Question

Find the missing frequencies f₁ and f₂ in the table given below, it being given that the mean of the given frequency distribution is 50.

Class	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100	Total
Frequency	17	f1	32	f2	19	120

Answer 1

1. Formulate the first equation using total frequency

The total sum of the frequencies is given as 120. Let’s add the given frequencies together:

Σf_i = 17 + f₁ + 32 + f₂ + 19 = 120

Simplify by adding the known numbers (17 + 32 + 19 = 68):

68 + f₁ + f₂ = 120

f₁ + f₂ = 120 – 68

f₁ + f₂ = 52   ...(Equation 1)

2. Determine class marks (x_i) and Formulate the mean equation

The class mark (x_i) is the midpoint of each class interval, calculated as `("Lower Limit" + "Upper Limit")/2`.

Class Interval	Frequency (fi)	Midpoint (xi)	Product (fi × xi)
0 – 20	17	10	170
20 – 40	f1	30	30f1
40 – 60	32	50	1600
60 – 80	f2	70	70f2
80 – 100	19	90	1710
Total	Σfi = 120		Σfixi = 3480 + 30f1 + 70f2

The formula for the mean `(barx)` is:

`barx = (sumf_ix_i)/(sumf_i)`

Given that the mean `(barx)` is 50:

`50 = (3480 + 30f_1 + 70f_2)/120`

Multiply both sides by 120:

6000 = 3480 + 30f₁ + 70f₂

30f₁ + 70f₂ = 6000 – 3480

30f₁ + 70f₂ = 2520

Divide the entire equation by 10 to simplify:

3f₁ + 7f₂ = 252   ...(Equation 2)

3. Solve the system of linear qquations

Multiply Equation 1 by 3 to line up the f₁ coefficients:

3(f₁ + f₂) = 3(52)

⇒ 3f₁ + 3f₂ = 156   ...(Equation 3)

Subtract Equation 3 from Equation 2:

(3f₁ + 7f₂) – (3f₁ + 3f₂) = 252 – 156

4f₂ = 96

f₂ = `96/4`

f₂ = 24

Substitute f₂ = 24 back into Equation 1 to find f₁:

f₁ + 24 = 52

f₁ = 52 – 24

f₁ = 28

The value of the missing frequency f₁ is 28 and f₂ is 24.