SOLUTION: Among U.S. households, 24% have telephone answering machines. If a telemarketing campaign involves 2,500 households, find the probability that more than 650 have answering machines

Question 69967: Among U.S. households, 24% have telephone answering machines. If a telemarketing campaign involves 2,500 households, find the probability that more than 650 have answering machines.
I get .2389
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
I get 0.0096 or less than 1%.
Converting 650 to a z-score I get:
z(650)=[650-600]/sqrt(2500*0.24*0.76)=2.34
which means that 650 is 2.34 standard
deviations to the right of the mean which
is 600.
The probability of the number being greater
than 650 is very small just as the probability
of z being greater than 2.34 is very small.
Hope this helps.
Cheers,
Stan H.
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