Question 840248
In a certain city of several million people, 7.7% of the adults are unemployed. If a random sample of 300 adults in this city is selected, approximate the probability that more than 26 in the sample are unemployed. 
Binomial Problem with n = 300 and p(unemployed) = 0.077
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mean = np = 300*0.077 = 23.1
std = sqrt(300*0.077*0.923) = 4.62
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Use the normal approximation to the binomial with a correction for continuity.
P(25.5 <= x <= 300.5)
z(25.5) = (25.5-23.1)/4.62 = 0.5195
z(301) = (300.5-23.1)/4.62 = 60.04
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P(0.5195<= z <=60.04) = normalcdf(0.5195,60) = 0.3017
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Cheers,
Stan H.
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