Marks obtained by 40 students in a short assessment is given below, where a and b are two missing data :

Marks	No. of students
5	6
6	a
7	16
8	13
9	b

If the mean of the distribution is 7.2, find a and b.

We construct the following table:

Given, total no. of students = 40 and mean = 7.2

∴ 35 + a + b = 40
⇒ a + b = 5
⇒ a = 5 - b      …..(i)

$\text{Mean} = \dfrac{∑f$ixi}{∑f_i} \\[1em] \therefore 7.2 = \dfrac{246 + 6a + 9b}{40} \\[1em] \Rightarrow 288 = 246 + 6a + 9b \\[1em] \Rightarrow 288 - 246 = 6a + 9b \\[1em] \Rightarrow 6a + 9b = 42

Putting value of a from Eq (i)

$\Rightarrow 6(5 - b) + 9b = 42 \\[1em] \Rightarrow 30 - 6b + 9b = 42 \\[1em] \Rightarrow 30 + 3b = 42 \\[1em] \Rightarrow 3b = 12 \\[1em] \Rightarrow b = 4.$

a = 5 - b = 5 - 4 = 1.

Hence, the value of a = 1 and b = 4.