1

### JEE Main 2019 (Online) 9th January Evening Slot

A data consists of n observations : x1, x2, . . . . . . ., xn.

If     $\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n$    and

$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n,$

then the standard deviation of this data is :
A
2
B
$\sqrt 5$
C
5
D
$\sqrt 7$

## Explanation

$\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9n$

$\Rightarrow \sum\limits_{i = 1}^n {x_i^2} + 2\sum\limits_{i = 1}^n {{x_i}} + n = 9n\,\,\,\,\,...\,(1)$

$\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n$

$\Rightarrow \sum\limits_{i = 1}^n {x_i^2} - 2\sum\limits_{i = 1}^n {{x_i}} + n = 5n\,\,\,\,\,...\,(2)$

Performing (1) + (2), we get

$2\sum\limits_{i = 1}^n {x_i^2} + 2n = 14n$

$\sum\limits_{i = 1}^n {x_i^2} = 6n$

Performing (1) $-$ (2), we get

$\Rightarrow 4\sum\limits_{i = 1}^n {{x_i}} = 4n$

$\Rightarrow$$\Rightarrow \sum\limits_{i = 1}^n {{x_i}} = n$

S.D($\sigma$)$= \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {\overline x } \right)}^2}}$

$\sigma$ $= \sqrt {{{6n} \over n} - \left( 1 \right)}$

$\sigma$ $= \sqrt 5$
2

### JEE Main 2019 (Online) 10th January Morning Slot

In a class of 140 students numbered 1 to 140, all even numbered students opted Mathematics course, those whose number is divisible by 3 opted Physics course and those whose number is divisible by 5 opted Chemistry course. Then the number of students who did not opt for any of the three courses is
A
42
B
102
C
1
D
38

## Explanation

Let n(A) = number of students opted mathematic = 70,

n(B) = number of studens opted Physics = 46,

n(C) = number of students opted Chemistry = 28,

n(A $\cap$ B) = 23,

n(B $\cap$ C) = 9,

n(A $\cap$ C) = 14,

n(A $\cap$ B $\cap$ C) = 4,

Now n(A $\cup$ B $\cup$ C)

= n(A) + n(B) + n(C) $-$ n(A $\cap$ B) $-$ n(B $\cap$ C)

$-$ n(A $\cap$ C) + n(A $\cap$ B $\cap$ C)

= 70 + 46 + 28 $-$ 23 $-$ 9 $-$ 14 + 4 = 102

So number of students not opted for any course

Total $-$ n(A $\cup$ B $\cup$ C)

= 140 $-$ 102 = 38
3

### JEE Main 2019 (Online) 10th January Morning Slot

The mean of five observations is 5 and their variance is 9.20. If three of the given five observations are 1, 3 and 8, then a ratio of other two observations is -
A
6 : 7
B
10 : 3
C
4 : 9
D
5 : 8

## Explanation

Let two observations are x1 & x2

mean = ${{\sum {{x_i}} } \over 5} = 5$

$\Rightarrow 1 + 3 + 8 + {x_1} + {x_2} = 25$

$\Rightarrow {x_1} + {x_2} = 13$      . . . . (1)

variance $\left( {{\sigma ^2}} \right)$ = ${{\sum {x_i^2} } \over 5} - 25 = 9.20$

$\Rightarrow$  ${\sum {x_i^2 = 171} }$

$\Rightarrow$  $x_1^2 + x_2^2 = 97$      . . . . . (2)

$\Rightarrow$(x1 + x2)2 $-$ 2x1x2 = 97

$\Rightarrow$ 169 - 2x1x2 = 97

or   x1x2 = 36

$\therefore$  x1 : x2 = 4 : 9
4

### JEE Main 2019 (Online) 10th January Evening Slot

If mean and standard deviation of 5 observations x1, x2, x3, x4, x5 are 10 and 3, respectively, then the variance of 6 observations x1, x2, ….., x5 and –50 is equal to
A
582.5
B
507.5
C
586.5
D
509.5

## Explanation

$\overline x = 10 \Rightarrow \sum\limits_{i = 1}^5 {{x_i} = 50}$

S.D.  $= \sqrt {{{\sum\limits_{i = 1}^5 {x_i^2} } \over 5} - {{\left( {\overline x } \right)}^2}} = 8$

$\Rightarrow \,\sum\limits_{i = 1}^5 {{{\left( {{x_i}} \right)}^2}} = 109$

variance $= \,\,{{\sum\limits_{i = 1}^5 {{{\left( {{x_i}} \right)}^2}} + {{\left( { - 50} \right)}^2}} \over 6} - \left( {\sum\limits_{i = 1}^5 {{{{x_i} - 50} \over 6}} } \right)$

$= \,\,507.5$