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.
- ABoth 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.
- CA is true but R is false.
- DA is false but R is true.
Answer
View full question & answer→(a) Both A and R are true and R is the correct explanation of A .
Explanation: Assertion: 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,
$\begin{array}{l}\bar{y}=\frac{1}{n} \sum_{i=1}^n y_i=\frac{1}{n} \sum_{i=1}^n\left(x_i+a\right) \\ =\frac{1}{n}\left[\sum_{i=1}^n x_i+\sum_{i=1}^n a\right] \\ =\frac{1}{n} \sum_{i=1}^n x_i+\frac{n a}{n}=\bar{x}+a\end{array}$
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.
Explanation: Assertion: 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,
$\begin{array}{l}\bar{y}=\frac{1}{n} \sum_{i=1}^n y_i=\frac{1}{n} \sum_{i=1}^n\left(x_i+a\right) \\ =\frac{1}{n}\left[\sum_{i=1}^n x_i+\sum_{i=1}^n a\right] \\ =\frac{1}{n} \sum_{i=1}^n x_i+\frac{n a}{n}=\bar{x}+a\end{array}$
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.