1. **State the problem:** We want to find the probability that at least 240 out of 280 new employees will still be employed after one year, given the probability of any one employee staying is 0.82. 2. **Identify the distribution:** This is a binomial probability problem where $n=280$ and $p=0.82$. 3. **Use normal approximation to the binomial:** Since $n$ is large, we approximate the binomial distribution with a normal distribution with mean $\mu = np$ and variance $\sigma^2 = np(1-p)$. Calculate mean: $$\mu = 280 \times 0.82 = 229.6$$ Calculate variance: $$\sigma^2 = 280 \times 0.82 \times (1-0.82) = 280 \times 0.82 \times 0.18 = 41.328$$ Calculate standard deviation: $$\sigma = \sqrt{41.328} \approx 6.43$$ 4. **Apply continuity correction:** We want $P(X \geq 240)$, so use $P(X \geq 239.5)$ for the normal approximation. 5. **Calculate the z-score:** $$z = \frac{239.5 - 229.6}{6.43} = \frac{9.9}{6.43} \approx 1.54$$ 6. **Find the probability:** Using standard normal tables or a calculator, find $P(Z \geq 1.54)$. From tables, $P(Z \geq 1.54) = 1 - P(Z \leq 1.54) \approx 1 - 0.9382 = 0.0618$. 7. **Final answer:** The approximate probability that at least 240 employees remain after one year is about 0.0618 or 6.18%.