Question 965755
We need to use the binomial probability formula to solve this problem
Probability(P) (k successes in n trials) = (combination of n taken k at a time) * p^k * q^(n-k), where n = 10, p = .05, q=.95
a) k = 1, then
P = (10! / (1! * (10-1)!)) * .05^1 * .95^(10-1) = 0.315124705 approx 0.3151
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b) P ( at least 1 is sports fan ) = 1 - P ( no sports fan)
P ( no sports fan) means k = 0
P ( no sports fan) = (10! / (0! * (10-0)!)) * .05^0 * .95^(10-0) = 0.598736939 and P ( at least 1 is sports fan ) = 1 - 0.598736939 = 0.401263061 approx 0.4013
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c) P ( no sports fan) means k = 0
P ( no sports fan) = (10! / (0! * (10-0)!)) * .05^0 * .95^(10-0) = 0.598736939 approx 0.5987
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d) P ( at most 1 student is sports fan ) = P ( no sports fan ) + P ( exactly 1 sports fan)
P ( at most 1 student is sports fan ) = 0.598736939 + 0.315124705 = 0.913861644 approx 0.9139
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e) P ( more than 1 is a sports fan ) = 1 - P ( at most 1 student is sports fan )
P ( more than 1 is a sports fan ) = 1 - 0.913861644 = 0.086138356 approx 0.0861
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f) E(X) = n*p = 100 * 0.05 = 5, where E(X) is expected value of X
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g) V(X) = (std dev)^2 = npq, where V(X) is the variance of X
therefore
std dev = square root (100* 0.05 * 0.95) = 2.179449472 approx 2.18
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