Please answer correctly. i'll leave a like In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. In addition, thereis a weight limit of 2,700 pounds. Assume that the average weight of students, faculty, and staff on campus is 165 pounds,that the standard deviation is 31 pounds, and that the distribution of weights of individuals on campus is approximatelynormal. Suppose a random sample of 16 persons from the campus will be selected.(a) What is the mean of the sampling distribution of X?(b) What is the standard deviation of the sampling distribution of X?(c) What mean weights (in pounds) for a sample of 16 people will result in the total weight exceeding the weight limit of2,700 pounds?(d) What is the probability that a random sample of 16 people will exceed the weight limit?Step 1(a) What is the mean of the sampling distribution of X?We are given that the average weight of students, faculty, and staff at a college is μ = 165 pounds, that the standarddeviation iso = 31 pounds, and that the distribution of weights of individuals on campus is approximately normal.We are asked to determine the mean for the sampling distribution of the sample mean, x, for a random sample of sizen = 16.Recall the general property of the sampling distribution x: the mean is μ = μ. Therefore, μ- = 165✓Step 2(b) What is the standard deviation of the sampling distribution of X?σ- =Recall the general property of the sampling distribution X: the standard deviation isWe are given that thestandard deviation is = 31 pounds and the random sample has size n = 16. Use these values to calculate the mean forthe sampling distribution of the sample mean.Step 331 ✔Step 5√167.75✔Step 4We determined that the weight capacity is exceeded if 16x ≥ 2,700.In other words, if x 168.75exceeded.31(c) What mean weights (in pounds) for a sample of 16 people will result in the total weight exceeding the weight limitof 2,700 pounds?Since the elevator holds a maximum of 16 people and the average weight of each person is x, the weight capacity of2,700 pounds will be exceeded when 16 ✔16 x ≥ 2,700.7.75H=165- = 7.75(d) What is the probability that a random sample of 16 people will exceed the weight limit?P(x ≥ b) =P(X ≥ 168.75) ==An elevator at a college has a posted weight limit of 2,700 pounds, and a maximum capacity of 16 people. We determinedthat if the sample mean x ≥ 168.75 for a particular sample of 16 people, then the weight capacity will be exceeded. We areasked to determine the probability that a random sample of 16 people will exceed the weight limit.We previously determined the mean and standard deviation of the sampling distribution of X.- P(Z = b =#=)pzzWe are given that the distribution of weights of individuals on campus is approximately normal, so the sampling distributionof X is approximately normal also. We can calculate the desired probability by standardizing. Recall the standardizationformula.= 1 -In other words, we need to find the following. (Round your answer to four decimal places.)6168.75 ✔168.75 - 1650.4839=- P(Z ≤ 0.1916 X0168.75 for a particular sample of 16 people, then the weight capacity will be7.75P(X 168.75) = 1 - P(Z

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Please answer correctly. i'll leave a like In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. In addition, thereis a weight limit of 2,700 pounds. Assume that the average weight of students, faculty, and staff on campus is 165 pounds,that the standard deviation is 31 pounds, and that the distribution of weights of individuals on campus is approximatelynormal. Suppose a random sample of 16 persons from the campus will be selected.(a) What is the mean of the sampling distribution of X?(b) What is the standard deviation of the sampling distribution of X?(c) What mean weights (in pounds) for a sample of 16 people will result in the total weight exceeding the weight limit of2,700 pounds?(d) What is the probability that a random sample of 16 people will exceed the weight limit?Step 1(a) What is the mean of the sampling distribution of X?We are given that the average weight of students, faculty, and staff at a college is μ = 165 pounds, that the standarddeviation iso = 31 pounds, and that the distribution of weights of individuals on campus is approximately normal.We are asked to determine the mean for the sampling distribution of the sample mean, x, for a random sample of sizen = 16.Recall the general property of the sampling distribution x: the mean is μ = μ. Therefore, μ- = 165✓Step 2(b) What is the standard deviation of the sampling distribution of X?σ- =Recall the general property of the sampling distribution X: the standard deviation isWe are given that thestandard deviation is = 31 pounds and the random sample has size n = 16. Use these values to calculate the mean forthe sampling distribution of the sample mean.Step 331 ✔Step 5√167.75✔Step 4We determined that the weight capacity is exceeded if 16x ≥ 2,700.In other words, if x > 168.75exceeded.31(c) What mean weights (in pounds) for a sample of 16 people will result in the total weight exceeding the weight limitof 2,700 pounds?Since the elevator holds a maximum of 16 people and the average weight of each person is x, the weight capacity of2,700 pounds will be exceeded when 16 ✔16 x ≥ 2,700.7.75H=165- = 7.75(d) What is the probability that a random sample of 16 people will exceed the weight limit?P(x ≥ b) =P(X ≥ 168.75) ==An elevator at a college has a posted weight limit of 2,700 pounds, and a maximum capacity of 16 people. We determinedthat if the sample mean x ≥ 168.75 for a particular sample of 16 people, then the weight capacity will be exceeded. We areasked to determine the probability that a random sample of 16 people will exceed the weight limit.We previously determined the mean and standard deviation of the sampling distribution of X.- P(Z = b =#=)pzzWe are given that the distribution of weights of individuals on campus is approximately normal, so the sampling distributionof X is approximately normal also. We can calculate the desired probability by standardizing. Recall the standardizationformula.= 1 -In other words, we need to find the following. (Round your answer to four decimal places.)6168.75 ✔168.75 - 1650.4839=- P(Z ≤ 0.1916 X0168.75 for a particular sample of 16 people, then the weight capacity will be7.75P(X 168.75) = 1 - P(Z < 0.4839)= 1-90.191610 0.8086165.XX0.48390.4839Step 6Use technology to find the cumulative probability associated with a particular z-score, rounding the result to four decimalplaces.

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