1. **Problem Statement:** Find the probability of getting between 2 and 5 heads inclusive in 9 tosses of a fair coin. 2. **Part (i): Using the Binomial Distribution** - Number of trials $n = 9$. - Probability of success (head) in each trial $p = 0.5$. - We want $P(2 \leq X \leq 5)$ where $X \sim \text{Binomial}(9, 0.5)$. 3. The probability mass function (PMF) of $X$ is: $$P(X=k) = \binom{9}{k} (0.5)^k (0.5)^{9-k} = \binom{9}{k} (0.5)^9$$ 4. Calculate: $$P(2 \leq X \leq 5) = \sum_{k=2}^5 \binom{9}{k} (0.5)^9$$ 5. Calculate each term: - $\binom{9}{2} = 36$ - $\binom{9}{3} = 84$ - $\binom{9}{4} = 126$ - $\binom{9}{5} = 126$ 6. Sum: $$P = (36 + 84 + 126 + 126) (0.5)^9 = 372 \times \frac{1}{512} \approx 0.7266$$ 7. **Part (ii): Using the Normal Approximation to the Binomial** - The binomial distribution $X \sim \text{Binomial}(n=9, p=0.5)$ can be approximated by normal $Y \sim N(\mu, \sigma^2)$ with: $$\mu = np = 9 \times 0.5 = 4.5$$ $$\sigma = \sqrt{np(1-p)} = \sqrt{9 \times 0.5 \times 0.5} = 1.5$$ 8. Apply continuity correction: We want $P(2 \leq X \leq 5) \approx P(1.5 < Y < 5.5)$ 9. Standardize to find z-scores: $$z_1 = \frac{1.5 - 4.5}{1.5} = -2$$ $$z_2 = \frac{5.5 - 4.5}{1.5} = \frac{1}{1.5} = 0.6667$$ 10. Using standard normal table or calculator: $$P(1.5 < Y < 5.5) = P(-2 < Z < 0.6667) = \Phi(0.6667) - \Phi(-2)$$ - $\Phi(0.6667) \approx 0.7475$ - $\Phi(-2) = 1 - \Phi(2) \approx 1 - 0.9772 = 0.0228$ 11. Therefore: $$P \approx 0.7475 - 0.0228 = 0.7247$$ 12. **Final answers:** - Using binomial distribution: $\boxed{0.7266}$ - Using normal approximation: $\boxed{0.7247}$ The normal approximation is quite close to the exact binomial probability.