The variance of $20$ observations is $5 .$ If each observation is multiplied by $2,$ find the new variance of the resulting observations.

Vedclass pdf generator app on play store
Vedclass iOS app on app store

Let the observations be $x_{1}, x_{2}, \ldots, x_{20}$ and $\bar{x}$ be their mean. Given that variance $=5$ and $n=20 .$ We know that

Variance   $\left( {{\sigma ^2}} \right) = \frac{1}{n}\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \bar x} \right)}^2}} $

i.e., $5 = \frac{1}{{20}}\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \bar x} \right)}^2}} $

or    $\sum\limits_{i = 1}^{20} {{{\left( {{x_i} - \bar x} \right)}^2}}  = 100$        .......$(1)$

If each observation is multiplied by $2,$ and the new resulting observations are $y_{i},$ then

$y_{i}=2 x_{i} \text { i.e., } x_{i}=\frac{1}{2} y_{i}$

Therefore $\bar y = \frac{1}{n}\sum\limits_{i = 1}^{20} {{y_i}}  = \frac{1}{{20}}\sum\limits_{i = 1}^{20} {2{x_i} = 2.\frac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}} } $

i.e.  $\bar{y}=2 \bar{x} \quad$ or $\quad \bar{x}=\frac{1}{2} \bar{y}$

Substituting the values of $x_{i}$ and $\bar{x}$ in $(1),$ we get

${\sum\limits_{i = 1}^{20} {\left( {\frac{1}{2}{y_i} - \frac{1}{2}\bar y} \right)} ^2} = 100$ i.e.,  $\sum\limits_{i = 1}^{20} {{{\left( {{y_i} - \bar y} \right)}^2} = 400} $

Thus the variance of new observations $=\frac{1}{20} \times 400=20=2^{2} \times 5$

Similar Questions

Let the observations $\mathrm{x}_{\mathrm{i}}(1 \leq \mathrm{i} \leq 10)$ satisfy the equations, $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)=10$ and $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)^{2}=40$ If $\mu$ and $\lambda$ are the mean and the variance of the observations, $\mathrm{x}_{1}-3, \mathrm{x}_{2}-3, \ldots ., \mathrm{x}_{10}-3,$ then the ordered pair $(\mu, \lambda)$ is equal to :

  • [JEE MAIN 2020]

If $\mathop \sum \limits_{i = 1}^9 \left( {{x_i} - 5} \right) = 9$ and $\mathop \sum \limits_{i = 1}^9 {\left( {{x_i} - 5} \right)^2} = 45,$ then the standard deviation of the $9$ items  ${x_1},{x_2},\;.\;.\;.\;,{x_9}$ is :

  • [JEE MAIN 2018]

The mean of $5$ observations is $4.4$ and their variance is $8.24$. If three observations are $1, 2$ and $6$, the other two observations are

The mean and variance of $7$ observations are $8$ and $16$ respectively. If one observation $14$ is omitted a and $b$ are respectively mean and variance of remaining $6$ observation, then $a+3 b-5$ is equal to $..........$.

  • [JEE MAIN 2023]

The mean of the numbers $a, b, 8,5,10$ is $6$ and their variance is $6.8$. If $M$ is the mean deviation of the numbers about the mean, then $25\; M$ is equal to

  • [JEE MAIN 2022]