1. **Problem Statement:** We have a production process with output per hour normally distributed with mean $\mu=92.0$ and standard deviation $\sigma=3.6$. A sample of size $n=4$ is taken. We need to find: (a) Mean of the sampling distribution of the sample mean. (b) Variance of the sampling distribution of the sample mean. (c) Standard error of the sampling distribution of the sample mean. (d) Probability that the sample mean exceeds 93.0 units. 2. **Formulas and Rules:** - The sampling distribution of the sample mean $\bar{X}$ for a sample size $n$ from a normal distribution $N(\mu, \sigma^2)$ is also normal with: - Mean: $\mu_{\bar{X}} = \mu$ - Variance: $\sigma^2_{\bar{X}} = \frac{\sigma^2}{n}$ - Standard error (SE): $\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}$ - To find the probability $P(\bar{X} > 93.0)$, convert to the standard normal variable $Z$: $$Z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}}$$ Then use standard normal tables or a calculator. 3. **Calculations:** (a) Mean of sampling distribution: $$\mu_{\bar{X}} = \mu = 92.0$$ (b) Variance of sampling distribution: $$\sigma^2_{\bar{X}} = \frac{\sigma^2}{n} = \frac{3.6^2}{4} = \frac{12.96}{4} = 3.24$$ (c) Standard error: $$\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{3.6}{2} = 1.8$$ (d) Probability $P(\bar{X} > 93.0)$: Calculate $Z$: $$Z = \frac{93.0 - 92.0}{1.8} = \frac{1.0}{1.8} \approx 0.5556$$ Using standard normal distribution tables or calculator: $$P(Z > 0.5556) = 1 - P(Z \leq 0.5556) \approx 1 - 0.7107 = 0.2893$$ 4. **Final answers:** - (a) Mean = 92.0 - (b) Variance = 3.24 - (c) Standard error = 1.8 - (d) Probability sample mean exceeds 93.0 = 0.2893