question archive According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green
According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green
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According to Masterfoods, the company that manufactures M&M's, 12% of peanut M&M's are brown, 15% are yellow, 12% are red, 23% are blue, 23% are orange and 15% are green. You randomly select six peanut M&M's from an extra-large bag of the candies. (Round all probabilities below to four decimal places: your answer should look like 0.1234, not 0.1234444 or 12.34%.) Compute the probability that exactly four of the six M&M's are yellow: Compute the probability that four or five of the six M&M's are yellow: [:1 Compute the probability that at most four of the six M&M's are yellow: :] Compute the probability that at least four of the six M&M's are yellow: If you repeatedly select random samples of six peanut M&M's, on average how many do you expect to be yellow? With what standard deviation? M&M's
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Probability that exactly four are yellow = 0.0055
Probability that four or five are yellow = 0.0059
Probability that at most four are yellow = 0.9996
Probability that at least four are yellow = 0.0059
Average number of yellow = 0.9 = 1
Standard deviation = 0.8746
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