The distribution of prices for home sales in a certain New Jersey county is skewed to the right with a mean of $290,000 and a standard deviation of $145,000. Suppose Mrs. McCann selects a simple random sample of 100 home sales from this (very large) population. What is the probability that the mean of the sample is greater than $325,000
Question
Answer:
The probability that the mean of the sample is greater than $325,000
[tex]P( X > 3,25,000) = P( Z >2.413) = 0.008[/tex]
Step-by-step explanation:
Step(i):-
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
[tex]Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } } = \frac{325000-290000}{\frac{145000}{\sqrt{100} } } = 2.413[/tex]
Step(ii):–
The probability that the mean of the sample is greater than $325,000
[tex]P( X > 3,25,000) = P( Z >2.413)[/tex]
= 0.5 – A(2.413)
= 0.5 – 0.4920
= 0.008
Final answer:-
The probability that the mean of the sample is greater than $325,000
[tex]P( X > 3,25,000) = P( Z >2.413) = 0.008[/tex]