Step 1

Consider \(\displaystyle{p}={0.40}\) and \(\displaystyle{n}={7}\)

\(\displaystyle{E}{\left(\overline{{{p}}}\right)}={p}\)

\(\displaystyle={0.40}\)

The expected value of \(\displaystyle\overline{{{p}}}\) is 0.40

The expected value of \(\displaystyle\overline{{{p}}}\) is the mean of the sampling distribution of the sample proportion.

Step 2

The standard error of \(\displaystyle\overline{{{p}}}\) is calculated as follows:

\(\displaystyle\sigma_{{\overline{{{p}}}}}=\sqrt{{{\frac{{{0.40}{\left({1}-{0.40}\right)}}}{{{100}}}}}}\)

\(\displaystyle=\sqrt{{{\frac{{{0.40}\times{0.60}}}{{{100}}}}}}\)

\(\displaystyle\sqrt{{{0.0024}}}\)

\(\displaystyle={0.0490}\)

The standard error of \(\displaystyle\overline{{{p}}}\) is 0.0490.

The standard error of \(\displaystyle\overline{{{p}}}\) is obtained by dividing of the product of pp and \(\displaystyle{\left({1}−{p}\right)}\) by the sample size n and then taking square root.

Step 3

Check whether the sampling distribution of \(\displaystyle\overline{{{p}}}\) is approximately normal or not.

Obtain the value of \(\displaystyle{n}{p}\).

\(\displaystyle{n}{p}={100}{\left({0.40}\right)}\)

\(\displaystyle={40}{\left[{<}{5}\right]}\)

Obtain the value of \(\displaystyle{n}{\left({1}-{p}\right)}\)

\(\displaystyle{n}{\left({1}-{p}\right)}={100}{\left({1}-{0.40}\right)}\)

\(\displaystyle={60}{\left[{<}{5}\right]}\)

Since the values of \(\displaystyle{n}{p}\) and \(\displaystyle{n}{\left({1}-{p}\right)}\) is greater than 5, the sampling distribution of \(\displaystyle\overline{{{p}}}\) is approximately normal.

The mean of the sampling distribution of the sample proportion is the population proportion \(\displaystyle{p}\) , which is 0.40.

The standard deviation of the sampling distribution of the sample proportion is 0.0490.

The sampling distribution of \(\displaystyle\overline{{{p}}}\) has mean and standard deviation is 0.40 and 0.0490, respectively.

The general condition for the normality of the sampling distribution of the sample proportion is satisfied. The requirement for the sampling distribution of \(\displaystyle\overline{{{p}}}\) to be approximately normal since npnp and \(\displaystyle{n}{\left({1}-{p}\right)}\) are greater than 5.

By the central limit theorem, the mean of the sampling distribution is same as proportion of the population distribution for the large sample size. The variance of the sampling distribution is obtained by taking the ratio of \(\displaystyle{p}{\left({1}-{p}\right)}\) and the sample size.

Step 4

According to the central limit theorem, the sampling distribution of sample proportion \(\displaystyle\overline{{{p}}}\) shows the probability distribution for the sample proportion.

The sampling distribution of sample proportion \(\displaystyle\overline{{{p}}}\) shows the probability distribution for the sample proportion.

The sampling distribution of the sample proportion is approximately normal when \(\displaystyle{n}\geq{30}\) using central limit theorem. It is important to obtain the probabilities about the sample proportions.