Question 57673: I been working out this problem but I'm not sure if my answer is right.
Among U.S households, 24% have telephone answering machines, if a telemarketin campaign involves 2500 households, find the probability that more than 650 have answering machines.
My solution:
mean =2500 x .24 =600
std. d =sq.root of 2500 x.24 x.76 =sq.root of 450 = 21
650-600/21= 2.38 = .9813 ?
Thanks a lot!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! I been working out this problem but I'm not sure if my answer is right.
Among U.S households, 24% have telephone answering machines, if a telemarketin campaign involves 2500 households, find the probability that more than 650 have answering machines.
My solution:
mean =2500 x .24 =600
std. d =sq.root of 2500 x.24 x.76 =sq.root of 450 = 21
650-600/21= 2.38 = .9813 ?
Thanks a lot!
-----------------
Everything looks good until you get to
"650-600/21= 2.38 = .9813 ?"
You calculated the z-score corresponding to 650
and got 2.38.
The Probability you are looking for is
Prob (X>650)=Prob(z>2.38)
That probability is 0.00865...
Cheers,
Stan H.
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