1

### JEE Main 2019 (Online) 11th January Morning Slot

The outcome of each of 30 items was observed; 10 items gave an outcome ${1 \over 2}$ – d each, 10 items gave outcome ${1 \over 2}$ each and the remaining 10 items gave outcome ${1 \over 2}$+ d each. If the variance of this outcome data is ${4 \over 3}$ then |d| equals :
A
${2 \over 3}$
B
${{\sqrt 5 } \over 2}$
C
${\sqrt 2 }$
D
2

## Explanation

Variance is independent of region. So we shift the given data by ${1 \over 2}$.

so,   ${{10{d^2} + 10 \times {0^2} + 10{d^2}} \over {30}} - {\left( 0 \right)^2} = {4 \over 3}$

$\Rightarrow$  d2 $=$ 2 $\Rightarrow$ $\left| d \right| = \sqrt 2$
2

### JEE Main 2019 (Online) 12th January Morning Slot

If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is :
A
31
B
50
C
51
D
30

## Explanation

$\sum\limits_{i = 1}^{50} {\left( {{x_i} - 30} \right) = 50}$

$\sum {{x_i}} = 50 \times 30 = 50$

$\sum {{x_i}} = 50 + 50 + 30$

Mean $= \overline x = {{\sum {{x_i}} } \over n} = {{50 \times 30 + 50} \over {50}}$

$= 30 + 1 = 31$
3

### JEE Main 2019 (Online) 12th January Evening Slot

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 :
A
1
B
7
C
3
D
5

## Explanation

mean $\overline x$ = 4, $\sigma$2 = 5.2, n = 5, . x1 = 3 x2 = 4 = x3

$\sum {{x_i}} = 20$

x4 + x5 = 9 . . . . . . (i)

${{\sum {x_i^2} } \over x} - {\left( {\overline x } \right)^2} = \sigma \Rightarrow \sum {x_i^2} = 106$

$x_4^2 + x_5^2 = 65$ . . . . . .(ii)

Using (i) and (ii) (x4 $-$ x5)2 = 49

$\left| {{x_4} - {x_5}} \right| = 7$
4

### JEE Main 2019 (Online) 8th April Morning Slot

The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is :
A
40
B
48
C
49
D
45

## Explanation

Given mean ($\mu$) = 8

variance (${\sigma ^2}$) = 16

No of observations (N) = 7

Let the two unknown observation = x and y

We know,

${\sigma ^2} = {{\sum {x_i^2} } \over N} - {\mu ^2}$ = 16

$\Rightarrow$ ${{{2^2} + {4^2} + {{10}^2} + {{12}^2} + {{14}^2} + {x^2} + {y^2}} \over 7} - {\left( 8 \right)^2}$ = 16

$\Rightarrow$ x2 + y2 = 100 ........(1)

We know,

$\mu$ = ${{\sum {{x_i}} } \over N}$ = 8

$\Rightarrow$ ${{2 + 4 + 10 + 12 + 14 + x + y} \over 7}$ = 8

$\Rightarrow$ x + y = 14 ...........(2)

As (x + y)2 = x2 + y2 + 2xy

$\Rightarrow$ (14)2 = 100 + 2xy

$\Rightarrow$ 196 = 100 + 2xy

$\Rightarrow$ xy = 48