Question

The Wrigley Company, manufacturer of Skittles brand fruit candies, is considering adding a new flavor to the "rainbow" of flavors in a Skittles bag: passion peach. To test if America will like the new flavor, Skittles will do a taste test random sample of size n = 140 at a Peoria grocery store. In all of the other taste tests done so far in the United States in other markets, the proportion of the population who like the new flavor = 0.69. Use Table 1.

a.	Calculate the expected value and the standard error for the sampling distribution of the sample proportion of tasters in Peoria who will like the new flavor. (Round "expected value" to 2 decimal places and "standard deviation" to 4 decimal places.)

Expected value = .69
Standard error = .0391

b.	What is the probability that the sample proportion of the tasters in Peoria who like the new flavor is between 0.60 and 0.80? (Round final answer to 4 decimal places.)

Probability

c.	What is the probability that the sample proportion or the tasters in Peoria who like the new flavor is less than 0.60? (Round final answer to 4 decimal places.)

Probability

Answer 1

Solution:

Given that,

n = 140

= 0.69

1 - = 1 - 0.69 = 0.31

So,

a ) =   = 0.69

Expected value = 0.69

=   ( 1 - ) / n

=   0.69 * 0.31 / 140

=0.0391

= 0.0391

Standard error = 0.0391

b ) p ( 0.60 < < 0.80 )

p ( 0.60 - 0.69 / 0.0391 ) < ( -    / ) < ( 0.80 - 0.69 / 0.0391 )

p ( - 0.09 / 0.0391 < z < 0.11 / 0.0391 )

p ( - 2.30 < z < 2.81 )

p ( z < 2.81 ) - p ( z < - 2.30 )

Using z table

= 0.9975 - 0.0107

= 0.9868

Probability = 0.9868

c ) p ( < 0.60 )

p ( -    / ) < ( 0.60 - 0.69 / 0.0391 )

p ( z < - 0.09 / 0.0391 )

p ( z < - 2.30 )

Using z table

= 0.0107

Probability = 0.0107