If the mean of the following frequency distribution is 65.6,find the missing frequency f1 and f2 by using assumed mean method
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There was one mistake please send the table first for doing this question
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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
65.6=
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↑
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