MCQ 11 Mark
Assertion $(A):$ If each of the observations $x_1, x_2, \ldots, x_n$ is increased by $a$, where $a$ is a negative or positive number, then the variance remains unchanged.
Reason $(R):$ Adding or subtracting a positive or negative number to $($or from$)$ each observation of a group does not affect the variance.
Reason $(R):$ Adding or subtracting a positive or negative number to $($or from$)$ each observation of a group does not affect the variance.
- ✓Both $A$ and $R$ are true and $R$ is the correct explanation of $A.$
- BBoth $A$ and $R$ are true but $R$ is not the correct explanation of $A$.
- C$A$ is true but $R$ is false.
- D$A$ is false but $R$ is true.
Answer
View full question & answer→Correct option: A.
Both $A$ and $R$ are true and $R$ is the correct explanation of $A.$
Let $\overline{ x }$ be the mean of $x _1, x _2 \ldots, x _{ n }$.
Then, variance is given by
If a is added to each observation, the new observations will be
$y_i=x_i+a$
Let the mean of the new observations be $\overline{ y }$.
Then,
$\bar{y}=\frac{1}{n} \sum_{i=1}^n y_i=\frac{1}{n} \sum_{i=1}^n\left(x_i+a\right)$
$=\frac{1}{n}\left[\sum_{i=1}^n x_i+\sum_{i=1}^n a\right]$
$=\frac{1}{n} \sum_{i=1}^n x_i+\frac{n a}{n}=\bar{x}+a$
i.e. $\bar{y}=\bar{x}+a \ldots (ii)$
Thus, the variance of the new observations is
$\sigma_2^2=\frac{1}{n} \sum_{i=1}^n\left(y_i-\bar{y}\right)^2=\frac{1}{n} \sum_{i=1}^n\left(x_i+a-\bar{x}-a\right)^2$ $($using Eqs. $(i)$ and $(ii))$
$=\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2=\sigma_1^2$
Thus, the variance of the new observations is same as that of the original observations.
Reason: We may note that adding $($or subtracting$)$ a positive number to $($or from$)$ each observation of a group does not affect the variance.
Then, variance is given by
If a is added to each observation, the new observations will be
$y_i=x_i+a$
Let the mean of the new observations be $\overline{ y }$.
Then,
$\bar{y}=\frac{1}{n} \sum_{i=1}^n y_i=\frac{1}{n} \sum_{i=1}^n\left(x_i+a\right)$
$=\frac{1}{n}\left[\sum_{i=1}^n x_i+\sum_{i=1}^n a\right]$
$=\frac{1}{n} \sum_{i=1}^n x_i+\frac{n a}{n}=\bar{x}+a$
i.e. $\bar{y}=\bar{x}+a \ldots (ii)$
Thus, the variance of the new observations is
$\sigma_2^2=\frac{1}{n} \sum_{i=1}^n\left(y_i-\bar{y}\right)^2=\frac{1}{n} \sum_{i=1}^n\left(x_i+a-\bar{x}-a\right)^2$ $($using Eqs. $(i)$ and $(ii))$
$=\frac{1}{n} \sum_{i=1}^n\left(x_i-\bar{x}\right)^2=\sigma_1^2$
Thus, the variance of the new observations is same as that of the original observations.
Reason: We may note that adding $($or subtracting$)$ a positive number to $($or from$)$ each observation of a group does not affect the variance.