Question

According to the U.S. Bureau of Labor Statistics, 20% of all people 16 years of age or older do volunteer work. In this age group, women volunteer slightly more than men, with 22% of women volunteering and 19% of men volunteering. (Round all z values to 2 decimal places. Round all intermediate calculations and answers to 4 decimal places.)
1. What is the probability of randomly sampling 140 women 16 years of age or older and getting 35 or more who do volunteer work?p =According to the U.S. Bureau of Labor Statistics,2. What is the probability of getting 21 or fewer from this group?p =According to the U.S. Bureau of Labor Statistics,3. Suppose a sample of 300 men and women 16 years of age or older is selected randomly from the U.S. population. What is the probability that the sample proportion of those who do volunteer work is between 18% and 25%?p =According to the U.S. Bureau of Labor Statistics,Round all z values to 2 decimal places. Round all intermediate calculations and answer to 4 decimal places.The tolerance is +/- 0.0005.

Answer 1

Answer:

1. P(X≥35) = 0.0183

2. P(X≤21) = 0.0183

3. P(0.18<p<0.25) = 0.7915

Step-by-step explanation:

We have the proportion for women: pw=0.22, and the proportion for men: pm=0.19.

1. We have a sample of 140 woman and we have to calculate the probability of getting 35 or more who do volunteer work.

This is equivalent to a proportion of

$p=X/n=35/140=0.25$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.22*0.78}{300}}\\\\\\ \sigma_p=\sqrt{0.0006}=0.0239$

We calculate the z-score as:

$z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.25-0.22}{0.0239}=\dfrac{0.03}{0.0239}=0.8198$

Then, the probability of having 35 women or more who do volunteer work in this sample of 140 women is:

$P(X>35)=P(p>0.25)=P(z>2.0906)=0.0183$

2. We have to calculate the probability of having 21 or fewer women in the group who do volunteer work.

The proportion is now:

$p=X/n=21/140=0.15$

We can calculate then the z-score as:

$z=\dfrac{p-p_w}{\sigma_p}=\dfrac{0.15-0.2}{0.0239}=\dfrac{-0.05}{0.0239}=-2.0906$

Then, the probability of having 21 women or less who do volunteer work in this sample of 140 women is:

$P(X$

3. For the sample with men and women, we use the proportion for both, which is π=0.2.

The sample size is n=300.

Then, the standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.2*0.8}{300}}\\\\\\ \sigma_p=\sqrt{0.0005}=0.0231$

We can calculate the z-scores for p1=0.18 and p2=0.25:

$z_1=\dfrac{p_1-\pi}{\sigma_p}=\dfrac{0.18-0.2}{0.0231}=\dfrac{-0.02}{0.0231}=-0.8660\\\\\\z_2=\dfrac{p_2-\pi}{\sigma_p}=\dfrac{0.25-0.2}{0.0231}=\dfrac{0.05}{0.0231}=2.1651$

We can now calculate the probabilty of having a proportion within 0.18 and 0.25 as:

$P=P(0.18$