According to Master oods, the company that manufactures M&M;\'s, 12% of peanut M

Question

According to Master oods, the company that manufactures M&M;'s, 12% of peanut M&M; s are brown 15 % are yel 12% are r 23 are blu 2 are orange and 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 four of the six M&M;'s are orange. Compute the probability that four or five of the six M&M;'s are orange. Compute the probability that at most four of the six M&M;'s are orange. Compute the probability that at least four of the six M&M;'s are orange. If you repeatedly select random samples of six peanut M&M;'s, on average how many do you expect to be orange? (Round your answer to two decimal places.) orange M&M;'s With what standard deviation? (Round your answer to two decimal places.) orange M&M;'s

Explanation / Answer

Ans:

Probability of orange,p=0.23

Binomail distribution with n=6,p=0.23

P(x=r)=6Cr*0.23r*(1-0.23)6-r

P(x=4)=0.0249

P(x=4 or 5)=0.0249+0.0030=0.0279

P(x<=4)=1-P(x=5)-P(x=6)=1-0.0030-0.0001=1-0.0031=0.9969

P(x>=4)=0.0249+0.0030+0.001=0.0280

Average number of Orange MM=np=6*0.23=1.38

Standard deviation=sqrt(6*0.23*(1-0.23))=1.03

x BINOMDIST(x,6,0.23,false) 0 0.2084 1 0.3735 2 0.2789 3 0.1111 4 0.0249 5 0.0030 6 0.0001

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