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

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 60 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 60 candies? O The distribution is binomial and cannot be approximated O The distribution is approximated by a normal distribution with mean 0.14 and standard deviation 0.86. O The distribution is approximated by a normal distribution with mean 14 and standard deviation 0.0448. The distribution is approximated by a normal distribution with mean 0.14 and standard deviation 0.0448. O The distribution is approximated by a normal distribution with mean 8.4 and standard deviation 840 (b) what is the probability that the sample percentage of brown candies is less than 16%? (Round your answer to four decimal places.) 0.6736 (c) what is the probability that the sample percentage exceeds 20%? (Round your answer to four decimal places.) 0901 (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 0.05 upper limit 0.23

Explanation / Answer

b)

c)

for normal distribution z score =(p-p)/p here population proportion=     p= 0.140 sample size       =n= 60 std error of proportion=p=(p*(1-p)/n)= 0.0448 probability = P(X<0.16) = P(Z<0.45)= 0.6724

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