Question

In a population of 210​ women, the heights of the women are normally distributed with a mean of 64.4 inches and a standard deviation of 2.9 inches. If 36 women are selected at​ random, find the mean mu Subscript x overbar and standard deviation sigma Subscript x overbar of the population of sample means. Assume that the sampling is done without replacement and use a finite population correction factor. Round to two decimal places.

Answer 1

Given Information:

Population size = N = 210

Mean height of women population = μ = 64.4 in.

Standard deviation of height of women population = σ = 2.9 in.

Sample size = n = 36

Required Information:

Sample mean = μx = ?

Sample standard deviation = σx = ?

Answer:

Sample mean = μx = 64.4 in.

Sample standard deviation = σx = 0.44 in.

Step-by-step explanation:

We are given a Normal Distribution, which is a continuous probability distribution and is symmetrical around the mean. The shape of this distribution is like a bell curve and most of the data is clustered around the mean. The area under this bell shaped curve represents the probability.

The mean of the sample will be same as population mean that is

μ = μx = 64.4 in

Whereas the sample standard deviation is given by

$\sigma_{x} = \frac{\sigma}{\sqrt{n} } \cdot \sqrt{\frac{N-n}{N-1} }$

Where σ is the standard deviation of height of women population, N is the population size and n is the sample size.

The term $\sqrt{\frac{N-n}{N-1} }$ is known as finite population correction factor and is used since the sampling is done without replacement.

$\sigma_{x} = \frac{2.9}{\sqrt{36} } \cdot \sqrt{\frac{210-36}{210-1} }\\\sigma_{x} = 0.483 \cdot \sqrt{\frac{174}{209} }\\\sigma_{x} = 0.483 \cdot 0.912\\\sigma_{x} = 0.44$

Therefore, the standard deviation of sample is 0.44 in.