Question

The amounts of time employees of a telecommunications company have worked for the company are normally distributed with a mean of 5.6 years and a standard deviation of 1.9 years. Random samples of size 18 are drawn from the population and the mean of each sample is determined. Find the mean and the standard error.

Answer 1

Answer:

a) 5.6 years

b) 0.4478

Step-by-step explanation:

We are given the following in the question:

Mean, μ = 5.6 years

Standard Deviation, σ = 1.9 years

Sample size, n = 18

We are given that the distribution of amounts of time is a bell shaped distribution that is a normal distribution.

a) Mean of each sample.

The best estimator of the sample mean is the population mean itself. Thus, we have:

$\mu_{x} = \mu = 5.6$

Thus, the sample mean is 5.6 years.

b) Standard error

Formula:

$S.E = \dfrac{\sigma}{\sqrt{n}}$

Putting values, we have.

$S.E = \dfrac{1.9}{\sqrt{18}} = 0.4478$

Thus, the standard error is 0.4478.