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 14%. Suppose you randomly select a package of milk chocolate M&M'S that contains 55 candies and determine the proportion of brown candies in the package.

( (b) What is the probability that the sample percentage of brown candies is less than 20%? (Round your answer to four decimal places.)

(c) What is the probability that the sample percentage exceeds 20%? (Round your answer to four decimal places.)

Explanation / Answer

p=0.14 and n=55

here std error =(p(1-p)/n)1/2 =0.0468

b) probability that the sample percentage of brown candies is less than 20% =P(X<0.2)

=P(Z<((|phat-p|-1/2n)/std error ) (note here 1/2n is continutiy correction factor)

=P(Z<(0.2-0.14-1/110)/0.0468)=P(Z<1.0881)=0.8617

c)   probability that the sample percentage exceeds 20% =P(X>0.2)=1-P(X<0.2)=1-0.8617 =0.1383

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