The United Nations stores statistics about all the live births in the United States. The latest data concerns the birthweights of infants born in 2015. The μ is 3056 grams and the σ is 514 grams. Calculate the birthweight range for 95% of babies born in 2015.

Since we know that 95% represents ±2σ from the mean μ, we use the following for 2 above the mean: z=x−μσ +2=x−3056/514 1028=x−3056 x=4084 Now for 2 below the mean: −2=x−3056514 −1028=x−3056 x=2028 So, 95% of babies born are within 2028 to 4084 grams.

Question A data set has a mean of 45 and a standard deviation of 10. What is the probability that a randomly selected data value is less than 65?

Solution First, we must determine how many standard deviations 65 is from the mean. We know that σ=10, so 45+10+10=65. Therefore 65 is two standard deviations above than the mean. Based on the Empirical Rule, 95% of the data is within two standard deviations of the mean. However, we want to know the probability of the data values being less than 65, not the probability of the data falling within two standard deviations of the mean. Let's take a look at that graph again. The black arrow is pointing to where 65 would be, two standard deviations above the mean. All of the data that lies below that value would be "less than 65." So, we need to use the symmetric sections of the normal distribution graph, from the Empirical Rule, to evaluate what percentage of data lies below 65. Since the normal distribution is symmetric, we can see that only 2.5% of the data lies above 65. (2.35%+0.15%=2.5%) Since all of the data together would be 100%, we can subtract the 2.5% from 100% to find the remaining portion (below 65%). 100%−2.5%=97.5%, so 97.5% of the data is less than 65.

A data set has a mean of 45 and a standard deviation of 10. What is the probability that a randomly selected data value is less than 65?

Q scores are normally distributed with a mean score of 100 (μ) and a standard deviation of 15 (σ). Use the Empirical Rule to find the data that is within 1, 2, and 3 standard deviations of the mean.

Solution Here is a graph depicting the Empirical Rule, with the mean, 100, in the middle of the graph. Each of the 68−95−99.7 percentages are labeled on the distribution. Because we know the standard deviation is 15, each standard deviation from the mean is either 15 above or below that value. Data within one standard deviation of the mean: all scores between 85 and 115 (blue section on the graph) Data within two standard deviations of the mean: all scores between 70 and 130 (yellow section on the graph) Data within three standard deviations of the mean: all scores between 55 and 145 (green section on the graph)

Question Suppose it takes a child between 0.5 and 4 minutes to eat a donut. What is the probability that a different child eats a donut in greater than 2 minutes, given that the child has already been eating the donut for more than 1.5 minutes?

Solution Let X = the time, in minutes, it takes a child. Then X∼U(0.5,4). We know this is conditional probability because there is a given condition: "given that the child has already been eating the donut for more than 1.5 minutes." Since we know the child has already been eating the donut for more than 1.5 minutes, we are no longer starting at a=0.5 minutes. The starting point is 1.5 minutes. This condition affects: the probability density function (height) the interval (base) So, the probability density function is calculated using the new interval. 1.5 to 4. f(x)=1/4−1.5=25 This is theheightof the desired area. The base is the distance of the interval that we are looking for, from 2 to 4.

Suppose x has a normal distribution with μ=50 and σ=6. Use the Empirical Rule to find the data that is within 1, 2, and 3 standard deviations of the mean.

Solution The mean is 50, and the standard deviation is 6. So, the data that lies within one standard deviation of 50 (between −1σ and 1σ) will be the data that lies in the range that is 6 units less than 50 and 6 units more than 50. So, The values 50-6=44 and 50+6=56 are within one standard deviation of the mean. About 68% of the x-values lie between 44 and 56. The data that lies within two standard deviations of 50 (between −2σ and 2σ) will be the data that lies in the range that is (6)(2)=12 units less than 50 and (6)(2)=12 units more than 50. So, The values 50-12=38 and 50+12=62 are within two standard deviations of the mean. About 95% of the x-values lie between 38 and 62. The data that lies within three standard deviations of 50 (between −3σ and 3σ) will be the data that lies in the range that is (6)(3)=18 units less than 50 and (6)(3)=18 units more than 50. So, The values 50-18=32 and 50+18=68 are within three standard deviations of the mean. About 99.7% of the x-values lie between 32 and 68.

After collecting the data, Ms. Jones finds that the heights of her forty students is normally distributed with a mean of 165 centimeters and standard deviation of 7 centimeters . Which of the following gives the probability that a randomly selected student has a height of greater than 158 centimeters?

Solution We know that σ=7, and 165−7=158. Therefore 158 is one standard deviation less than the mean. Based on the Empirical Rule, 68% of the data is within one standard deviation of the mean. However, we want to know the probability of the data values being greater than 158 centimeters, not the probability of the data falling within one standard deviation of the mean. We know that the upper half of the data (above 158) represents 50% of the data. We also know that, by the Empirical Rule, one standard deviation less than the mean represents 34% of the data (half of 68). So, to find the probability of the heights of students being greater than 158 centimeters, we need to add the upper half (50%) to this section of one standard deviation less than the mean (34%). 50%+34%=84%, so 84% of the student's heights are greater than 158 centimeters.

Question After collecting the data, Ms. Jones finds that the heights of her forty students is normally distributed with a mean of 165 centimeters and standard deviation of 7 centimeters . Which of the following gives the probability that a randomly selected student has a height of greater than 158 centimeters?

Example 1 Question: At a university, 30% of the incoming freshmen elect to enroll in a statistics course offered by the university. Approximating the binomial distribution with a normal distribution, find the probability that of 600 randomly selected incoming freshmen, at least 200 have elected to enroll in the course.

Annie averages 23 stock sales per month with a standard deviation of 4 stocks. Suppose Annie's stock sales per month are normally distributed. Let X= the number of stock sales per month. Then X∼N(23,4). Round your answers to TWO decimal places.

Suppose Annie has 34 stock sales in March. The z-score when x=34 is 2.75. This z-score tells you that x=34 is 2.75 standard deviations to the right of the mean, which is 23 The z-score can be found using this formula: z=x−μσ=34−234=114=2.75 The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. So, selling 34 stocks is 2.75 standard deviations away from the mean. A positive value of z means that that the value is above (or to the right of) the mean, which was given in the problem: μ=23 stock sales per month.

Hugo averages 59 sales per ad with a standard deviation of 10.5 sales per ad. Suppose Hugo's sales per ad are normally distributed. Let X= the number of sales per ad. Then X∼N(59,10.5). Round your answers to THREE decimal places.

Suppose Hugo has 48 sales per ad on Wednesday. The z-score when x=48 is −1.048. This z-score tells you that x=48 is 1.048 standard deviations to the left of the mean, which is 59 The z-score can be found using this formula: z=x−μσ=48−59/10.5=−11/10.5≈−1.048 The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. So, having 48 sales is 1.048 standard deviations away from the mean. A negative value of z means that that the value is below (or to the left of) the mean, which was given in the problem: μ=59 sales per ad.

University A averages 58 students per course with a standard deviation of 10.5 students per course. Suppose University A's students per course are normally distributed. Let X= the number of students per course. Then X∼N(58,10.5). Round your answers to THREE decimal places. Suppose University A has 85 students in their business course. The z-score when x=85 is ___. This z-score tells you that x=85 is ___ standard deviations to the right of the mean, which is _____

Suppose University A has 85 students in their business course. The z-score when x=85 is 1. This z-score tells you that x=85 is 2.571 standard deviations to the right of the mean, which is 58 Correct answers:2.571, 2.571, 58 The z-score can be found using this formula: z=x−μσ=85−58/10.5=27/10.5≈2.571 The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. So, having 85 students is 2.571 standard deviations away from the mean. A positive value of z means that that the value is above (or to the right of) the mean, which was given in the problem: μ=58 students per course.

These three parts can be seen in the graph below

The dark blue sections show the 68% of data that falls within one standard deviation of the mean. The medium blue sections (along with the dark blue section in between) show the 95% of data that falls within two standard deviations of the mean. The light blue sections (along with other sections in between) show the 99.7% of data that falls within three standard deviations of the mean. The z-scores for −1σ and 1σ are −1 and 1, respectively. The z-scores for −2σ and 2σ are −2 and 2, respectively. The z-scores for −3σ and 3σ are −3 and 3, respectively.

Question Determine the area under the standard normal curve that lies to the right of the z-score of 1.11. Round to nearest ten-thousandth.

The ones and tenths digits of 1.11 corresponds to "1.1" in the second row of the table. The hundreth digit corresponds to the second column of the table "0.01." Go to where the row and column meet. It shows that the area to the left of the z-score 1.11 is 0.8665. However, we want the area to the right. Use the Complement Rule: 1−0.8665=0.1335

The symmetric property of normally distributed data can be used to estimate the probability of a value within a range of data. Because the normal distribution curve is symmetric, we can think of the Empirical Rule graph split up into equal parts to the right and the left of the mean. The graph below represents this symmetric breakdown of the Empirical Rule.

The pink section represents the 68% of the data that lies within one standard deviation of the mean. This section is split in half, with 34% to the left of the mean and 34% to the right of the mean. The smaller red sections (marked with 13.5%), along with the inner pink sections, represent the 95% of the data that lies within two standard deviations of the mean. (If we add the four sections together, we will get 95%:13.5%+34%+34%+13.5%=95%.) The small blue sections (marked with 2.35%), along with the four inner sections, represent the 99.7% of the data that lies within three standard deviations of the mean. (If we add the sections together, we will get 99.7%:2.35%+13.5%+34%+34%+13.5%+2.35%=99.7%.) *Remember: 50% of the data lie above the mean, and 50% of the data lie below the mean in normal distributions. So, if we have a normally distributed data set, with a mean of 10.5, then we can say that 50% of the data is greater than 10.5.

Porter averages 21 housing sales per week with a standard deviation of 4.5 houses. Suppose Porter's sales per week are normally distributed. Let X= the number of sales per week. Then X∼N(21,4.5). Round your answers to THREE decimal places. Suppose Porter sells 14 houses on week 1. The z-score when x=14 is . This z-score tells you that x=14 is standard deviations to the left of the mean, which is

The z-score can be found using this formula: z=x−μσ=14−21/4.5=−7/4.5≈−1.556 The z-score tells you how many standard deviations the value x is above (to the right of) or below (to the left of) the mean, μ. So, selling 14 houses is 1.556 standard deviations away from the mean. A negative value of z means that that the value is below (or to the left of) the mean, which was given in the problem: μ=21 sales per week.

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