Question 232876
Suppose household incomes in Indiana Co have a mean of $48,000 and a standard deviation of $18,000. What is the probability that a random sample of 100 households with a mean income below $45,000?
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z(45000) = (45000-48000)/[18000/(sqrt(100)]
= -3000/(1800)
= -1.66667
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P(x < 45000) = P(z < -1.66667) = 0.04779,,,

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Cheers,
Stan H.