1

### 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
2

### 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)

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

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

$= \sqrt {5 - 1}$

$=$ $\sqrt 4$

$=2$
3

### JEE Main 2018 (Online) 15th April Morning Slot

The mean of set of 30 observations is 75. If each observation is multiplied by a non-zero number $\lambda$ and then each of them is decreased by 25, their mean remains the same. Then $\lambda$ is equal to :
A
${1 \over 3}$
B
${2 \over 3}$
C
${4 \over 3}$
D
${10 \over 3}$

## Explanation

As mean is a linear operation, so if each observation is multiplied by $\lambda$ and decreased by 25 then the mean becomes 75$\lambda $$-25. According to the question, 75\lambda - 25 = 75 \Rightarrow \lambda = {4 \over 3}. 4 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 15th April Evening Slot If the mean of the data : 7, 8, 9, 7, 8, 7, \lambda , 8 is 8, then the variance of this data is : A {7 \over 8} B 1 C {9 \over 8} D 2 ## Explanation \overline x = {{7 + 8 + 9 + 7 + 8 + 7 + \lambda + 8} \over 8} = 8 \Rightarrow$$\,\,\,$ ${{54 + \lambda } \over 8}$ = 8  $\Rightarrow$  $\lambda$ = 10

Now variance = $\sigma$2

= ${{{{\left( {7 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2} + {{\left( {9 - 8} \right)}^2} + {{\left( {7 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2} + {{\left( {7 - 8} \right)}^2} + {{\left( {10 - 8} \right)}^2} + {{\left( {8 - 8} \right)}^2}} \over 8}$

$\Rightarrow$  $\sigma$2 = ${{1 + 0 + 1 + 1 + 0 + 1 + 4 + 0} \over 8}$ = ${8 \over 8}$ = 1

Hence, the variance is 1.