A coaching centre grouped the scores of 50 students in a practice test, out of 100…

Step 1. The mean. $\sum f x = 6(10) + 9(30) + 15(50) + 12(70) + 8(90) = 60 + 270 + 750 + 840 + 720 = 2640$. So $\bar{x} = \dfrac{2640}{50} = 52.8$ marks.

Step 2. The distances. The class marks are $42.8, 22.8, 2.8, 17.2$ and $37.2$ away from the mean.

Step 3. Multiply by the frequencies. $6 \times 42.8 = 256.8$, then $205.2$, $42$, $206.4$ and $297.6$. The sum is $1008$.

Step 4. The mean deviation. $\text{MD} = \dfrac{1008}{50} = 20.16$ marks.

Step 5. The coefficient. $\dfrac{20.16}{52.8} = 0.382$.