1. **State the problem:** We need to determine for which binomial distributions the normal distribution is not a reasonable approximation. 2. **Recall the rule:** The normal approximation to the binomial distribution is reasonable if both $np \geq 10$ and $n(1-p) \geq 10$. 3. **Calculate for each distribution:** - For $n=75$, $p=0.11$: $$np = 75 \times 0.11 = 8.25$$ $$n(1-p) = 75 \times 0.89 = 66.75$$ Since $np = 8.25 < 10$, normal approximation is **not** reasonable. - For $n=50$, $p=0.4$: $$np = 50 \times 0.4 = 20$$ $$n(1-p) = 50 \times 0.6 = 30$$ Both are $\geq 10$, so normal approximation is reasonable. - For $n=40$, $p=0.8$: $$np = 40 \times 0.8 = 32$$ $$n(1-p) = 40 \times 0.2 = 8$$ Since $n(1-p) = 8 < 10$, normal approximation is **not** reasonable. - For $n=40$, $p=0.12$: $$np = 40 \times 0.12 = 4.8$$ $$n(1-p) = 40 \times 0.88 = 35.2$$ Since $np = 4.8 < 10$, normal approximation is **not** reasonable. 4. **Conclusion:** The normal distribution is not a reasonable approximation for: - $n=75$, $p=0.11$ - $n=40$, $p=0.8$ - $n=40$, $p=0.12$ Only for $n=50$, $p=0.4$ is the normal approximation reasonable.