1. **State the problem:** We want to find the probability that exactly 16 out of 250 workers use public transportation, given that 5% of workers use it. 2. **Identify the distribution:** The number of workers using public transportation, $X$, follows a binomial distribution with parameters $n=250$ and $p=0.05$. 3. **Use normal approximation:** For large $n$, the binomial distribution can be approximated by a normal distribution with mean $\mu = np$ and variance $\sigma^2 = np(1-p)$. Calculate mean and standard deviation: $$\mu = 250 \times 0.05 = 12.5$$ $$\sigma = \sqrt{250 \times 0.05 \times 0.95} = \sqrt{11.875} \approx 3.447$$ 4. **Apply continuity correction:** To find $P(X=16)$, approximate $P(15.5 < X < 16.5)$ using the normal distribution. 5. **Convert to standard normal variable $Z$:** $$Z = \frac{X - \mu}{\sigma}$$ Calculate: $$Z_1 = \frac{15.5 - 12.5}{3.447} \approx 0.87$$ $$Z_2 = \frac{16.5 - 12.5}{3.447} \approx 1.16$$ 6. **Find probabilities from standard normal table:** $$P(15.5 < X < 16.5) \approx P(0.87 < Z < 1.16) = P(Z < 1.16) - P(Z < 0.87)$$ Using standard normal CDF values: $$P(Z < 1.16) \approx 0.8770$$ $$P(Z < 0.87) \approx 0.8078$$ 7. **Calculate final probability:** $$P(X=16) \approx 0.8770 - 0.8078 = 0.0692$$ **Answer:** The probability that exactly 16 workers use public transportation is approximately **0.0692**.