1

### JEE Main 2018 (Online) 16th April Morning Slot

The mean and the standarddeviation(s.d.) of five observations are9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :
A
0
B
1
C
2
D
4

## Explanation

Here mean = $\overline x$ = 9

$\Rightarrow$   $\overline x$ = ${{\sum {{x_i}} } \over n}$ = 9

$\Rightarrow$  ${\sum {{x_i}} }$ = 9 $\times$ 5 = 45

Now, standard deviation = 0

$\therefore\,\,\,$ all the five terms are same i.e.; 9

Now for changed observation

${\overline x _{new}}$ = ${{36 + {x_5}} \over 5} = 10$

$\Rightarrow$   x5 = 14

$\therefore\,\,\,$ $\sigma$new = $\sqrt {{{\sum {{{\left( {{x_i} - {{\overline x }_{new}}} \right)}^2}} } \over n}}$

= $\sqrt {{{4{{\left( {9 - 10} \right)}^2} + {{\left( {14 - 10} \right)}^2}} \over 5}}$ = 2
2

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

5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is :
A
16
B
22
C
20
D
18

## Explanation

Average height of 5 students,

$\overline x = {{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5} = 150$

$\Rightarrow \,\,\,\sum\limits_{i = 1}^5 {{x_i}} = 750$

We know,

Variance $\left( \sigma \right) = {{\sum {x_i^2} } \over 5} - {\left( {\overline x } \right)^2}$

given that,

${{\sum {x_i^2} } \over 5} - {\left( {150} \right)^2} = 18$

$\Rightarrow \,\,\,\sum {x_i^2} = 112590$

Height of new student, x6 $=$ 156 cm

New average height  $\left( {{{\overline x }_{new}}} \right) = {{750 + 156} \over 6} = 151$

New variance   $= {{\,\sum\limits_{i = 1}^6 {x_i^2} } \over 6} - {\left( {{{\overline x }_{new}}} \right)^2}$

$= {{112590 + {{\left( {156} \right)}^2}} \over 6} - {\left( {151} \right)^2}$

$= 22821 - 22801$

$= 20$
3

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

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

$\sigma$ $= \sqrt 5$
4

### 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

NEET