Question

in a memory test, the test subjects are given a large number and are asked to memorize it. historical records show that 80% of test subjects pass the test. to pass the test, a subject must exactly repeat all the digits in the number after two hours. a random sample of 625 people to take the memory test is going to be chosen. let $widehat{p}$ be the proportion of people in the sample who pass the test. answer the following. (if necessary, consult a list of formulas.) (a) find the mean of $widehat{p}$. (b) find the standard deviation of $widehat{p}$. (c) compute an approximation for $p(widehat{p}>0.76)$, which is the probability that more than 76% of the people in the sample pass the test. round your answer to four decimal places.

Explanation:

Step1: Recall mean formula for sample proportion

The mean of the sample - proportion $\hat{p}$ is given by $\mu_{\hat{p}}=p$, where $p$ is the population proportion. Given $p = 0.80$ and $n=625$. So, $\mu_{\hat{p}}=0.80$.

Step2: Recall standard - deviation formula for sample proportion

The standard deviation of the sample proportion is $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$. Substitute $p = 0.80$ and $n = 625$ into the formula: $\sigma_{\hat{p}}=\sqrt{\frac{0.8\times(1 - 0.8)}{625}}=\sqrt{\frac{0.8\times0.2}{625}}=\sqrt{\frac{0.16}{625}}=\frac{0.4}{25}=0.016$.

Step3: Standardize the value

We want to find $P(\hat{p}>0.76)$. First, standardize $\hat{p}$ using the formula $z=\frac{\hat{p}-\mu_{\hat{p}}}{\sigma_{\hat{p}}}$. Substitute $\hat{p}=0.76$, $\mu_{\hat{p}} = 0.80$, and $\sigma_{\hat{p}}=0.016$ into the formula: $z=\frac{0.76 - 0.80}{0.016}=\frac{-0.04}{0.016}=-2.5$. Then $P(\hat{p}>0.76)=P(Z>-2.5)$. Since $P(Z > - 2.5)=1 - P(Z\leqslant-2.5)$. From the standard - normal table, $P(Z\leqslant-2.5)=0.0062$. So $P(Z>-2.5)=1 - 0.0062 = 0.9938$.

Answer:

(a) $0.80$
(b) $0.016$
(c) $0.9938$