JEE Main 12 Jan 2019 Evening Question 15
Question: The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are 3, 4 and 4; then the absolute value of the difference of the other two observations is [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A) 3
B) 1
C) 7
D) 5
Show Answer
Answer:
Correct Answer: C
Solution:
- Let the other two observations are $ x _1 $ and $ x _2 $
$ \therefore $ Mean $ (\overline{x})=\frac{3+4+4+x _1+x _2}{5}=4 $
$ \Rightarrow $ $ x _1+x _2=9 $ -(i)
And variance $ ({{\sigma }^{2}})=\frac{\sum\limits _{{}}^{{}}{x_i^{2}}}{n}-{{(\overline{x})}^{2}} $
$ \Rightarrow $ $ 5.20=\frac{9+16+16+x_1^{2}+x_2^{2}}{5}-16 $
$ \Rightarrow $ $ \frac{41+x_1^{2}+x_2^{2}}{5}=21.2\Rightarrow x_1^{2}+x_2^{2}=106-41 $
$ \Rightarrow $ $ x_1^{2}+x_2^{2}=65 $ …(ii)
Using (i) and (ii), we get $ {{(x _1+x _2)}^{2}}=81\Rightarrow 65+2x _1x _2=81 $
$ \Rightarrow $ $ 2x _1x _2=16 $
$ \therefore $ $ {{(x _1-x _2)}^{2}}=65-16=49 $
$ \Rightarrow $ $ |x _1-x _2|=7 $
