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6 March, 09:14

Incomes in a certain town are strongly right-skewed with mean $36,000 and standard deviation $7000. a random sample of 75 households is taken. what is the probability the sample mean is greater than $37,000? 0.1075 0 0.4432

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  1. 6 March, 12:41
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    Since the sample is greater than 10, we can approximate this binomial problem with a normal distribution.

    First, calculate the z-score:

    z = (x - μ) / σ = (37000 - 36000) / 7000 = 0.143

    The probability P (x > 37000$) = 1 - P (x < 37000$),

    therefore we need to look up at a normal distribution table in order to find

    P (z < 0.143) = 0.55567

    And

    P (x > 37000$) = 1 - 0.55567 = 0.44433

    Hence, there is a 44.4% probability that the sample mean is greater than $37,000.
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