In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. Further more, there is a weight limit of 2500 pounds. Assume the average weight of students, faculty and staff on campus is 150 pounds, the standard devaiton is 27 pounds and that the distribution is approximately normal, If a random sample of 16 persons from the campus is taken. What is the standard deviation of the sampling distribution
Are you talking about the standard error of the mean (SEm)?
SEm = SD/√n
To find the standard deviation of the sampling distribution, we need to use the formula for the standard deviation of a sample mean:
Standard deviation of the sampling distribution = standard deviation of the population / square root of the sample size
Given:
Standard deviation of the population (σ) = 27 pounds
Sample size (n) = 16
We can now calculate the standard deviation of the sampling distribution using the provided values:
Standard deviation of the sampling distribution = 27 / √16
Calculating the square root of 16:
√16 = 4
Dividing 27 by 4:
Standard deviation of the sampling distribution = 27 / 4 = 6.75 pounds
Therefore, the standard deviation of the sampling distribution is approximately 6.75 pounds.
To find the standard deviation of the sampling distribution, we need to use the formula:
Standard deviation of the sampling distribution = (Standard deviation of the population) / √(Sample size)
In this case, the standard deviation of the population is given as 27 pounds, and the sample size is 16 persons. So, plugging in these values into the formula:
Standard deviation of the sampling distribution = 27 / √16
To simplify, the square root of 16 is 4:
Standard deviation of the sampling distribution = 27 / 4
Calculating this, we get:
Standard deviation of the sampling distribution ≈ 6.75 pounds