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
![Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } } = \frac{325000-290000}{\frac{145000}{\sqrt{100} } } = 2.413](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7Bx%5E%7B-%7D-mean%20%7D%7B%5Cfrac%7BS.D%7D%7B%5Csqrt%7Bn%7D%20%7D%20%7D%20%20%3D%20%5Cfrac%7B325000-290000%7D%7B%5Cfrac%7B145000%7D%7B%5Csqrt%7B100%7D%20%7D%20%7D%20%20%3D%202.413)
<u><em>Step(ii):</em></u>-
The probability that the mean of the sample is greater than $325,000
![P( X > 3,25,000) = P( Z >2.413)](https://tex.z-dn.net/?f=P%28%20X%20%3E%203%2C25%2C000%29%20%3D%20P%28%20Z%20%3E2.413%29)
= 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