1. **Problem statement:** We are given that the average age (mean) of bank managers is $\mu = 40$ years, and the standard deviation is $\sigma = 5$ years. The ages are normally distributed. We want to find the probability that a randomly selected bank manager's age is between 35 and 46 years. 2. **Formula and concept:** For a normal distribution, the probability that a value $X$ lies between $a$ and $b$ is given by: $$P(a < X < b) = P\left(\frac{a - \mu}{\sigma} < Z < \frac{b - \mu}{\sigma}\right)$$ where $Z$ is a standard normal variable with mean 0 and standard deviation 1. 3. **Calculate the Z-scores:** $$Z_1 = \frac{35 - 40}{5} = \frac{-5}{5} = -1$$ $$Z_2 = \frac{46 - 40}{5} = \frac{6}{5} = 1.2$$ 4. **Find the probabilities from the standard normal distribution table:** - $P(Z < -1) = 0.1587$ - $P(Z < 1.2) = 0.8849$ 5. **Calculate the probability between 35 and 46:** $$P(35 < X < 46) = P(-1 < Z < 1.2) = P(Z < 1.2) - P(Z < -1) = 0.8849 - 0.1587 = 0.7262$$ 6. **Interpretation:** There is approximately a 72.62% chance that a randomly selected bank manager's age is between 35 and 46 years. **Final answer:** $$\boxed{0.7262}$$