Question 957269
For the city of Quahog, R.I., assume the the mean household income is normally distributed with μ = $46,000 and σ = $5,060. If 35 households are randomly selected, determine the probability that their mean income is greater than $45,828.94.
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z(45828.94) = (45828.94-46,000)/[5060/sqrt(35)] = -0.2
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P(x-bar > 45828.94) = P(z > -0.2) = normalcdf(-0.2,100) = 0.5793
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Cheers,
Stan H.