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The variance of $20$ observations is $5 .$ If each observation is multiplied by $2,$ find the new variance of the resulting observations.
Solution
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
Find the mean and variance for the following frequency distribution.
Classes | $0-30$ | $30-60$ | $60-90$ | $90-120$ | $120-150$ | $50-180$ | $180-210$ |
$f_i$ | $2$ | $3$ | $5$ | $10$ | $3$ | $5$ | $2$ |
Calculate the mean, variance and standard deviation for the following distribution:
Class | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ | $80-90$ | $90-100$ |
$f_i$ | $3$ | $7$ | $12$ | $15$ | $8$ | $3$ | $2$ |