Question

Answer all parts please

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 57 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 57 candies? The distribution is approximated by a normal distribution with mean 6.8 and standard deviation 684. The distribution is binomial and cannot be approximated. The distribution is approximated by a normal distribution with mean 12 and standard deviation 0.0430. The distribution is approximated by a normal distribution with mean 0.12 and standard deviation 0.88. o The distribution is approximated by a normal distribution with mean 0.12 and standard deviation 0.0430. (b) What is the probability that the sample percentage of brown candies is less than 20%? (Round your answer to four decimal places.) 0.9464x (c) What is the probability that the sample percentage exceeds 21%? (Round your answer to four decimal places.) 0.0179x (d) Within what range would you expect the sample proportion to lie about 95% of the time? (Round your answers to two decimal places.) lower limit 0.206 upper limit 0.034

Answer 1

for query in above, comment.

Debine broun Candies the puk oge Ther einomia5T,12) LA ST * 20-, 12 21-12 PSP-12 o182 (d) z'O3S7 L ower L'm:t = 112-ペos x'0A30 K430 -'20A3 1 2