Question

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 $\hat{p}=$= 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

a. $0.20$

b. Standard deviation is $0.0730$.

c. Approximate to normal, skewed to right.

Step-by-step solution

Given Information

It is given that $p=20\%=0.20$

$n=30$

Mean of Sampling Distribution

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

${\mu }_{\hat{p}}=p=20\%=0.20$

Standard Deviation

We know that ${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}$

Also, as calculated ${\mu }_{\hat{p}}=p=20\%=0.20$

Hence, ${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.20(1-0.20)}{30}}=0.0730$

Standard deviation is$0.0730$

Shape of Sampling Distribution

Normality condition is $np\ge 10\text{and}n(1-p)\ge 10$

$np=30(0.20)=6<10$

$n(1-n)=30(1-0.20)=24\ge 10$

As population is close to $0$.

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