# CLASS-9MEAN & MEDIAN - UNGROUPED FREQUENCY DISTRIBUTION - DIRECT METHOD

MEAN & MEDIAN- UNGROUPED FREQUENCY DISTRIBUTION-

DIRECT METHOD-

Let, y₁, y₂, y₃,…………, yₑ of a variable y, occurring with frequencies f₁, f₂, f₃,…………, fₑ respectively.

Then the mean of these given observation is given by –

(f₁y₁ + f₂y₂ +………..+ fₑyₑ)           ∑  fᵢ yᵢ

Mean, ӯ = --------------------------- = -----------

(f₁ + f₂ + ……………+ fₑ)              ∑  fᵢ

Example.1) The ages of 50 students in a class given below – Find the mean age of class.

Ans.) As per the given condition, we should prepare the table given below  - Ʃ fᵢyᵢ           761

So, mean age = --------- = -------- =  15.22 years   (Ans.)

Ʃ fᵢ            50

Example.2) If the mean of the following frequency distribution is 24.5. Then find the value of z. Ans.) We prepare the table as under – Ʃ fᵢyᵢ

As per the given condition, mean age = ----------- =  22.5

Ʃ fᵢ

1360 + 25 Z

Or, ----------------- =  22.5

65 + Z

Or, 1360 + 25 Z = 22.5 X (65 + Z)

Or, 1360 + 25 Z =  1462.5 + 22.5 Z

Or, 25 Z – 22.5 Z = 1462.5 – 1360

Or,        - 2.5 Z = 102.5

Or,               Z = 41

Hence the obtained value of Z = 41     (Ans.)

Example.3) The mean of the following distribution is 16.5 Find the missing frequencies of P & Q

Ans.) We prepare the table given below – Here,  Ʃ fᵢ = 100 + P + Q.

But,  Ʃ fᵢ =  180

So,  100 + P + Q =  180

So,   P + Q  = 80 …………………….(i)

Ʃ fᵢyᵢ            1624 + 15 P + 20 Q

Also, mean = ----------- = ----------------------

Ʃ fᵢ                100 + P + Q

But, as per the given condition the mean is given 16.5

1624 + 15 P + 20 Q

So, ------------------------  =  16.5

100 + P + Q

1624 + 20 P + 20 Q – 5P

So, ---------------------------- =  16.5

100 + P + Q

1624 + 20 (P + Q) – 5P

So,  ------------------------- =  16.5

100 + (P + Q)

Now, we will substitute the value of (i), and we get –

1624 + (20 X 80) – 5P

So,  ------------------------  =  16.5

100 + 80

So,   1624 + 1600 – 5P = (16.5 X 180)

So,     3224 – 5P = 2970

So,     5P = 3224 – 2970 = 254

So,     5P = 254

So,      P = 50.8

Now we will substitute the value of P in (i), and we get –

P + Q  =  80

Or,   50.8 + Q = 80

Or,   Q = 80 – 50.8 = 29.2

Hence,  P = 50.8, and Q = 29.2    (Ans.)