Question

If the mean of the following frequency distribution is 188, find the missing frequencies x and y, if the sum of all frequencies is 100.

Class	0 – 80	80 – 160	160 – 240	240 – 320	320 – 400
Frequency	20	25	x	y	10

Answer 1

1. Identify given parameters

From the problem description, we are given:

Total number of frequencies (Σf_i) = 100

Mean of the distribution `(barx) = 188`

2. Form frequency equation

The sum of all frequencies can be written as the sum of known values plus the unknown variables x and y:

20 + 25 + x + y + 10 = 100

55 + x + y = 100

x + y = 45   ...(Equation 1)

3. Calculate class marks

The class mark (x_i) for each interval is obtained by calculating the midpoint `("Lower Limit" + "Upper Limit")/2`:

Class Interval	Frequency (fi)	Class Mark (xi)	fixi
0 – 80	20	40	800
80 – 160	25	120	3000
160 – 240	x	200	200x
240 – 320	y	280	280y
320 – 400	10	360	3600
Total	100		7400 + 200x + 280y

4. Form mean equation

The formula for the mean of a grouped frequency distribution is:

`barx = (sumf_ix_i)/(sumf_i)`

Substitute the known values into the formula:

`188 = (7400 + 200x + 280y)/100`

18800 = 7400 + 200x + 280y

200x + 280y = 18800 – 7400

200x + 280y = 11400

Divide the entire equation by 20 to simplify:

10x + 14y = 570   ...(Equation 2)

5. Solve linear equations

Multiply Equation 1 by 10 to make coefficients match:

10x + 10y = 450   ...(Equation 3)

Subtract Equation 3 from Equation 2 to eliminate x:

(10x + 14y) – (10x + 10y) = 570 – 450

4y = 120

y = 30

Substitute y = 30 back into Equation 1:

x + 30 = 45

x = 15