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
Total Frequency	17	f1	32	f2	19	120

For equal class intervals, we will solve by finding mid points of these classes using direct method.

We have got

Mean = 50 and N = 120 (as given in the question)

Σf_i = 68 + f₁ + f₂ and Σf_ix_i = 3480 + 30f₁ + 70f₂

∵ mean is given by

⇒ (∵ given: mean is 50)

⇒ 3400 + 50f₁ + 50f₂ = 3480 + 30f₁ + 70f₂

⇒ 50f₁ – 30f₁ + 50f₂ – 70f₂ = 3480 – 3400

⇒ 20f₁ – 20f₂ = 80

⇒ f₁ – f₂ = 4 …(i)

As given in the question, frequency(Σf_i) = 120

And as calculated by us, frequency (Σf_i) = 68 + f₁ + f₂

Equalizing them, we get

68 + f₁ + f₂ = 120

⇒ f₁ + f₂ = 120 – 68 = 52

⇒ f₁ + f₂ = 52 …(ii)

We will now solve equations (i) and (ii), adding them we get

(f₁ + f₂) + (f₁ – f₂) = 52 + 4

⇒ 2f₁ = 56

⇒ f₁ = 56/2

⇒ f₁ = 28

Substitute f₁ = 28 in equation (ii),

28 + f₂ = 52

⇒ f₂ = 52 – 28

⇒ f₂ = 24

Thus, f₁ = 28 and f₂ = 24.