Question: The Intelligence Quotient (IQ) test scores for adults are normally distributed with a mean of 100 and a standard deviation of 15. What is the probability we could select a sample of 50 adults and find the mean of this sample is less than 95?
1. **State the problem:** We want to find the probability that the sample mean IQ score of 50 adults is less than 95, given that the population mean $\mu = 100$ and population standard deviation $\sigma = 15$.
2. **Formula and rules:** For a sample mean, the sampling distribution is normal with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$, where $n$ is the sample size.
3. Calculate the standard error (SE) of the mean:
$$SE = \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{50}} = \frac{15}{7.071} \approx 2.121$$
4. Find the $z$-score for the sample mean $\bar{x} = 95$:
$$z = \frac{\bar{x} - \mu}{SE} = \frac{95 - 100}{2.121} = \frac{-5}{2.121} \approx -2.357$$
5. Use the standard normal distribution table or calculator to find the probability $P(Z < -2.357)$.
From the table, $P(Z < -2.357) \approx 0.0091$.
6. **Interpretation:** There is about a 0.91% chance that the sample mean IQ of 50 adults is less than 95.
**Final answer:** $0.0091$