1. **State the problem:** We want to find the binomial probability $P(X=20)$ where $n=42$ trials, probability of success $p=0.5$, and $X$ is the number of successes. 2. **Binomial probability formula:** $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. 3. **Calculate $P(X=20)$:** $$P(X=20) = \binom{42}{20} (0.5)^{20} (0.5)^{22} = \binom{42}{20} (0.5)^{42}$$ 4. **Calculate the combination:** $$\binom{42}{20} = \frac{42!}{20! \times 22!}$$ Using a calculator or software, $\binom{42}{20} = 269128937220$ approximately. 5. **Calculate the probability:** $$P(X=20) = 269128937220 \times (0.5)^{42}$$ Since $(0.5)^{42} = \frac{1}{2^{42}} \approx 2.273736754 \times 10^{-13}$, $$P(X=20) \approx 269128937220 \times 2.273736754 \times 10^{-13} = 0.0612$$ 6. **Check if normal approximation is appropriate:** The normal approximation to the binomial is appropriate if both $np$ and $n(1-p)$ are greater than 5. Here, $np = 42 \times 0.5 = 21$ and $n(1-p) = 21$, both greater than 5, so normal approximation is valid. 7. **Normal approximation parameters:** Mean: $$\mu = np = 21$$ Standard deviation: $$\sigma = \sqrt{np(1-p)} = \sqrt{42 \times 0.5 \times 0.5} = \sqrt{10.5} \approx 3.2404$$ 8. **Apply continuity correction:** We approximate $P(X=20)$ by $P(19.5 < X < 20.5)$ in the normal distribution. 9. **Calculate z-scores:** $$z_1 = \frac{19.5 - 21}{3.2404} = \frac{-1.5}{3.2404} \approx -0.463$$ $$z_2 = \frac{20.5 - 21}{3.2404} = \frac{-0.5}{3.2404} \approx -0.154$$ 10. **Find probabilities from standard normal table:** $$P(z < -0.154) \approx 0.439$$ $$P(z < -0.463) \approx 0.322$$ 11. **Calculate approximate probability:** $$P(19.5 < X < 20.5) = P(z < -0.154) - P(z < -0.463) = 0.439 - 0.322 = 0.117$$ 12. **Compare results:** Exact binomial probability: $0.0612$ Normal approximation: $0.117$ The normal approximation overestimates the probability in this case. **Final answer:** $$P(X=20) \approx 0.0612$$