1. **Problem statement:** You roll a fair die $n=115$ times and count the number of times $X$ the top side shows a 6. 2. **Distribution of $X$:** $X$ follows a Binomial distribution with parameters $n=115$ and $p=\frac{1}{6} \approx 0.1666666667$. 3. **Mean and standard deviation of $X$:** Mean: $$\mu = np = 115 \times 0.1666666667 = 19.1666666705$$ Standard deviation: $$\sigma = \sqrt{np(1-p)} = \sqrt{115 \times 0.1666666667 \times (1-0.1666666667)} = \sqrt{115 \times 0.1666666667 \times 0.8333333333} = \sqrt{15.97222222} = 3.996525126$$ 4. **Distribution of sample proportion $\hat{p}$:** The sample proportion $\hat{p} = \frac{X}{n}$ is approximately Normal by the Central Limit Theorem for large $n$. Mean of $\hat{p}$: $$\mu_{\hat{p}} = p = 0.1666666667$$ Standard deviation of $\hat{p}$: $$\sigma_{\hat{p}} = \frac{\sigma}{n} = \frac{3.996525126}{115} = 0.0347497837$$ 5. **Probability that $\hat{p}$ is between 0.13 and 0.22:** Convert to $Z$-scores: $$Z_1 = \frac{0.13 - 0.1666666667}{0.0347497837} = -1.054092553$$ $$Z_2 = \frac{0.22 - 0.1666666667}{0.0347497837} = 1.544331066$$ Using standard Normal CDF values: $$P(Z < 1.544331066) = 0.9385$$ $$P(Z < -1.054092553) = 0.1461$$ Probability between 0.13 and 0.22: $$0.9385 - 0.1461 = 0.7924$$ 6. **Probability that $\hat{p} \geq 0.2261$:** Calculate $Z$-score: $$Z = \frac{0.2261 - 0.1666666667}{0.0347497837} = 1.707$$ Probability: $$P(Z \geq 1.707) = 1 - P(Z < 1.707) = 1 - 0.9569 = 0.0431$$ **Final answers:** (a) Mean = 19.1666666705, Standard deviation = 3.996525126 (b) The distribution of $\hat{p}$ is approximately Normal with mean $\mu_{\hat{p}} = 0.1666666667$ and standard deviation $\sigma_{\hat{p}} = 0.0347497837$ (c) Probability $0.13 \leq \hat{p} \leq 0.22$ is approximately 0.7924 (d) Probability $\hat{p} \geq 0.2261$ is approximately 0.0431