The mean and standard deviation of six observations are $8$ and $4,$ respectively. If each observation is multiplied by $3,$ find the new mean and new standard deviation of the resulting observations.
The mean and standard deviation of six observations are $8$ and $4,$ respectively. If each observation is multiplied by $3,$ find the new mean and new standard deviation of the resulting observations.
Let the observations be $x_{1}, x_{2}, x_{3}, x _{4}, x_{5} ,$ and $x_{6}$
It is given that mean is $8$ and standard deviation is $4$
Mean, $\bar{x}=\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}}{6}=8$ .......$(1)$
If each observation is multiplied by $3$ and the resulting observations are $y_{i},$ then
$y_{1}=3 x_{1}$ i.e., $x_{1}=\frac{1}{3} y_{1},$ for $i=1$ to $6$
New Mean, $\bar{y}=\frac{y_{1}+y_{2}+y_{3}+y_{4}+y_{5}+y_{6}}{6}$
$=\frac{3\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}+x_{6}\right)}{6}$
$=3 \times 8$ .......[ Using $(1)$ ]
$=28$
Standard deviation, $\sigma = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^6 {{{\left( {{x_1} - \bar x} \right)}^2}} } $
$\therefore {\left( 4 \right)^2} = \frac{1}{6}\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \bar x} \right)}^2}} $
$\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \bar x} \right)}^2}} = 96$ ........$(2)$
From $(1)$ and $(2),$ it can be observed that,
$\bar{y}=3 \bar{x}$
$\bar{x}=\frac{1}{3} \bar{y}$
Substituting the values of $x_{1}$ and $\bar{x}$ in $(2),$ we obtain
$\sum\limits_{i = 1}^6 {{{\left( {\frac{1}{3}{y_1} - \frac{1}{3}\bar y} \right)}^2} = 96} $
$ \Rightarrow \sum\limits_{i = 1}^6 {{{\left( {{y_1} - \bar y} \right)}^2} = 864} $
Therefore, variance of new observations $=\left(\frac{1}{6} \times 864\right)=144$
Hence, the standard deviation of new observations is $\sqrt{144}=12$
Similar Questions
Let $\mathrm{n}$ be an odd natural number such that the variance of $1,2,3,4, \ldots, \mathrm{n}$ is $14 .$ Then $\mathrm{n}$ is equal to ..... .
Let $\mathrm{n}$ be an odd natural number such that the variance of $1,2,3,4, \ldots, \mathrm{n}$ is $14 .$ Then $\mathrm{n}$ is equal to ..... .
- [JEE MAIN 2021]
In a series of $2n$ observation, half of them are equal to $'a'$ and remaining half observations are equal to $' -a'$. If the standard deviation of this observations is $2$ then $\left| a \right|$ equals
In a series of $2n$ observation, half of them are equal to $'a'$ and remaining half observations are equal to $' -a'$. If the standard deviation of this observations is $2$ then $\left| a \right|$ equals
- [JEE MAIN 2013]
Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in $A.P.$ and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then $8 \sigma^2$ is equal to
Let the six numbers $a_1, a_2, a_3, a_4, a_5, a_6$ be in $A.P.$ and $a_1+a_3=10$. If the mean of these six numbers is $\frac{19}{2}$ and their variance is $\sigma^2$, then $8 \sigma^2$ is equal to
- [JEE MAIN 2023]
The mean and variance of $20$ observations are found to be $10$ and $4,$ respectively. On rechecking, it was found that an observation $9$ was incorrect and the correct observation was $11$. Then the correct variance is
The mean and variance of $20$ observations are found to be $10$ and $4,$ respectively. On rechecking, it was found that an observation $9$ was incorrect and the correct observation was $11$. Then the correct variance is
- [JEE MAIN 2020]
For the frequency distribution :
Variate $( x )$
$x _{1}$
$x _{1}$
$x _{3} \ldots \ldots x _{15}$
Frequency $(f)$
$f _{1}$
$f _{1}$
$f _{3} \ldots f _{15}$
where $0< x _{1}< x _{2}< x _{3}<\ldots .< x _{15}=10$ and
$\sum \limits_{i=1}^{15} f_{i}>0,$ the standard deviation cannot be
For the frequency distribution :
| Variate $( x )$ | $x _{1}$ | $x _{1}$ | $x _{3} \ldots \ldots x _{15}$ |
| Frequency $(f)$ | $f _{1}$ | $f _{1}$ | $f _{3} \ldots f _{15}$ |
where $0< x _{1}< x _{2}< x _{3}<\ldots .< x _{15}=10$ and
$\sum \limits_{i=1}^{15} f_{i}>0,$ the standard deviation cannot be
- [JEE MAIN 2020]