1

### JEE Main 2016 (Online) 10th April Morning Slot

The mean of 5 observations is 5 and their variance is 124. If three of the observations are 1, 2 and 6 ; then the mean deviation from the mean of the data is :
A
2.4
B
2.8
C
2.5
D
2.6

## Explanation

Let 5 observations are x1, x2, x3, x4, x5

given,   x1 = 1, x2 = 2, x3 = 6

Mean = 5

$\therefore$   Mean$\left( {\overline x } \right)$ = ${{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5}$ = 5

$\Rightarrow$   1 + 2 + 6 + x4 + x5 = 25

$\therefore$   x4 + x5 = 16

$\Rightarrow$   (x4 $-$ 5) + (x5 $-$ 5) + 10 = 16

$\Rightarrow$   (x4 $-$ 5) + (x5 $-$ 5) = 6

$\therefore$   Mean deviation about mean,

=   ${{\sum {\left| {{x_i} - \overline x } \right|} } \over n}$

=    ${{\left| {1 - 5} \right| + \left| {2 - 5} \right| + \left| {6 - 5} \right| + \left| {{x_4} - 5} \right| + \left| {{x_5} - 5} \right|} \over 5}$

=   ${{4 + 3 + 1 + 6} \over 5}$

=    ${{14} \over 5}$

=   2.8
2

### JEE Main 2017 (Online) 8th April Morning Slot

The mean age of 25 teachers in a school is 40 years. A teacher retires at the age of 60 years and a new teacher is appointed in his place. If now the mean age of the teachers in this school is 39 years, then the age (in years) of the newly appointed teacher is :
A
25
B
30
C
35
D
40

## Explanation

Mrean $\left( {\overline x } \right)$ = ${{{x_1} + {x_2}..... + {x_n}} \over n}$ = ${{\sum x } \over n}$

Here mean = 40 of 25 teachers

$\therefore\,\,\,$ 40 = ${{\sum x } \over {25}}$

$\Rightarrow $$\,\,\, \sum x = 40 \times 25 = 1000 After retireing of a 60 year old teacher, total age of 24 teachers, x1 + x2 + . . . . . .x24 = 1000 - 60 = 940 Now a new teacher of age A year is appointed. \therefore\,\,\, Now total age of this 25 teachers x1 + x2 + x3 + . . . . . + x25 = 940 + A \therefore\,\,\, mean age = {{940 + A} \over {25}} According to question, {{940 + A} \over {25}} = 39 \Rightarrow$$\,\,\,$ A = 35
3

### JEE Main 2017 (Online) 9th April Morning Slot

The sum of 100 observations and the sum of their squares are 400 and 2475, respectively. Later on, three observations, 3, 4 and 5, were found to be incorrect. If the incorrect observations are omitted, then the variance of the remaining observations is :
A
8.25
B
8.50
C
8.00
D
9.00
4

### JEE Main 2018 (Offline)

If $\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right)} = 9$ and

$\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}} = 45$, then the standard deviation of the 9 items
${x_1},{x_2},.......,{x_9}$ is
A
3
B
9
C
4
D
2

## Explanation

IMPORTANT POINT :-

When every number is added or subtracted by a fixed number then the standard Deviation remain unchanged.

so let ${x_i} - 5 = {y_i}$

So, new equation is $\sum\limits_{i = 1}^9 {{y_i}} = 9$

and $\sum\limits_{i = 1}^9 {y_i^2} = 45$

As, we know. Standard Deviation (S.D)

$= \sqrt {{{\sum\limits_{i = 1}^9 {y_i^2} } \over 9} - {{\left( {{{\sum\limits_{i = 1}^9 {yi} } \over 9}} \right)}^2}}$

$= \sqrt {{{45} \over 9} - {{\left( {{9 \over 9}} \right)}^2}}$

$= \sqrt {5 - 1}$

$=$ $\sqrt 4$

$=2$