Withdrawal symptoms may occur when a person using a painkiller suddenly stops using it. For a special type of painkiller, withdrawal symptoms occur in $4 \%$ of the cases.
A random sample of 2400 people who have stopped using the painkiller is going to be taken. Let $\hat{p}$ be the proportion of people in the sample who experience withdrawal symptoms.
Answer the following. (If necessary, consult a list of formulas.)
(a) Find the mean of $\hat{p}$.
(b) Find the standard deviation of $\hat{p}$.
(c) Compute an approximation for $P(\hat{p}> 0.03)$, which is the probability that more than $3 \%$ of the people in the sample experience withdrawal symptoms. Round your answer to four decimal places.
Answer
To find the probability that the proportion is greater than $0.03$, we can standardize and use the z-score formula $Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$. Substituting $\hat{p} = 0.03$, $p = 0.04$ and $n = 2400$, we get $Z = -2.5$. Looking up this value in a standard normal distribution table, we get the probability as $0.9938$. So, the approximation for $P(\hat{p}>0.03)$ is $\boxed{0.9938}$.
Step 1 ：The mean of a proportion is simply the probability of success, which is given as $4 \%$ or $0.04$. So, the mean of $\hat{p}$ is $\boxed{0.04}$.
Step 2 ：The standard deviation of a proportion is given by the formula $\sqrt{\frac{p(1-p)}{n}}$, where $p$ is the probability of success and $n$ is the sample size. Substituting $p = 0.04$ and $n = 2400$, we get the standard deviation of $\hat{p}$ as $\boxed{0.004}$.
Step 3 ：To find the probability that the proportion is greater than $0.03$, we can standardize and use the z-score formula $Z = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$. Substituting $\hat{p} = 0.03$, $p = 0.04$ and $n = 2400$, we get $Z = -2.5$. Looking up this value in a standard normal distribution table, we get the probability as $0.9938$. So, the approximation for $P(\hat{p}>0.03)$ is $\boxed{0.9938}$.
