7.3a Central Limit Theorem

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

mean(μX)= 32 n=47 So the mean of the sample sum distribution is (n)(μX) =(47)(32) =1504

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

0.093 Ux (mean)= 55 Std(new normal distribution)σx =σ/√n =5/√39 =.801 If you are using a calculator, input the parameters into the normalcdf( ) function as follows: normalcdf(55.2,55.4,55,5/√39) as the format for normalcdf function is normalcdf(lower value, upper value, μ, σn√). After pressing enter, the calculator displays a value closer to 0.0928.

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

0.098 the mean is (1630)(36)=58680 the standard deviation is (149)(√36)≈894 To find the probability using the Standard Normal Table, we find that the z-scores for the two values, 58260 and 58492, are −0.47 and −0.21 respectively, using the formula z=x−μ/σ. Using the Standard Normal Table, the area to the left of z=−0.47 is 0.319, and the area to the left of z=−0.21 is 0.417. 0.417−0.319=0.098, so the probability is about 10%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(58260, 58492, 58680, 149⋅36‾‾‾√), which gives us a result of 0.0975.

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

0.159 mean(Ux) = 22 STD σx =σ/√n =2/√100 =.2 To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(−10^10, 21.8, 22, 2/√100), which gives us a result of 0.159.

Suppose weights, in pounds, of dogs in a city have an unknown distribution with mean 33 and standard deviation 4 pounds. A sample of size n=41 is randomly taken from the population and the mean is taken. What is the probability that the resulting mean is more than 32.6 pounds?

0.739 Ux (mean)= 33 Std(new normal distribution)σx =σ/√n =4/√41 =.625 The problem is asking for the probability the sample mean will be more than 32.6. P(x>32.6). To compute this, we will compute the z−score of 32.6. P(x>32.6) use formula: z=x−μ/σ =P(Z>32.6−33/.625) =P(Z<−.640) Using the Standard Normal Table, the area to the left of z=−0.640 is 0.2611. Remember that the area to the right of z=−0.640 will be the complement of this, 1−0.2611=0.7389 So the probability is about 74%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(32.6, 10^10, 33, 4/√41), which gives us a result of 0.7390.

Suppose speeds of vehicles traveling on a highway have an unknown distribution with mean 62 and standard deviation 7 miles per hour. A sample of size n=44 is randomly taken from the population and the mean is taken. What is the probability that the resulting mean is less than 62.7 miles per hour?

0.746 mean distribution is 62 standard deviation is 7/√44≈1.055 find the z-score for the value 62.7 is 0.664, using the formula z=x−μ/σ. Using the Standard Normal Table, the area to the left of z=0.664 is 0.7454, so the probability is about 75%. To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(−10^10, 62.7, 62, 7/√44), which gives us a result of 0.7464.

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

845.61 The original standard deviation is 113, and the sample size is 56. So, the standard deviation of the distribution of sample sums is: (σX)(√n)=(113)(√56)≈845.61

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

=(n)(μX) =(65)(51) =3315

In 2016, the CDC estimated the mean weight for a U.S. man over 20 years old was 195.7 pounds with a standard deviation of 68 pounds. If samples of 30 men are randomly drawn from the U.S. population and each sample is put onto an airplane, what is the standard deviation of the weight on an airplane? Assume the airplanes all weigh the same when they are empty.

=(σX)(√n) =(68)(√30) ≈372.5

An unknown distribution has a mean of 13 and a standard deviation of 4.5. Samples, with size n=30, were randomly drawn from a population. Using the Central Limit Theorem for Sums, what is the mean of the distribution created by sums?

=390 (n)=30 (μX)=13 mean of the distribution created by sums: =(n)(μX) =(30)(13) =390

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

P(x<21.7 inches) = 0.2661 mean(Ux) = 22 STD σx =σ/√n =3/√39 =.480 The problem is asking for the probability the sample mean will be less than 21.7 inches. P(x<21.7 inches) To find the probability using a calculator, we can put the values into the normalcdf(-10^10, 21.7, 22, (3/√39) gives us a result of 0.2661.

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

P(x>11.3)=0.1714 Ux (mean)= 11 Std(new normal distribution)σx =σ/√n =2/√40 The problem is asking for the probability the sample mean will be more than 11.3. P(x>11.3). To find the probability using a calculator, we can put the values into the normalcdf() function as: normalcdf(11.3, 10^10, 11, 2/√40), which gives us a result of 0.1714.

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

Std(new normal distribution)σx =σ/√n =34/√45 =5.068420749 =5.07

The finishing time for cyclists in a race have an unknown distribution with mean 150 and standard deviation 15 minutes. A sample, with size n=66, is randomly drawn from the population and the values are added together. Using the Central Limit Theorem for Sums, what is the mean for the sample sum distribution?

The Central Limit Theorem for Sums states that the mean of the normal distribution of sums is equal to the mean of the original distribution multiplied by the sample size ((n)(μX)). So the mean of the sample sum distribution is (n)(μX)=(66)(150)=9900

What is the probability that the sample mean for a sample of size 36 will be less than 20? Use the results from above in your calculation and round your answer to the nearest percent. Use a calculator.

Ux (mean)= 29 Std(new normal distribution)σx =1.5 The problem is asking for the probability the sample mean will be less than 20. P(x<20). To compute this, we will compute the z−score of 20. P(x<20)=P(Z<20−29/1.5) =P(Z<−6.00) The lower bound is negative infinity which is represented by -9999, and the upper bound is −6.00. Now use a TI-83, TI-83 plus, or TI-84 calculator to find the probability. 1. Press 2nd and VARS for the DISTR menu. 2. Press 2 for normalcdf(. 3. Enter −9999,−6,0,1), and press ENTER. The answer is P(Z<−6.00)=9.9012×10−10≈0%.

Phyllis, a college business student wants to study how many students in her class study for less than 20 hours per week. Past data reveals that the hours college students study per week has a mean of 29 with standard deviation 9 hours. The normal distribution for the population is shown by the dotted black line. She plans to take a random sample of 36 such college students and will calculate the mean hours studying to compare to the known hours studying. Compute the the mean and standard deviation of the sampling distribution of sample means for a sample of size 36. Round your answers to the nearest tenth.

Ux (mean)= 29 Std(new normal distribution)σx =σ/√n =9/√36 =1.5

The student test scores for a nationwide standardized test have an unknown distribution with mean 259 and standard deviation 38 points. A sample, with size n=55, was randomly drawn from the population. Using the Central Limit Theorem for Means, what is the mean for the sample mean distribution? Provide only the number with no units.

Ux(mean)=259 The Central Limit Theorem for Means equal to the mean of the original distribution, 259

A biologist studies the gestation period of wolves. Past data reveals that such gestation periods have a mean of 65 days with standard deviation 20 days. The normal distribution for the population is shown by the dotted black line. The biologist plans to take a random sample of 15 such wolves and will calculate the mean gestation period of the sample to compare to the known gestation periods. Compute the mean and standard deviation of the sampling distribution of sample means for a sample of size 15. Round your answers to the nearest tenth.

Ux(mean)=65 Std(new normal distribution)σx =σ/√n =20/√15 =5.16 =5.2

The speeds of vehicles traveling on a highway have an unknown distribution with mean 71 and standard deviation 4 miles per hour. A sample, with size n=37, is randomly drawn from the population and the sum is taken. What is the probability that the sum is between 2606 and 2612 miles per hour?

the mean (n)(μX) =(37)(71) =2627 the standard deviation is (σX)(√n) =(4)(√37) =24.33 Fine z-score for 2606: z=x−μ/σ =2606-2627/24.33 =-0.86 Fine z-score for 2612: z=x−μ/σ =2612-2627/24.33 =-0.62 Using the Standard Normal Table: the area to the left of z=-0.86 is 0.1949 the area to the left of z=-0.62 is 0.2676 0.2676-0.1949=.0727 so the probability is about 7%. To find the probability using a calculator, we can put the values into the normalcdf gives us a result of 0.0747.

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

the mean (n)(μX) =(40)(243) =9720 the standard deviation is (σX)(√n) =(34)(√40) =215.03 To find the probability using a calculator, we can put the values into the normalcdf(-10^10,9522,9720, (34)(√40)) gives us a result of 0.179.