Question

The average weekly unemployment benefit in Montana is $272. Suppose that the benefits are normally distributed with a standard deviation of $43. A random sample of 8 benefits is chosen in Montana. What is the probability that the mean for this sample is greater than $299?

Answer 1

From the given question we can extract all the necessary parameters to enable us to find the solution to the question.

We would therefore have the following parameters

`\begin{gathered} \text{observed }value=X=299 \\ S\tan darddeviation=\sigma=43 \\ \text{sample}=n=8 \\ \text{average}=\mu=272 \end{gathered}`

This would be inserted into the formula given for the z score below

`z=\frac{x-\mu}{\frac{\sigma}{\sqrt[]{n}}}`

This would then be written as

`\begin{gathered} z=\frac{299-272}{\frac{43}{\sqrt[]{8}}} \\ z=1.78 \end{gathered}`

We look up the z score on the probability table to get 0.9625. We then subtract from 1 to get the answer

`p(z>1.78)=0.0375`

ANSWER=0.0375