To find the mean and standard deviation of the distribution of sample proportions for a random sample of size [tex]\( n = 604 \)[/tex] from a population with a population proportion [tex]\( p = 0.33 \)[/tex], we can proceed as follows:
### Part (a)
Mean of the Distribution of Sample Proportions
The mean ([tex]\(\mu_{\bar{p}}\)[/tex]) of the distribution of sample proportions is equal to the population proportion [tex]\( p \)[/tex].
So,
[tex]\[
\mu_{\bar{p}} = p = 0.33
\][/tex]
Therefore,
[tex]\[
\mu_{\bar{p}} = 0.33
\][/tex]
### Part (b)
Standard Deviation of the Distribution of Sample Proportions
The standard deviation ([tex]\(\sigma_{\bar{p}}\)[/tex]) of the distribution of sample proportions is calculated using the formula:
[tex]\[
\sigma_{\bar{p}} = \sqrt{\frac{p(1 - p)}{n}}
\][/tex]
First, we substitute the given values into the formula:
[tex]\[
\sigma_{\bar{p}} = \sqrt{\frac{0.33 \cdot (1 - 0.33)}{604}}
\][/tex]
Simplifying the expression inside the square root first,
[tex]\[
0.33 \cdot (1 - 0.33) = 0.33 \cdot 0.67 = 0.2211
\][/tex]
Then divide by the sample size [tex]\( n \)[/tex]:
[tex]\[
\frac{0.2211}{604} \approx 0.000366
\][/tex]
Taking the square root of the result:
[tex]\[
\sigma_{\bar{p}} = \sqrt{0.000366} \approx 0.0191
\][/tex]
Rounding to 2 decimal places,
[tex]\[
\sigma_{\bar{p}} \approx 0.02
\][/tex]
Therefore,
[tex]\[
\sigma_{\bar{p}} = 0.02
\][/tex]
### Summary
a. The mean of the distribution of sample proportions ([tex]\(\mu_{\bar{p}}\)[/tex]) is:
[tex]\[
\mu_{\bar{p}} = 0.33
\][/tex]
b. The standard deviation of the distribution of sample proportions ([tex]\(\sigma_{\bar{p}}\)[/tex]) is:
[tex]\[
\sigma_{\bar{p}} = 0.02
\][/tex]
