A transport office studies how long 70 passengers waited for a bus at one stop on a busy morning…

Step 1. The mean. $\sum f x = 6(15) + 11(25) + 22(35) + 18(45) + 9(55) + 4(65) = 90 + 275 + 770 + 810 + 495 + 260 = 2700$. So $\bar{x} = \dfrac{2700}{70} = 38.57$ minutes.

Step 2. The median. The running totals are $6, 17, 39$, and $\dfrac{N}{2} = 35$ is first reached in the class $30$ to $40$. Then $\text{median} = 30 + \dfrac{35 - 17}{22} \times 10 = 30 + 8.18 = 38.18$ minutes.

Step 3. The mode. The largest frequency is $22$, in the class $30$ to $40$. Then $\text{mode} = 30 + \dfrac{22 - 11}{44 - 11 - 18} \times 10 = 30 + \dfrac{11}{15} \times 10 = 37.33$ minutes.

Step 4. The rule of thumb. $3 \times 38.18 - 2 \times 38.57 = 114.55 - 77.14 = 37.4$ minutes.