Question

If the mean of the following frequency distribution is 65.6,find the missing frequency f1 and f2 by using assumed mean method

Answer 1

There was one mistake please send the table first for doing this question

Answer 2

Answer:

f₁ = 50 and  fixi = 2260+60f₁+100f₂

5+8+f₁ +20+f₂+2=50   { add all the value of fi}

35+f₁+f₂=50

• f₁+f₂=15 ...(1)

mean=65.6

$mean= \frac{fixi}{fi}$

65.6=$\frac{2260+60f_{1} +100f_{2} }{50}$

65.6 × 50 = 2260+60f₁+100f₂

3280 =2260+60f₁+100f₂

3280- 2260=60f₁+100f₂

1020=60f₁+100f₂

canceling,

• 3f₁+5f₂=51 ...(2)

now multiply eq. (1) by 3

• 3f₁+3f₂=45...(3)

now we can subtract eq. (3) from eq. (2)

3f₁+5f₂=51

3f₁+3f₂=45

2f₂=6 =3      

now put the value of f₂=3 in eq. (3)

3f₁+3f₂=45

3f₁+3×3=45

3f₁+9=45

3f₁=45 - 9

3f₁=36

f₁=12

                                 

DO IT WITH DIRECT METHOD , ITS VERY EASY↑