SOLUTION: 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 r

Algebra ->  Statistics  -> Binomial-probability -> SOLUTION: 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 r      Log On


   



Question 1166570: 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; i.e. your answer should look like 0.1234, not 0.1234444 or 12.34%.)
Compute the probability that exactly three of the six M&M’s are blue.
Compute the probability that three or four of the six M&M’s are blue.
Compute the probability that at most three of the six M&M’s are blue.
Compute the probability that at least three of the six M&M’s are blue.
If you repeatedly select random samples of six peanut M&M’s, on average how many do you expect to be blue? (Round your answer to two decimal places.)
blue M&M’s
With what standard deviation? (Round your answer to two decimal places.)

Answer by ikleyn(53309) About Me  (Show Source):
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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;
i.e. your answer should look like 0.1234, not 0.1234444 or 12.34%.)
(a) Compute the probability that exactly three of the six M&M’s are blue.
(b) Compute the probability that three or four of the six M&M’s are blue.
(c) Compute the probability that at most three of the six M&M’s are blue.
(d) Compute the probability that at least three of the six M&M’s are blue.
(e) If you repeatedly select random samples of six peanut M&M’s, on average how many do you expect to be blue?
(Round your answer to two decimal places.)
(f) With what standard deviation? (Round your answer to two decimal places.)
~~~~~~~~~~~~~~~~~~~~~~~~~

(a)  In this case, we have a binomial distribution with n=6 trials, k=3 successful trials.
     p=0.23 probability of the individual success in each trial.  Apply the standard formula for this probability

        P = C%5B6%5D%5E3%2A0.23%5E3%2A%281-0.23%29%5E3 = %28%286%2A5%2A4%29%2F%281%2A2%2A3%29%29%2A0.23%5E3%2A0.77%5E3 = 0.1111 (rounded).   ANSWER



(b)  In this case

        P = P(3) + P(4) = C%5B6%5D%5E3%2A0.23%5E3%2A0.77%5E3 + C%5B6%5D%5E4%2A0.23%5E4%2A0.77%5E2 = 

                        = 20%2A0.23%5E3%2A0.77%5E3+%2B+15%2A0.23%5E4%2A0.77%5E2 = 0.1360  (rounded).    ANSWER



(c)  Continue in the same style

        P = P(0) + P(1) + P(2) + P(3) = . . . = 0.972  (rounded).    ANSWER


     To replace monotonic calculations, you may use very convenient online calculator
     https://stattrek.com/online-calculator/binomial/



(d)  Continue in the same style

        P = P(3) + P(4) + P(5) + P(6) = 1 - ( P(0) + P(1) + P(2) ) = 1 - 0.8609 = 0.1391  (rounded).    ANSWER


     To replace monotonic calculations, you may use very convenient online calculator
     https://stattrek.com/online-calculator/binomial/



(e)  Mathematical expectation is  n*p = 6*0.23 = 1.38 blue peanuts.



(f)  With the standard deviation  SD = sqrt%28n%2Ap%2A%281-p%29%29 = sqrt%286%2A0.23%2A%281-0.23%29%29 = 1.0308  (rounded).    ANSWER

Solved: all questions are answered.