From the following frequency distribution table, find the values of `f_(1)` and `f_(12)`. The mean of the edats is 166.4 and the sum of the observation is 50. Take assumed mean 165.

Explanation

`bar(x)=A+bar(d)`
`:.166.4=165+bar(d):.bar(d)=1.4`
`bar(d)=(sigmaf_(i)d_(i))/(sigmaf_(i))`
`=(80-10f_(1)+10f_(2))/(33+f_(1)+f_(2))`
`:.1.4=(80-10f_(1)+10f_(2))/(33+f_(1)+f_(2))`
`:.1.4xx33+1.4f_(1)+1.4f_(2)=80-10f_(1)+10f_(2)`
`:.46.2+1.4f_(1)+1.4f_(2)=80-10f_(1)+10f_(2)`
`:.1.4f_(1)+1.4f_(2)+10f_(1)-10f_(2)=80-46.2`
`:.11.4f_(1)-8.6f_(2)=33.8`
`:.114f_(1)-86f_(2)=338`..... (Multiplying both the sides by 10)........1
Also `33+f_(1)+f_(2)=50`
`:.f_(1)+f_(2)=17`.............2
`:.86f_(1)+86f_(2)=1462`.....(Multiplying both the sides by 86)...........3
Adding equations 1 and 3
`{:(114f_(1)- 86f_(2)=338" "...(1)),(86f_(1)+86f_(2)=1462" "...(3)),(bar(200f_(1)" "=1800)" "therefore f_(1)=9):}`
Substituting `f_(1)=9` in equatioin 2
`9+f_(2)=17 :.f_(2)=17-9:.f_(2)=8`
Ans. The values of `f_(1)` and `f_(2)` are 9 and 8 respectively.