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 51 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 25%? (Round your answer to four decimal places.) (c) What is the probability that the sample percentage exceeds 31%? (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.) lower limit_____ upper limit_____

Explanation / Answer

p=0.14 and n=51

here std error =(p(1-p)/n)1/2 =0.0486

a) probability that the sample percentage of brown candies is less than 25% =P(X<0.25)=P(Z<(0.25-0.14)/0.0486)

=P(Z<2.2636)=0.9882

b) P(X>0.31)=1-P(X<0.31)=1-P(Z<(0.31-0.14)/0.0486)=1-P(Z<3.4988)=1-0.9998 =0.0002

c) for 95% CI ; critical value of z =1.96

therefore 95% confidence interval =sample proportion -/+z*std error

lower limit =0.0448 ~ 0.04

upper limit =0.2352 ~ 0.24

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.