ALTA - CH 6 - THE NORMAL DISTRIBUTION

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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 belo

A normal distribution is observed from the number of customers retained per company's locations for a certain pharmaceutial company. If the mean is 16 customers and the standard deviation is 2 customers, what is the probability that a randomly selected location, the company retained between 12 and 20 customers?

$\text{probability=}95\text{%}$probability=95% Notice that 12 customers and 20 customers are each 2 standard deviations away from the mean. Using the Empirical Rule, there is a 95% probability that the company's customers retained is within 2 standard deviations of 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?

$\text{probability=}97.5\text{%}$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.

How To How to Find Probability of a Normal Distribution With a Calculator

1. Go into 2nd DISTR. 2. Press 2:normalcdf. 3. Enter the correct values in the parentheses, separated by commas. The syntax for the instructions are as follows: normalcdf(lower value, upper value, mean, standard deviation). 4.Press Enter. The value that displays will be the area between the lower and the upper bounds that we designated in the normalcdf() function.

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.

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?

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.

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?

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.

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.

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.

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?

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?

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

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

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?

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. A normal curve is over a horizontal axis and is centered on mu. The area under the curve between mu and mu plus sigma is 34 percent of the data. The area under the curve between mu and mu minus sigma is 34 percent of the data. The area under the curve between mu plus sigma and mu plus 2 sigma is 13.5 percent of the data. The area under the curve between mu minus sigma and mu minus 2 sigma is 13.5 percent of the data. The area under the curve between mu plus 2 sigma and mu plus 3 sigma is 2.35 percent of the data. The area under the curve between mu minus 2 sigma and mu minus 3 sigma is 2.35 percent of the data. The area under the curve to the right of mu plus 3 sigma is 0.15 percent of the data. The area under the curve to the left of mu minus 3 is 0.15 percent of the data. A line under the graph from the left side of the curve to mu plus 2 sigma labeled All the data less than 65. 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.

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, fewer than 200 have elected to enroll in the course.

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. A normal curve is over a horizontal axis. Vertical line segments extend from the horizontal axis to the curve at 199.5 and 200. The area to the left of 199.5 is shaded. 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.

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$$. This z-score tells you that x=48 is 2$$ standard deviations to the left of the mean, which is 3$$ Correct answers:1$-1.048$−1.0482$1.048$1.0483$59$59 The z-score can be found using this formula: z=x−μσ=48−5910.5=−1110.5≈−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.

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 time taken to assemble a car in a certain plant is a random variable having a normal distribution with a mean of 20 hours and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in a period of time less than 19.5 hours?

The mean is μ=20 and the standard deviation is σ=2. 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 19.5 for X, 20 for Mean, 2 for Standard_dev, and TRUE for Cumulative. Thus, the answer, rounded to four decimal places, is 0.4013.

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 scores of a test are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability that an individual's score will be greater than 105?

We can use the calculator to determine this probability using the normalcdf() function. After opening the function on the calculator, we need to input the lower and upper values, the mean, and the standard deviation. Lower Value: Since we know that we are looking for scores greater than 105, this is our lower value. Upper Value: Since we are not given an upper value, we can use an extremely high number for our upper value. The Empirical Rule tells us that over 99.7% of the data falls within 3 standard deviations of the mean, which means that MOST of the data is less than 145 (which is 3σ's above the mean). So, we can choose a number much greater than 145, and that will suffice as an upper value. Let's choose 1000. Mean: 100 Standard Deviation: 15 So, the function should look like: normalcdf (105,1000,100,15) After pressing ENTER, we get the value 0.3694 (or about 36.9%), which is the probability that an individual's score will be greater than 105. NOTE: We can use the same strategy for the lower value, as we did with the upper value here

The z-score of 4 tells us that x=−3 is four standard deviations to the right of the mean. The standard deviation is 2.5. What is the mean?

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=(−3−μ)2.5 10=−3−μ 13=−μ −13=μ We can think of this conceptually as well. We know that x=−3 is four standard deviations to the right of the mean, and each standard deviation is 2.5. So four standard deviations is (4)(2.5)=10 units. So, now we know that −3 is 10 units to the right of the mean. (In other words, the mean is 10 units to the left of x=−3.) Ten units to the left of −3 is −13, so −13 is the mean.

50% of full-time workers are saving for retirement at companies in a small city. 96 full-time workers are randomly selected. Let X be the number of full-time workers who are saving for retirement. What normal distribution best approximates X? Round to one decimal place if entering a decimal answer below.

1$48$48 2$4.9$4.9 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, p=q=12, and n=96. So we find that a good approximation of the distribution of X is μ=12(96)=48. And σ is σ=npq−−−√=(96)(12)(12)−−−−−−−−−−−−√=24−−√≈4.9 So the best approximation is N(48,4.9).

Karen scores an average of 18 points a game, and her points per game are normally distributed. Suppose Karen scores 12 points in a game, and this value has a z-score of −3. What is the standard deviation?

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−μ)σ −3=(12−18)σ −3=−6σ −3σ=−6 σ=2 We can think of this conceptually as well. We know that the z-score is −3, which tells us that x=12 is three standard deviations to the left of the mean, 18. So we can think of the distance between 12 and 18, which is 6. This distance is broken up into three standard deviations, so 63=2. So 2 is the standard deviation.

If z follows the standard normal distribution, find P(z<2). Remember that σ from −2 to 2 is 0.95.

0.975 We know that the area from −2 standard deviations to +2 standard deviations is 0.95. We have to add the rest of the left-tailed area, which is half of the probability of the remaining 5%. P(z<2)=0.95+(12)(0.05) =0.95+0.025=0.975

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.

1$2.75$2.75 2$2.75$2.75 3$23$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.

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?

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.

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

The time taken to assemble a car in a certain plant is a random variable having a normal distribution of mean 20 hours and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in a period of time between 20 and 22 hours?

The mean is μ=20 and the standard deviation is σ=2. 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 22 for X, 20 for mean, 2 for Standard_dev, and TRUE for Cumulative, all for the higher value of X. Thus, the answer, rounded to four decimal places, is 0.8413. 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 20 for X, 20 for mean, 2 for Standard_dev, and TRUE for Cumulative, all for the lower value of X. Thus, the answer is 0.5. Now subtract: 0.8413−0.5=0.3413. Thus, the probability that a car can be assembled at this plant in a period of time between 20 and 22 hours is 0.3413.

Find the area to the right of the z-score 1.40 and to the left of the z-score 1.58 under the standard normal curve.

$0.0237$0.0237 First, we need to find the probabilities at each of the z-scores from the Standard Normal Table. Then we will subtract one of the areas from the other to determine the portion of the area in between. From the table, we find that P(Z<1.58)=0.9429 and P(Z<1.4)=0.9192. We can take the area to the left of z=1.4 and subtract it from the area to the left of z=1.58, to find what is left in between. P(1.4<Z<1.58)=P(Z<1.58)−P(Z<1.4)=0.9429−0.9192=0.0237 So the area between the z-scores of 1.4 and 1.58 is 0.0237.

A company market researcher is reviewing the number of premiums of each of its products. Product B has a z-score of −1.36 and Product H has a z-score of −1.21. Determine the area under the standard normal curve that lies to the right of the z-score −1.36 and to the left of the z-score −1.21.

$0.0262$0.0262 First, we need to find the probabilities at each of the z-scores from the Standard Normal Table. Then we will subtract one of the areas from the other to determine the portion of the area in between. From the table, we find that P(Z<−1.21)=0.1131 and P(Z<−1.36)=0.0869. We can take the area to the left of z=−1.36 and subtract it from the area to the left of z=−1.21, to find what is left in between. P(−1.36<Z<−1.21)=P(Z<−1.21)−P(Z<−1.36)=0.1131−0.0869=0.0262 So the area between the z-scores of −1.36 and −1.21 is 0.0262. So the probability that other products have a number of premiums between the z-scores of −1.36 and −1.21 is 0.0262.

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.

$0.1151$0.1151 The mean is μ=44, and the standard deviation is σ=5. The probability that Mike can travel, without refueling, more than 50 miles is shown below. A normal curve is over a horizontal axis and is divided into 3 regions. Vertical line segments extend from the horizontal axis to the curve at the mean, 44 and at 50. The right region is shaded. 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.

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.

$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.

$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.

Find the area under the standard normal curve between the z-scores −0.94 and −0.76.

$0.0500$0.0500 First, we need to find the probabilities at each of the z-scores from the Standard Normal Table. Then we will subtract one of the areas from the other to determine the portion of the area in between. From the table, we find that P(Z<−0.76)=0.2236 and P(Z<−0.94)=0.1736. We can take the area to the left of z=−0.94 and subtract it from the area to the left of z=−0.76, to find what is left in between.P(−0.94<Z<−0.76)=P(Z<−0.76)−P(Z<−0.94)=0.2236−0.1736=0.0500So the area between the z-scores of −0.94 and −0.76 is 0.0500.

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.

$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.)

Determine the area under the standard normal curve that lies to the right of the z-score of 1.44. (Use a four-digit decimal. No rounding necessary.)

$0.0749$0.0749 We can look at a portion of the Standard Normal Table below. The ones and tenths digits of 1.44 corresponds with the last row of the table shown here, "1.4." The hundredths digit corresponds to the fifth column of the table, "0.04." Finding where the row and column meet, we can see that the area to the left of the z-score, 1.44, is 0.9251. However, we want to know the area to the right of the z-score, which is P(Z>1.44). So, we need to use the Complement Rule. If 0.9251 is the area to the left, we can subtract this value from 1, to find the area to the right. P(Z>1.44)=1−0.9251=0.0749 So the area to the right of the z-score of 1.44 is 0.0749.

Determine the area under the standard normal curve that lies to the right of the z-score 0.05 and to the left of the z-score 0.25.

$0.0788$0.0788 First, we need to find the probabilities at each of the z-scores from the Standard Normal Table. Then we will subtract one of the areas from the other to determine the portion of the area in between. From the table, we find that P(Z<0.25)=0.5987 and P(Z<0.05)=0.5199. We can take the area to the left of z=0.05 and subtract it from the area to the left of z=0.25, to find what is left in between. P(0.05<Z<0.25)=P(Z<0.25)−P(Z<−0.05)=0.5987−0.5199=0.0788 So the area between the z-scores of 0.05 and 0.25 is 0.0788.

Determine the area under the standard normal curve that lies to the right of the z-score −0.97 and to the left of the z-score −0.59.

$0.1116$0.1116 First, we need to find the probabilities at each of the z-scores from the Standard Normal Table. Then we will subtract one of the areas from the other to determine the portion of the area in between. From the table, we find that P(Z<−0.59)=0.2776 and P(Z<−0.97)=0.166. We can take the area to the left of z=−0.97 and subtract it from the area to the left of z=−0.59, to find what is left in between.P(−0.97<Z<−0.59)=P(Z<−0.59)−P(Z<−0.97)=0.2776−0.1660=0.1116So the area between the z-scores of −0.97 and −0.59 is 0.1116.

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.

$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.)

As stated in the previous problem, in a small town, 50% of single family homes have a front porch. 48 single family houses are randomly selected. If X represents the number of single family homes with a porch, then the distribution of X can be approximated with a normal distribution, N(24,3.5), where the mean (μ) is 24 and standard deviation (σ) is 3.5.Using this approximation, find the probability that 27 or 28 single family homes will have a porch. You may use the portion of the Standard Normal Table below.

$0.14$0.14 We approximate the binomial with a normal distribution to determine the probability of 27 or 28 single family homes having a porch. We must choose values just to the left of 27 and to the right of 28. So, we can choose 26.5 and 28.5, and compute the area between these two values.First, find the z-scores that correspond with the values 26.5 and 28.5. Using the formula z=x−μσ, we find that the z-scores are approximately 0.71 and 1.29, which correspond with 26.5 and 28.5 respectively.Using the Standard Normal Table, the area to the left of z=1.29 is 0.9015, and the area to the left of z=0.71 is 0.7611. The area between these two scores is the difference 0.9015−0.7611, which is 0.1404. So, 0.1404 or about 14% is the probability of 27 or 28 single family homes having a porch from 48 randomly selected single family homes in a small town. (Using a calculator, we find the area in between 26.5 and 28.5 using normalcdf (26.5,28.5,24,3.5), which is 0.1383.)

As in the previous problem, 96 full-time workers are selected in a small city. If X is the number of full-time workers who are saving for retirement, then the distribution of X can be approximated with a normal distribution, N(48,4.9), where the mean (μ) is 48 and standard deviation (σ) is 4.9.Using this approximation, find the probability that 52, 53, or 54 full-time workers are saving for retirement. You may use the portion of the Standard Normal Table below. Round the final answer to two decimal places.

$0.15$0.15 We approximate the binomial with a normal distribution to determine the probability 52, 53, or 54 full-time workers are saving for retirement. We must choose values just to the left of 52 and to the right of 54. So, we can choose 51.5 and 54.5, and compute the area between these two values.First, find the z-scores that correspond with the values 51.5 and 54.5. Using the formula z=x−μσ, we find that the z-scores are approximately 0.71 and 1.33, which correspond with 51.5 and 54.5 respectively.Using the Standard Normal Table, the area to the left of z=1.33 is 0.9082, and the area to the left of z=0.71 is 0.7611. The area between these two scores is the difference 0.9082−0.7611, which is 0.1471. So, 0.1471 or about 15% is the probability that 52, 53, or 54 full-time workers are saving for retirement from 96 randomly selected full-time workers in a small city. (Using a calculator, we find the area in between 51.5 and 54.5 using normalcdf (51.5,54.5,48,4.9), which is 0.1452.)

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.

$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.

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.

$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.

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.

$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.

Find the area to the left of the z-score −0.40 under the standard normal curve. z−0.4−0.3−0.2−0.10.00.000.34460.38210.42070.46020.50000.010.34090.37830.41680.45620.50400.020.33720.37450.41290.45220.50800.030.33360.37070.40900.44830.51200.040.33000.36690.40520.44430.51600.050.32640.36320.40130.44040.51990.060.32280.35940.39740.43640.52390.070.31920.35570.39360.43250.52790.080.31560.35200.38970.42860.53190.090.31210.34830.38590.42470.5359 Use the value(s) from the table above.

$0.3446$0.3446 We can look at the portion of the Standard Normal Table given in the problem. The ones and tenths digits of −0.40 corresponds with the row of the table labeled "−0.3." The hundredths digit corresponds to the column of the table labeled "0.10." Finding where the row and column meet, we can see that the area to the left of the z-score, −0.40, is 0.3446. This also is the probability, P(Z<−0.40).

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.

$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.

$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 local store is reviewing their sales compared to all local area stores. The products' sales of this local store have a z-score of 0.67. Determine the area under the standard normal curve that lies to the left of the z-score of 0.67.

$0.7486$0.7486 We can look at the portion of the Standard Normal Table given in the problem. The ones and tenths digits of 0.67 corresponds with the row of the table labeled "0.6." The hundredths digit corresponds to the column of the table labeled "0.07." Finding where the row and column meet, we can see that the area to the left of the z-score, 0.67, is 0.7486. This also is the probability, P(Z<0.67). So the probability that other stores' products' sales are to the left or less than, z-score, 0.67, is 0.7486.

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

$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.

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.

$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.

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$126 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.

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?

$15\min$15min z=x−μσ 1=70−55σ 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.

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.

$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: z444σσ=x−μσ=180−112σ=68σ=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 684=17. So 17 is the standard deviation.

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.

$179$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: z452−179179=x−μσ=231−μ13=231−μ=−μ=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.

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.

$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.

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?

$40\ \min$40 min z=x−μσ −1.5=x−6416 −24=x−64 x=40 A banking executive takes 40 minutes to commute to work in Chicago.

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.

$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.

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?

$49\min$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. z111μ=x−μσ=60−μ11=60−μ=49 The mean commute time, in Miami, is 49 minutes.

John's points per pinball game are normally distributed with a standard deviation of 20 points. If John scores 562 points, and the z-score of this value is 3, then what is his mean points in a game? Do not include the units in your answer. For example, if you found that the mean is 510 points, you would enter 510.

$502$502 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: z360−502502=x−μσ=562−μ20=562−μ=−μ=μ We can think of this conceptually as well. We know that the z-score is 3, which tells us that x=562 is three standard deviations to the right of the mean, and each standard deviation is 20. So three standard deviations is (3)(20)=60 points. So, now we know that 562 is 60 units to the right of the mean. (In other words, the mean is 60 units to the left of x=562.) So the mean is 562−60=502.

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.

$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.

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.

$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−864=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.

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.

$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.

$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.

$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.

$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.

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.

$P\left(120\le X\le153\right)=0.6004$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.

$P\left(3,000<X<4,800\right)=0.6225$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.

First, we need to find the probabilities at each of the z-scores from the Standard Normal Table. Then we will subtract one of the areas from the other to determine the portion of the area in between. P(X>x) = 1 - P(X<x) From the table, we find that P(Z<−0.59)=0.2776 and P(Z<−0.97)=0.166. We can take the area to the left of z=−0.97 and subtract it from the area to the left of z=−0.59, to find what is left in between.P(−0.97<Z<−0.59)=P(Z<−0.59)−P(Z<−0.97)=0.2776−0.1660=0.1116So the area between the z-scores of −0.97 and −0.59 is 0.1116.

$\text{ 99.7% of the expenses lie between }\ 58\ \text{ and }88.$ 99.7% of the expenses lie between 58 and 88. The Empirical Rule says that 99.7% of the data lies within three standard deviations of the mean. The standard deviation is 5. So, the data that lie within three standard deviations of 73 (between −3σ and 3σ) will be the data that lie in the range that is (5)(3)=15 units less than the mean (73) and more than the mean (73). So, the values 73−15=58 and 73+15=88 are within three standard deviations of the mean. About 99.7% of the x-values lie between 58 and 88.

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?

$\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.

A local purse shop's prices are normally distributed with a mean price of 85 (μ) and a standard deviation of 4 (σ). Using the Empirical Rule, about 68% of the prices lie between which two values?

$\text{68% of the prices lie between }\ 81\ \text{ and }89.$68% of the prices lie between 81 and 89. The Empirical Rule says that 68% of the data lies within one standard deviation of the mean. The standard deviation is 4. So, the data that lie within one standard deviation of 85 (between −σ and σ) will be the data that lie in the range that is (4)(1)=4 units less than the mean (85) and more than the mean (85). So, the values 85−4=81 and 85+4=89 are within one standard deviation of the mean. About 68% of the x-values lie between 81 and 89.

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?

$\text{This means that x=1,000 grams is }4\text{ standard deviation(s) below or to the left of the mean, μ=3056 grams.}$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=2056514 z=−4 So, x=1,000 grams is 4 standard deviations below or to the left of the mean.

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

$\text{probability=}0.15\text{%}$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?

$\text{probability=}2.5\text{%}$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?

$\text{probability=}2.5\text{%}$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?

$\text{probability=}84\text{%}$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.

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?

$\text{probablity=}84\text{%}$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.

A survey of 500 motorists, in New York City, found they had a mean commute time of 73 minutes, the longest time for a city. The standard deviation was 20 minutes. What is the z-score for a Wall Street worker who drives for an hour and a half?

$z=0.85$z=0.85 z=x−μσ z=90−7320 z=17/20 z=0.85 A commute time of an hour and half or 90 minutes is 0.85 standard deviations above or to the right of the mean.

As in the previous problem, a fair coin is flipped 60 times. If X is the number of heads, then the distribution of X can be approximated with a normal distribution, N(30,3.9), where the mean (μ) is 30 and standard deviation (σ) is 3.9. Using this approximation, find the probability of flipping 34 or 35 heads. You may use the portion of the Standard Normal Table below.

0.10 We approximate the binomial with a normal distribution to determine the probability of flipping 34 or 35 heads. We must choose values just to the left of 34 and to the right of 35. So, we can choose 33.5 and 35.5, and compute the area between these two values. First, find the z-scores that correspond with the values 33.5 and 35.5. Using the formula z=x−μσ, we find that the z-scores are approximately 0.90 and 1.41, which correspond with 33.5 and 35.5 respectively. Using the Standard Normal Table, the area to the left of z=1.41 is 0.9207, and the area to the left of z=0.90 is 0.8159. The area between these two scores is the difference 0.9207−0.8159, which is 0.1048. So, 0.1048 or about 10% is the probability of flipping 34 or 35 heads in 60 tosses. (Using a calculator, we find the area in between 33.5 and 35.5 using normalcdf (33.5,35.5,30,3.9), which is 0.1055.)

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 Round the final answer to two decimal places.

0.19 $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.)

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?

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=7σ σ=1

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.

1$-1.556$−1.556 2$1.556$1.556 3$21$21 The z-score can be found using this formula: z=x−μσ=14−214.5=−74.5≈−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.

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.

1$10$10 2$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=12, and n=20. So we find that a good approximation of the distribution of X is μ=12(20)=10. 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.

1$14$14 2$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).

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.

1$16$16 2$2.8$2.8 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).

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.

1$2.571$2.571 2$2.571$2.571 3$58$58 The z-score can be found using this formula: z=x−μσ=85−5810.5=2710.5≈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.

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.

1$2028$2028 2$4084$4084 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−3056514 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.

A baby boutique is launching a new line of baby clothes. For assurance in the infant's size, the designers want to review the standard for infant clothing. In 2014, the CDC found a sample mean weight for infants between 0 and 2 months of age to be 5.4 pounds with a standard deviation of 0.9 pounds. Assuming infants' weights follow a normal distribution with this mean and standard deviation, we want to find the probability that an infant's weight will be greater than 4 pounds. If we were using a calculator, what values would we input into the normalcdf function?

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. A normal curve is over a horizontal axis and is centered on 100. The area under the curve between 85 and 115 is 68 percent of the data. The area under the curve between 70 and 130 is 95 percent of the data. The area under the curve between 55 and 145 is 99.7 percent of the data. 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)

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.

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. A normal curve is over a horizontal axis. Vertical line segments extend from the horizontal axis to the curve at 200 and 200.5. The entire region below the curve and to the left of the line segment at 200.5 is shaded. 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.

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. A normal curve is over a horizontal axis and is divided into 3 regions. Vertical line segments extend from the horizontal axis to the curve at 199.5, 200 and 200.5. The area between 199.5 and 200.5 is shaded. 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.

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, more than 200 have elected to enroll in the course.

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.

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?

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).

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?

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).

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?

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.

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.

ALTA - CH 6 - THE NORMAL DISTRIBUTION

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