Question

Use the Central Limit Theorem to find the mean and standard error of the mean of the indicated samplingg distribution.

The amounts of time employees of a telecommunications company have worked for the company are normally distributed with a mean of 5.5 years and a standard deviation of 2.1 years. Random samples of size 17 are drawn from the population and the mean of each sample is determined.

a. 1.33 years, 2.1 years
b. 5.5 years, 0.12 years
c. 5.5 years, 0.51 years
d. 1.33 years, 0.51 years

Answer 1

c. 5.5 years, 0.51 years

Explanation:

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean
`\mu = p` and standard deviation
`s = \sqrt{(p(1-p))/(n)}`

Mean of 5.5 years and a standard deviation of 2.1 years.

This means that, for the population,
`\mu = 5.5, \sigma = 2.1`

Random samples of size 17.

This means that
`n = 17`

Use the Central Limit Theorem to find the mean and standard error of the mean of the indicated sampling distribution.

The mean is the same as the mean for the population, that is, 5.5 years.

The standard deviation is:

`s = (\sigma)/(√(n)) = (2.1)/(√(17)) = 0.51`

This means that the correct answer is given by option c.