1. **Problem statement:** Use the normal approximation to the binomial distribution to find the probability of getting more than 10 and fewer than 15 successes in 24 rolls. 2. **Formula and rules:** The binomial distribution $B(n,p)$ can be approximated by a normal distribution $N(\mu, \sigma^2)$ where: $$\mu = np$$ $$\sigma = \sqrt{np(1-p)}$$ This approximation is appropriate when $n$ is large and $p$ is not too close to 0 or 1. 3. **Given:** Number of trials $n=24$. The probability of success $p$ is not given explicitly, so we assume it is known or estimated from context. For demonstration, let's assume $p=0.5$ (fair coin-like scenario). 4. **Calculate mean and standard deviation:** $$\mu = 24 \times 0.5 = 12$$ $$\sigma = \sqrt{24 \times 0.5 \times 0.5} = \sqrt{6} \approx 2.449$$ 5. **Apply continuity correction:** We want $P(10 < X < 15)$ which is $P(11 \leq X \leq 14)$ for discrete binomial. Using continuity correction for normal approximation: $$P(10.5 < X < 14.5)$$ 6. **Convert to standard normal variable $Z$:** $$Z = \frac{X - \mu}{\sigma}$$ Calculate: $$Z_1 = \frac{10.5 - 12}{2.449} = \frac{-1.5}{2.449} \approx -0.612$$ $$Z_2 = \frac{14.5 - 12}{2.449} = \frac{2.5}{2.449} \approx 1.021$$ 7. **Find probabilities from standard normal table:** $$P(Z < 1.021) \approx 0.846$$ $$P(Z < -0.612) \approx 0.270$$ 8. **Calculate final probability:** $$P(10 < X < 15) = P(10.5 < X < 14.5) = P(-0.612 < Z < 1.021) = 0.846 - 0.270 = 0.576$$ **Final answer:** The probability of getting more than 10 and fewer than 15 successes in 24 rolls is approximately $0.576$.