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](https://tex.z-dn.net/?f=p%3DX%2Fn%3D35%2F140%3D0.25)
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](https://tex.z-dn.net/?f=%5Csigma_p%3D%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cdfrac%7B0.22%2A0.78%7D%7B300%7D%7D%5C%5C%5C%5C%5C%5C%20%5Csigma_p%3D%5Csqrt%7B0.0006%7D%3D0.0239)
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](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7Bp-p_w%7D%7B%5Csigma_p%7D%3D%5Cdfrac%7B0.25-0.22%7D%7B0.0239%7D%3D%5Cdfrac%7B0.03%7D%7B0.0239%7D%3D0.8198)
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](https://tex.z-dn.net/?f=P%28X%3E35%29%3DP%28p%3E0.25%29%3DP%28z%3E2.0906%29%3D0.0183)
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](https://tex.z-dn.net/?f=p%3DX%2Fn%3D21%2F140%3D0.15)
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](https://tex.z-dn.net/?f=z%3D%5Cdfrac%7Bp-p_w%7D%7B%5Csigma_p%7D%3D%5Cdfrac%7B0.15-0.2%7D%7B0.0239%7D%3D%5Cdfrac%7B-0.05%7D%7B0.0239%7D%3D-2.0906)
Then, the probability of having 21 women or less who do volunteer work in this sample of 140 women is:
![P(X](https://tex.z-dn.net/?f=P%28X%3C21%29%3DP%28p%3C0.15%29%3DP%28z%3C-2.0906%29%3D0.0183)
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](https://tex.z-dn.net/?f=%5Csigma_p%3D%5Csqrt%7B%5Cdfrac%7Bp%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cdfrac%7B0.2%2A0.8%7D%7B300%7D%7D%5C%5C%5C%5C%5C%5C%20%5Csigma_p%3D%5Csqrt%7B0.0005%7D%3D0.0231)
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](https://tex.z-dn.net/?f=z_1%3D%5Cdfrac%7Bp_1-%5Cpi%7D%7B%5Csigma_p%7D%3D%5Cdfrac%7B0.18-0.2%7D%7B0.0231%7D%3D%5Cdfrac%7B-0.02%7D%7B0.0231%7D%3D-0.8660%5C%5C%5C%5C%5C%5Cz_2%3D%5Cdfrac%7Bp_2-%5Cpi%7D%7B%5Csigma_p%7D%3D%5Cdfrac%7B0.25-0.2%7D%7B0.0231%7D%3D%5Cdfrac%7B0.05%7D%7B0.0231%7D%3D2.1651)
We can now calculate the probabilty of having a proportion within 0.18 and 0.25 as:
![P=P(0.18](https://tex.z-dn.net/?f=P%3DP%280.18%3Cp%3C0.25%29%3DP%28-0.8660%3Cz%3C2.1651%29%5C%5C%5C%5CP%3DP%28z%3C2.1651%29-P%28z%3C-0.8660%29%5C%5C%5C%5CP%3D0.9848-0.1933%5C%5C%5C%5CP%3D0.7915)