Question

labor force participation among women rose in the united states between 1975 and 2000 and has beeen declining ever since. according to the u.s. bureau of labor statistics, 56% of women were in the labor force in 2015.

1. if 50 working - age women are randomly selected, what is the probability that between 48% and 63% are in the labor force? round your answer to 4 decimal places.
2. if 190 working age women are randomly selected, what is the probability that between 48% and 63% are in the labor force? round your answer to 4 decimal places.
3. why did the probability increase?

the probability increased since the sample size increased and the sample proportions are more concentrated near the true labor force participation rate of 56%.
the probability increased since the sample size increased and the distribution of sample proportions is more spread out.

Explanation:

Step1: Identify the distribution

The sampling distribution of the sample - proportion $\hat{p}$ is approximately normal with mean $\mu_{\hat{p}}=p$ and standard deviation $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$, where $p = 0.56$ (population proportion of women in the labor - force).

Step2: Calculate the standard deviation for $n = 50$

For $n = 50$, $\sigma_{\hat{p}}=\sqrt{\frac{0.56\times(1 - 0.56)}{50}}=\sqrt{\frac{0.56\times0.44}{50}}\approx\sqrt{\frac{0.2464}{50}}\approx\sqrt{0.004928}\approx0.0702$.

Step3: Standardize the bounds for $n = 50$

For $\hat{p}_1 = 0.48$, $z_1=\frac{0.48 - 0.56}{0.0702}=\frac{- 0.08}{0.0702}\approx - 1.14$.
For $\hat{p}_2 = 0.63$, $z_2=\frac{0.63 - 0.56}{0.0702}=\frac{0.07}{0.0702}\approx0.997$.
$P(0.48<\hat{p}<0.63)=P(-1.14 < Z < 0.997)=\Phi(0.997)-\Phi(-1.14)=0.8400 - 0.1271 = 0.7129$.

Step4: Calculate the standard deviation for $n = 190$

For $n = 190$, $\sigma_{\hat{p}}=\sqrt{\frac{0.56\times(1 - 0.56)}{190}}=\sqrt{\frac{0.56\times0.44}{190}}\approx\sqrt{\frac{0.2464}{190}}\approx\sqrt{0.001297}\approx0.0360$.

Step5: Standardize the bounds for $n = 190$

For $\hat{p}_1 = 0.48$, $z_1=\frac{0.48 - 0.56}{0.0360}=\frac{-0.08}{0.0360}\approx - 2.22$.
For $\hat{p}_2 = 0.63$, $z_2=\frac{0.63 - 0.56}{0.0360}=\frac{0.07}{0.0360}\approx1.94$.
$P(0.48<\hat{p}<0.63)=P(-2.22 < Z < 1.94)=\Phi(1.94)-\Phi(-2.22)=0.9738 - 0.0132 = 0.9606$.

Answer:

1. $0.7129$
2. $0.9606$
3. The probability increased since the sample size increased and the sample proportions are more concentrated near the true labor - force participation rate of $56\%$.