An advertiser claims that the average percentage of brown M&M\'S candies in a pa

Question

An advertiser claims that the average percentage of brown M&M'S candies in a package of milk chocolate M&M'S is 12%. Suppose you randomly select a package of milk chocolate M&M'S that contains 51 candies and determine the proportion of brown candies in the package.

(a) What is the approximate distribution of the sample proportion of brown candies in a package that contains 51 candies?

A. The distribution is approximated by a normal distribution with mean 12 and standard deviation 0.0455.

B. The distribution is approximated by a normal distribution with mean 0.12 and standard deviation 0.0455.    

C. The distribution is binomial and cannot be approximated.

D. The distribution is approximated by a normal distribution with mean 6.1 and standard deviation 612.

E. The distribution is approximated by a normal distribution with mean 0.12 and standard deviation 0.88.

(b) What is the probability that the sample percentage of brown candies is less than 18%? (Round your answer to four decimal places.)


(c) What is the probability that the sample percentage exceeds 25%? (Round your answer to four decimal places.)


(d) Within what range would you expect the sample proportion to lie about 95% of the time? (Round your answers to two decimal places.)

Explanation / Answer

(a) = np = 51 * 0.12 = 6.12 and = (npq) = (51 * 0.12 * 0.88) = 2.32

We can use normal approximation to the binomial with = 6.12 and = 2.32

(b) x = 0.18 * 51 = 9.18

z = (9.18 = 6.12)/2.32 = 1.319

P(x < 9.18) = P(z < 1.319) = 0.9064

(c) x = 0.25 * 51 = 12.75

z = (12.75 - 6.12)/2.32 = 2.8578

P(x > 12.75) = P(z > 2.8578) = 0.0021

(d) z- scores which contain 95% of the data = ± 1.96

x1 = - z * = 6.12 - 1.96 * 2.32 = 1.57 and x2 = + z * = 6.12 + 1.96 * 2.32 = 10.67

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