In: Statistics and Probability
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:
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%