Question 861562
Assume a binomial probability distribution has p = .60 and n = 200.
a. What are the mean and standard deviation?
mean = np = 200*0.6 = 120
std: = sqrt(npq) =  sqrt(120*0.4) = sqrt(48) = 4sqrt(3)
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b. Is this situation one in which binomial probabilities can be approximated by the normal probability distribution?
Yes:: Both np and qp > 5
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c. What is the probability of 100 to 110 successes?
Find P(99.5 <= x <= 110.5)
z(99.5) = (99.5-120)/sqrt(48) = -2.95
z(110.5) = (110.5-120)/sqrt(48) = -1.37
P(99.5<= x <=110.5) = P(-2.95<= z <=-1.37) = 0.0838
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Use the same procedure on the next problem.

d. What is the probability of 130 or more successes?
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e. What is the advantage of using the normal probability distribution to approximate the binomial probabilities?
Check your textbook to discuss these advantages.
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Cheers,
Stan H.
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