1. **Problem Statement:** You roll a fair die $n=115$ times. Each roll is a success if the top side shows a 6, failure otherwise. Let $X$ be the number of successes (6s). 2. **Distribution of $X$:** $X$ follows a Binomial distribution with parameters $n=115$ and $p=\frac{1}{6} \approx 0.1666666667$ because the probability of rolling a 6 on a fair die is $\frac{1}{6}$. 3. **Mean and Standard Deviation of $X$:** The mean of a Binomial is $\mu = np$. The standard deviation is $\sigma = \sqrt{np(1-p)}$. Calculate: $$\mu = 115 \times 0.1666666667 = 19.1666666705$$ $$\sigma = \sqrt{115 \times 0.1666666667 \times (1 - 0.1666666667)} = \sqrt{115 \times 0.1666666667 \times 0.8333333333}$$ $$= \sqrt{15.97222222} = 3.996525$$ 4. **Distribution of the sample proportion $\hat{p}$:** The sample proportion $\hat{p} = \frac{X}{n}$. By the Central Limit Theorem, for large $n$, $\hat{p}$ is approximately Normal with mean and standard deviation: $$\mu_{\hat{p}} = p = 0.1666666667$$ $$\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.1666666667 \times 0.8333333333}{115}} = \sqrt{0.0012077922} = 0.034758$$ 5. **Probability that $\hat{p}$ is between 13% and 22%:** Convert percentages to decimals: 0.13 and 0.22. Standardize these values to $Z$-scores: $$Z_1 = \frac{0.13 - 0.1666666667}{0.034758} = \frac{-0.0366666667}{0.034758} = -1.0545$$ $$Z_2 = \frac{0.22 - 0.1666666667}{0.034758} = \frac{0.0533333333}{0.034758} = 1.5340$$ Using standard Normal distribution tables or a calculator: $$P(Z < 1.5340) = 0.9373$$ $$P(Z < -1.0545) = 0.1461$$ Therefore, $$P(0.13 < \hat{p} < 0.22) = 0.9373 - 0.1461 = 0.7912$$ 6. **Probability that $\hat{p} \geq 0.2261$ (observed proportion):** Calculate $Z$-score for $\hat{p} = 0.2261$: $$Z = \frac{0.2261 - 0.1666666667}{0.034758} = \frac{0.0594333333}{0.034758} = 1.7100$$ Find $P(Z \geq 1.7100)$: $$P(Z \geq 1.7100) = 1 - P(Z < 1.7100) = 1 - 0.9564 = 0.0436$$ **Final answers:** (a) Mean $= 19.1666666705$, Standard deviation $= 3.996525$ (b) Distribution of $\hat{p}$ is approximately Normal with mean $\mu_{\hat{p}} = 0.1666666667$ and standard deviation $\sigma_{\hat{p}} = 0.034758$ (c) Probability $P(0.13 < \hat{p} < 0.22) = 0.7912$ (d) Probability $P(\hat{p} \geq 0.2261) = 0.0436$