1. **Problem statement:** Emily hits 70% of her free throws. She had 25 free throws last week. 2. **Formulas and rules:** For a binomial distribution with parameters $n$ (number of trials) and $p$ (probability of success), - Expected value (mean): $$\mu = np$$ - Standard deviation: $$\sigma = \sqrt{np(1-p)}$$ 3. **Calculations for part (i):** - Given $n=25$, $p=0.7$ - Expected number of hits: $$\mu = 25 \times 0.7 = 17.5$$ - Standard deviation: $$\sigma = \sqrt{25 \times 0.7 \times 0.3} = \sqrt{5.25} \approx 2.29$$ 4. **Problem statement for part (ii):** Emily had 30 free throws yesterday. Find the probability she made between 5 and 20 hits inclusive. 5. **Approach:** Use binomial distribution $B(n=30, p=0.7)$ and calculate $$P(5 \leq X \leq 20) = \sum_{k=5}^{20} \binom{30}{k} (0.7)^k (0.3)^{30-k}$$ 6. **Calculation:** This sum is best computed using a binomial cumulative distribution function (CDF): $$P(5 \leq X \leq 20) = P(X \leq 20) - P(X \leq 4)$$ Using a calculator or software, - $P(X \leq 20) \approx 0.996$ - $P(X \leq 4) \approx 0.0000003$ Therefore, $$P(5 \leq X \leq 20) \approx 0.996 - 0.0000003 = 0.9959997 \approx 0.996$$ --- **Final answers:** - Expected hits last week: 17.5 - Standard deviation last week: 2.29 - Probability of 5 to 20 hits yesterday: approximately 0.996