1. **State the problem:** We want to find the probability of observing 97 or fewer recoveries out of 150 patients if the new drug is equally effective as the standard drug, which has a recovery rate of 75% within three days. 2. **Identify the distribution:** The number of recoveries follows a binomial distribution with parameters $n=150$ and $p=0.75$. 3. **Calculate mean and standard deviation:** - Mean $\mu = np = 150 \times 0.75 = 112.5$ - Standard deviation $\sigma = \sqrt{np(1-p)} = \sqrt{150 \times 0.75 \times 0.25} = \sqrt{28.125} \approx 5.3$ 4. **Use normal approximation:** Since $n$ is large, approximate the binomial with a normal distribution $N(112.5, 5.3^2)$. 5. **Apply continuity correction:** We want $P(X \leq 97)$, so use $P(X \leq 97.5)$ for the normal approximation. 6. **Calculate the z-score:** $$z = \frac{97.5 - 112.5}{5.3} = \frac{-15}{5.3} \approx -2.83$$ 7. **Find the probability:** Using standard normal tables or a calculator, $P(Z \leq -2.83) \approx 0.0023$. **Final answer:** The probability of observing 97 or fewer recoveries if the new drug is equally effective is approximately $0.0023$, or 0.23%. This is a very small probability, suggesting the new drug may be less effective than the standard.