Consider the binomial distribution with p = 0.17 and n = 10,000, where n is the

Question

Consider the binomial distribution with p = 0.17 and n = 10,000, where n is the number of trials and p is the probability of success. Complete parts a through c. a. What is the expected value of the number of successes? The expected value is What is the variance of the number of successes? The variance is b. Find the probability that the number of successes will exceed 1800. The probability is (Round to four decimal places as needed.) c. Would you expect the number of successes to exceed 6800? Explain. A. No, because 6800 is smaller than the expected value of the number of successes. B. Yes, because the corresponding probability is quite large. C. No, because the corresponding probability is too small. D. Yes, because 6800 is smaller than the expected value of the number of successes.

Explanation / Answer

a) expected value =np=0.17*10000=1700

variance =np(1-p) =1411

b)here std deviation =(1411)1/2 =37.5633

therefore probability that number of success excced 1800 =P(X<1800)=1-P(X<=1800)

=1-P(Z<(1800.5-1700)/37.5633)=1-P(Z<2.6755)=1-0.9963 =0.0037

c)option C is correct.

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