Answer to Question #97571 in Statistics and Probability for Juliet Beglaryan

Question

5. Based on the 2000 Census, the proportion of the California population aged 15 years old or older who are married is p = 0.524. Suppose n = 1000 persons are to be sampled from this population and the sample proportion of married persons ( ) is to be calculated. pˆa) What is the mean of the sampling distribution of ? pˆb) What is the standard deviation of the sampling distribution of ?pˆc) What is the approximate probability that less than 50% of the people in the sample are married?

Answer 1

Let $X=$ the sample proportion of items: $X\sim B(n, p)$ X\sim B(n, p).

The mean and variance for the binomial variate is

$$mean=\mu=E(X)=np, Var(X)=\sigma^2=np(1-p)$$

Given that $n=1000, p=0.524$. Then

$$mean=\mu=E(X)=np=1000(0.524)=524>10$$ $$Var(X)=\sigma^2=np(1-p)=1000(0.524)(1-0.524)=249.424$$

$$n(1-p)=1000(1-0.524)=476>10$$

We can use the normal model

$$N\big(p, \sqrt{{p(1-p) \over n}}\big)=N\big(0.524, \sqrt{{0.524(1-0.524) \over 1000}}\big)$$

to approximate the sampling distribution of $\hat{p}$

a) What is the mean of the sampling distribution of $\hat{p}$

$$mean=0.524$$

b) What is the standard deviation of the sampling distribution of $\hat{p}$

$$\sigma_{\hat{p}}=\sqrt{{0.524(1-0.524) \over 1000}}\approx0.0158$$

c) What is the approximate probability that less than 50% of the people in the sample are married?

$$P(\hat{p}<0.5)=P(z<{0.5-0.524 \over 0.0158})\approx P(Z<-1.5190)\approx0.0644$$