Answer:
The probability that the mean of the sample is greater than $325,000
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given the mean of the Population( )= $290,000
Standard deviation of the Population = $145,000
Given the size of the sample 'n' = 100
Given 'X⁻' be a random variable in Normal distribution
Let X⁻ = 325,000

<u><em>Step(ii):</em></u>-
The probability that the mean of the sample is greater than $325,000

= 0.5 - A(2.413)
= 0.5 - 0.4920
= 0.008
<u><em>Final answer:-</em></u>
The probability that the mean of the sample is greater than $325,000