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 devia

Question

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

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Thu Cúc 4 years 2021-08-09T11:12:16+00:00 1 Answers 35 views 0

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  1. minhkhue
    0
    2021-08-09T11:13:35+00:00
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    Answer:

    The probability that the mean of the sample is greater than $325,000

    P( X > 3,25,000) = P( Z >2.413) = 0.008

    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

    Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } } = \frac{325000-290000}{\frac{145000}{\sqrt{100} } } = 2.413

    Step(ii):–

    The probability that the mean of the sample is greater than $325,000

    P( X > 3,25,000) = P( Z >2.413)

                               = 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

    P( X > 3,25,000) = P( Z >2.413) = 0.008

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