Question 1206752: In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked more than once.
Give the distribution of X. (Enter exact numbers as integers, fractions, or decimals.)
How many pages do you expect to advertise footwear on them? (Round your answer to the nearest whole number.)
Is it probable that all twenty will advertise footwear on them? Why or why not? (Round your answer to two decimal places.)
What is the probability that fewer than ten will advertise footwear on them? (Round your answer to four decimal places.)
What is the probability that you only need to survey at most five pages in order to find one that advertises footwear on it? (Round your answer to four decimal places.)
How many pages do you expect to need to survey in order to find one that advertises footwear? (Round your answer to the nearest whole number.)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! X=binomial with n=20 and p=29/192
for each k=1,...20, p(X=k)= 20Ck*(29/192)^k*(163/192)^(20-k)
expected value is n*p=580/192 rounds to 3.
all 20 having it would be (29/192)^20=4 x 10^-17 or essentially 0.
can use binomialcdf with arguments 20, (29/192), and 9 for fewer than 10. Get 0.9997
for finding at most 5 pages is the cdf using 5 as the last argument and the value is 0.9308.
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