1. **Problem Statement:** We want to approximate the probability $P(X \geq x)$ for a binomial random variable $X$ using a normal distribution. 2. **Formula:** The normal approximation to the binomial uses the mean $\mu = np$ and standard deviation $\sigma = \sqrt{np(1-p)}$ where $n$ is the number of trials and $p$ is the probability of success. 3. **Normal Approximation:** We approximate $P(X \geq x)$ by $$P\left(Z \geq \frac{x - \mu}{\sigma}\right)$$ where $Z$ is a standard normal variable. 4. **Continuity Correction:** Since $X$ is discrete and $Z$ is continuous, apply a continuity correction: $$P(X \geq x) \approx P\left(Z \geq \frac{x - 0.5 - \mu}{\sigma}\right)$$ 5. **Steps to Calculate:** - Calculate $\mu = np$ - Calculate $\sigma = \sqrt{np(1-p)}$ - Compute the z-score with continuity correction: $z = \frac{x - 0.5 - \mu}{\sigma}$ - Find $P(Z \geq z)$ using standard normal tables or software. This method provides a good approximation when $n$ is large and $p$ is not too close to 0 or 1. **Final answer:** $$P(X \geq x) \approx P\left(Z \geq \frac{x - 0.5 - np}{\sqrt{np(1-p)}}\right)$$