Saved Bookmarks
| 1. |
Mean and variance of five observations are `4` and `5.2` respectively. If three of these observations are `3, 4, 4` then find absolute difference between the other two observations (A) `3` (B) `7` (C) `2` (D) `5`A. 1B. 7C. 5D. 3 |
|
Answer» Correct Answer - B Given mean `bar(x) = 4` variance `sigma^(2)=5.20` and number of observation n = 5 Let `x_(1)=3, x_(2)=4, x_(3)=4 and x_(4), x_(5)` be the five observations So, `sum_(i=1)^(5)x_(i)=5*bar(x)=5xx4=20` `rArr x_(1)+x_(2)+x_(3)+x_(4)+x_(5)=20` `rArr 3+4+4+x_(4)+x_(5)=20` `rArr x_(4)+x_(5)=9" ...(i)" ` Now, variance `sigma^(2)=(sum_(i=1)^(5)x_(i)^(2))/(5)-(bar(x))^(2)` `rArr (x_(1)^(2)+x_(2)^(2)+x_(3)^(2)+x_(4)^(2)+x_(5)^(2))/(5)-(4)^(2)=5.20` `rArr (9+16+16+x_(4)^(2)+x_(5)^(2))/(5)=16+5.20` `rArr 41+x_(4)^(2)+x_(5)^(2)=5xx21.20` `rArr x_(4)^(2)+x_(5)^(2)=106-41` `rArr x_(4)^(2)+x_(5)^(2)=65 " ...(ii)" ` `because (x_(4)+x_(5))^(2)=x_(4)^(2)+x_(5)^(2)+2x_(4)x_(5)` `therefore 81=65+2x_(4)x_(5) " " `[from Eqs. (i) and (ii)] `rArr 16=2x_(4)x_(5)` `rArr x_(4)x_(5)=8 " ...(iii)" ` Now, `(|x_(4)-x_(5)|)^(2)=x_(4)^(2)+x_(5)^(2)-2x_(4)x_(5)` `=65-16 " " `[from Eqs. (ii) and (iii)] `=49` `rArr |x_(4)-x_(5)|=7` |
|