Question

According to the M&M^® website, the average percentage of brown M&M^® candies in a package of milk chocolate M&Ms is 14%. (This percentage varies, however, among the different types of packaged M&Ms.) Suppose you randomly select a package of milk chocolate M&Ms that contains 57 candies and determine the proportion of brown candies in the package.

(b) What is the probability that the sample proportion of brown candies is less than 25%? (Round your answer to four decimal places.)


(c) What is the probability that the sample proportion exceeds 33%? (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:

Answer 1

Solution

Given that,

p = 0.14

1 - p = 1 - 0.14 = 0.86

n = 57

a) = p = 0.14

=  [p ( 1 - p ) / n] =   [(0.14 * 0.86) / 57 ] = 0.0460

b) P( < 0.25)

= P[( - ) / < (0.25 - 0.14) / 0.0460 ]

= P(z < 2.39)

Using z table,

= 0.9916

c) P( > 0.33) = 1 - P( < 0.33)

= 1 - P(( - ) / < (0.33 - 0.14) / 0.0460 )

= 1 - P(z < 4.13)

Using z table

= 1 - 1

= 0

d) Using standard normal table,

P( -z < Z < z) = 95%

= P(Z < z) - P(Z <-z ) = 0.95

= 2P(Z < z) - 1 = 0.95

= 2P(Z < z) = 1 + 0.95

= P(Z < z) = 1.95 / 2

= P(Z < z) = 0.975

= P(Z < 1.96) = 0.975

= z  ± 1.96

Using z-score formula,

  = z *   +  

  = -1.96 * 0.0460 + 0.14

  = 0.0498

Using z-score formula,

  = z *   +  

  = 1.96 * 0.0460 + 0.14

  = 0.2302

lower limit = 4.98%

upper limit = 23.02%