1. **Problem statement:** We are given a binomial distribution with parameters $n=100$ and $p=0.2$ (probability of defective chip). We want to find the probabilities for (i) at most 15 defective chips and (ii) exactly 15 defective chips using normal approximation. 2. **Formula and conditions:** The binomial distribution $B(n,p)$ can be approximated by a normal distribution $N(\mu, \sigma^2)$ where: $$\mu = np$$ $$\sigma = \sqrt{npq}$$ with $q=1-p$. The approximation is valid since $np=20 > 15$ and $nq=80 > 15$. 3. **Calculate mean and standard deviation:** $$\mu = 100 \times 0.2 = 20$$ $$\sigma = \sqrt{100 \times 0.2 \times 0.8} = \sqrt{16} = 4$$ 4. **Apply continuity correction:** - For "at most 15" (i.e., $X \leq 15$), use $X \leq 15.5$ for normal approximation. - For "exactly 15", use the interval $14.5 < X < 15.5$. 5. **Calculate Z-scores:** (i) For $X \leq 15.5$: $$Z = \frac{15.5 - 20}{4} = \frac{-4.5}{4} = -1.125$$ (ii) For $14.5 < X < 15.5$: $$Z_1 = \frac{14.5 - 20}{4} = -1.375$$ $$Z_2 = \frac{15.5 - 20}{4} = -1.125$$ 6. **Find probabilities from standard normal table:** (i) $P(X \leq 15) \approx P(Z \leq -1.125) = 0.1303$ (ii) $P(X = 15) \approx P(14.5 < X < 15.5) = P(-1.375 < Z < -1.125) = P(Z < -1.125) - P(Z < -1.375) = 0.1303 - 0.0840 = 0.0463$ **Final answers:** - (i) Probability at most 15 defective chips $\approx 0.1303$ - (ii) Probability exactly 15 defective chips $\approx 0.0463$