Question 484089: According to a February 2005 survey completed by the Pew Internet and American Life Project, about 36 million Americans, or 27% of Internet users, say they download either music or video files. Assuming the percentage is correct, use the normal approximation to the binomial to find the probability of the following in a survey of 2000 American Internet users.
a) At least 500 download music or video files
b) At least 575 download music or video files
c) At most 520 download music or video files
d) At most 555 download music or video files
I've been stumped by this question for awhile now. I really can't seem to figure out how to go about it. Please help!
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! According to a February 2005 survey completed by the Pew Internet and American Life Project, about 36 million Americans, or 27% of Internet users, say they download either music or video files. Assuming the percentage is correct, use the normal approximation to the binomial to find the probability of the following in a survey of 2000 American Internet users.
a) At least 500 download music or video files
mean = np = 2000*0.27 = 540
std = sqrt(npq) = sqrt(540*0.73) = 19.85
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P(500 <= x <= 2000) = P(499.5 <= x <= 2000.5)
z(499.5) = (499.5-540)/19.85 = -2.0403
z(2000.5) = (2000.5-540)/19.85 = 3.7058
Approx = P(-2.0403 < z < 3.7058) = 0.5807
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Use the similar approximation procedure for the following:
b) At least 575 download music or video files
c) At most 520 download music or video files
d) At most 555 download music or video files
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Cheers,
Stan H.
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