(a) What is the probability that there is at least 50% red M&Ms in the random sample? (b) What is the probability no more than one-third of the M&Ms in the random sample are red?

Question

51% of the M&Ms in a large bag of M&Ms are red. Suppose you take a random sample of 30 M&Ms from the large bag. (Round your answers to four decimal places.)

Answer 1

.51 x 30 = 15.3

use a binomial distribution to approximate a normal distribution

n=30

p =.51

q =.49

mean = np = 15.3

variance = npq= 30(.51)(.49) = 7.497

standard deviation = sqr7.497 = about 2.738065

(15-15.3)/2.738065 = -0.1096 = z score

use an online z score calculator or z tables

Pr(>50% at least 15 M&M's are red) = 50.0436%

(10-15.3)/2.738065 =-1.9357 = z score%

Pr(no more than 10) = 1-.97355 = .02645 = 2.645%