Answered

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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?

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 class=

Sagot :

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

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

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

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

This would then be written as

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

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

[tex]p(z>1.78)=0.0375[/tex]

ANSWER=0.0375

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