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Chapter 7: Q 35. (page 462)

Orange SkittlesThe makers of Skittles claim that 20% of Skittles candies are orange. Suppose this claim is true. You select a random sample of 30 Skittles from a large bag. Let p^== the proportion of orange Skittles in the sample.

a. Identify the mean of the sampling distribution of p^.

b. Calculate and interpret the standard deviation of the sampling distribution of p^.

Verify that the 10% condition is met.

Short Answer

Expert verified

a. 0.20

b. Standard deviation is 0.0730.

c. Approximate to normal, skewed to right.

Step by step solution

01

Given Information

It is given that p=20%=0.20

n=30

02

Mean of Sampling Distribution

The mean of sampling distribution of sample proportions is same to proportion of population.

渭p^=p=20%=0.20

03

Standard Deviation

We know that 蟽p^=p(1-p)n

Also, as calculated 渭p^=p=20%=0.20

Hence, 蟽p^=p(1-p)n=0.20(1-0.20)30=0.0730

Standard deviation is0.0730

04

Shape of Sampling Distribution

Normality condition is np鈮�10andn(1-p)鈮�10

np=30(0.20)=6<10

n(1-n)=30(1-0.20)=24鈮�10

As population is close to 0.

So, the sampling distribution of sample proportions will be skewed to the right.

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