1. The problem is to find the probability of getting between 2 and 5 heads inclusive in 9 tosses of a fair coin. This means finding $P(2 \leq X \leq 5)$ where $X$ is the number of heads. 2. (i) Using the binomial distribution: Here, $X \sim \text{Binomial}(n=9, p=0.5)$ because each toss is a Bernoulli trial with success probability $p=0.5$. 3. The probability mass function (PMF) for the binomial distribution is: $$P(X=k) = \binom{9}{k} (0.5)^k (0.5)^{9-k} = \binom{9}{k} (0.5)^9$$ 4. We want $P(2 \leq X \leq 5) = \sum_{k=2}^5 P(X=k) = (0.5)^9 \sum_{k=2}^5 \binom{9}{k}$. 5. Calculate the binomial coefficients: $\binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36$ $\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$ $\binom{9}{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$ $\binom{9}{5} = \frac{9 \times 8 \times 7 \times 6 \times 5}{5 \times 4 \times 3 \times 2 \times 1} = 126$ 6. Sum the coefficients: $36 + 84 + 126 + 126 = 372$. 7. Compute the probability: $$P(2 \leq X \leq 5) = 372 \times (0.5)^9 = 372 \times \frac{1}{512} \approx 0.7265625$$ 8. (ii) Using the normal approximation to the binomial: The binomial $X$ can be approximated by $Y \sim N(\mu, \sigma^2)$ with $\mu = np = 9 \times 0.5 = 4.5$ $\sigma = \sqrt{np(1-p)} = \sqrt{9 \times 0.5 \times 0.5} = \sqrt{2.25} = 1.5$ 9. Apply continuity correction: We approximate $P(2 \leq X \leq 5)$ by $$P(1.5 < Y < 5.5)$$ 10. Standardize to the standard normal variable $Z = \frac{Y - \mu}{\sigma}$: $$P\left(\frac{1.5 - 4.5}{1.5} < Z < \frac{5.5 - 4.5}{1.5}\right) = P(-2 < Z < 0.6667)$$ 11. Using standard normal tables or calculator, find: $$P(Z < 0.6667) \approx 0.7475$$ $$P(Z < -2) \approx 0.0228$$ 12. Therefore, $$P(-2 < Z < 0.6667) = 0.7475 - 0.0228 = 0.7247$$ 13. Final answers: - Binomial probability: $\boxed{0.7266}$ - Normal approximation probability: $\boxed{0.7247}$ Both methods provide very close results indicating the normal approximation is reasonable here.