Forty buses run these distances in one day. Use the assumed mean a = 425 to find the mean distance.

Step 1. The deviations. $d = x - 425$ is $-100, -50, 0, 50, 100$.

Step 2. The products. $fd$ is $-500, -450, 0, 400, 400$. The sum is $\sum fd = -150$.

Step 3. The mean. $\bar{x} = 425 + \dfrac{-150}{40} = 425 - 3.75 = 421.25\ \text{km}$.