Question: The population of weights of a particular fruit is normally distributed, with a mean of $359$ grams and a standard deviation of $16$ grams. If $8$ fruits are picked at random, then what is the weight of the sample mean will exceed $5\%$ of the time?
Hint: Since the problem asks for the value that the sample mean is greater than $5\%$ of the time, the area to use in the inverse normal function is the area to the left, which is: $100\%-5\%$.
Round your answer to the nearest gram.
1. **State the problem:** We have a population with mean $\mu=359$ grams and standard deviation $\sigma=16$ grams. We pick a sample of size $n=8$. We want to find the value $x$ such that the sample mean exceeds $x$ only $5\%$ of the time.
2. **Understand the sampling distribution:** The sample mean $\bar{X}$ of size $n$ is normally distributed with mean $\mu_{\bar{X}}=\mu=359$ and standard deviation $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{16}{\sqrt{8}}$.
3. **Calculate the standard deviation of the sample mean:**
$$\sigma_{\bar{X}}=\frac{16}{\sqrt{8}}=\frac{16}{2.8284}=5.6569$$
4. **Determine the z-score corresponding to the left area of $95\%$:** Since the sample mean exceeds $x$ only $5\%$ of the time, the area to the left is $95\%$ or $0.95$. From standard normal tables or using inverse normal function,
$$z=1.645$$
5. **Use the z-score formula to find $x$:**
$$z=\frac{x-\mu_{\bar{X}}}{\sigma_{\bar{X}}} \implies x=\mu_{\bar{X}} + z \cdot \sigma_{\bar{X}}$$
6. **Calculate $x$:**
$$x=359 + 1.645 \times 5.6569=359 + 9.31=368.31$$
7. **Round to the nearest gram:**
$$x \approx 368$$ grams
**Final answer:** The weight of the sample mean will exceed approximately $368$ grams only $5\%$ of the time.