1. **Problem statement:** We have a binomial random variable $X$ with parameters $n=30$ and $p=0.60$. We want to find using the normal approximation: (i) $P(X=14)$ (the probability that $X$ is exactly 14). (ii) $P(X<12)$ (the probability that $X$ is less than 12). 2. **Formula and rules:** The binomial distribution $X \sim Binomial(n,p)$ can be approximated by a normal distribution $Y \sim N(\mu, \sigma^2)$ where: $$\mu = np$$ $$\sigma = \sqrt{np(1-p)}$$ Because $X$ is discrete and $Y$ is continuous, we use the continuity correction: - For $P(X=k)$, approximate $P(k-0.5 < Y < k+0.5)$. - For $P(X < k)$, approximate $P(Y < k - 0.5)$. 3. **Calculate mean and standard deviation:** $$\mu = 30 \times 0.60 = 18$$ $$\sigma = \sqrt{30 \times 0.60 \times 0.40} = \sqrt{7.2} \approx 2.683$$ 4. **(i) Calculate $P(X=14)$ using normal approximation:** Apply continuity correction: $$P(X=14) \approx P(13.5 < Y < 14.5)$$ Convert to standard normal $Z$: $$Z = \frac{Y - \mu}{\sigma}$$ Calculate $Z$ values: $$Z_1 = \frac{13.5 - 18}{2.683} \approx -1.68$$ $$Z_2 = \frac{14.5 - 18}{2.683} \approx -1.31$$ Find probabilities from standard normal table or calculator: $$P(13.5 < Y < 14.5) = P(Z_1 < Z < Z_2) = \Phi(-1.31) - \Phi(-1.68)$$ Using standard normal CDF values: $$\Phi(-1.31) \approx 0.0951$$ $$\Phi(-1.68) \approx 0.0465$$ So, $$P(X=14) \approx 0.0951 - 0.0465 = 0.0486$$ 5. **(ii) Calculate $P(X < 12)$ using normal approximation:** Apply continuity correction: $$P(X < 12) = P(X \leq 11) \approx P(Y < 11.5)$$ Convert to standard normal: $$Z = \frac{11.5 - 18}{2.683} \approx -2.44$$ Find probability: $$P(Y < 11.5) = \Phi(-2.44) \approx 0.0073$$ **Final answers:** (i) $P(X=14) \approx 0.0486$ (ii) $P(X<12) \approx 0.0073$