1

### 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. 2 MCQ (Single Correct Answer) ### 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 3 MCQ (Single Correct Answer) ### 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 4 MCQ (Single Correct Answer) ### 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($\sigma$)$= \sqrt {{{\sum {x_i^2} } \over n} - {{\left( {\overline x } \right)}^2}}$

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

$\sigma$ $= \sqrt 5$