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 $