Question

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

Marks	5	6	7	8	9
Number of Students	6	a	16	13	b

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

Answer 1

Marks (x)	Number of students (f)	fx
5	6	30
6	a	6a
7	16	112
8	13	104
9	b	9b
	`sumf = 35 + a + b`	`sumfx = 246 + 6a + 9b`

It is given that the number of students is 40.

∴ 35 + a + b = 40

`=>` a + b – 5 = 0    ...(1)

Mean = `(sumfx)/(sumf)`

`=> (246 + 6a + 9b)/(35 + a + b) = 7.2`

`=>` 246 + 6a + 9b = 7.2(35 + a + b)

`=>` 246 + 6a + 9b = 252 + 7.2a + 7.2b

`=>` 0 = 252 – 246 + 7.2a – 6a + 7.2b – 9b

`=>` 6 + 1.2a – 1.8b = 0

`=>` 10 + 2a – 3b = 0   ...(2)

Solving equations (1) and (2), we have

5a – 5 = 0

`=>` a = 1

From (1), we have b = 4

Hence, the values of a and b are 1 and 4 respectively.

Answer 2

Mean = `(sumfx)/(sumf)` 

`=> 7.2 = (6 xx 5 + a xx 6 + 16 xx 7 + 13 xx 8 + b xx 9)/(6 + a + 16 + 13 + b)`

`=> 7.2 = (246 + 6a + 9b)/(35 + a + b)` 

`=>` 1.2 – 1.8b = –6  ...(i)  

Total number of students = 6 + a + 16 + 13 + b 

`=>` 40 = 35 + a + b 

`=>` a + b = 5  ...(ii) 

Multiple equation (ii) by 1.8 and add it to equation (i)

1.8a + 1.8b = 9
1.2a – 1.8b = –6 
   3a            = 3

`=>` a = 1 

Substituting = 1 in equation (ii) we get,

1 + b = 5 

`=>` b = 4

Answer 3

Marks (x)	No. of Students (f)	fx
5	6	30
6	a	6a
7	16	112
8	13	104
9	b	9b
Total	`sumf = 35 + a + b`	`sumfx = 246 + 6a + 9b`

Now, `sumf` = 40

35 + a + b = 40

a + b = 5    ...(1)

And `barX = (sumfx)/(sumf)`

`7.2 = (246 + 6a + 9b)/(40)`

`=>` 6a + 9b + 246 = 288

`=>` 6a + 9b = 42

`=>` 2a + 3b = 14    ...(2)

From (1) and (2), 

a = 1, b = 4