Find the variance of the following data: $6,8,10,12,14,16,18,20,22,24$
Find the variance of the following data: $6,8,10,12,14,16,18,20,22,24$
From the given data we can form the following Table The mean is calculated by step-deviation method taking $14$ as assumed mean. The number of observations is $n=10$
| ${x_i}$ | ${d_i} = \frac{{{x_i} - 14}}{2}$ |
Deviations orom mean $\left( {{x_i} - \bar x} \right)$ |
$\left( {{x_i} - \bar x} \right)$ |
| $6$ | $-4$ | $-9$ | $81$ |
| $8$ | $-3$ | $-7$ | $49$ |
| $10$ | $-2$ | $-5$ | $25$ |
| $12$ | $-1$ | $-3$ | $9$ |
| $14$ | $0$ | $-1$ | $1$ |
| $16$ | $1$ | $1$ | $1$ |
| $18$ | $2$ | $3$ | $9$ |
| $20$ | $3$ | $5$ | $25$ |
| $22$ | $4$ | $7$ | $49$ |
| $24$ | $5$ | $9$ | $81$ |
| $5$ | $330$ |
Therefore $Mean\,\,\bar x = $ assumed mean $ + \frac{{\sum\limits_{i = 1}^n {{d_i}} }}{n} \times h$
$ = 14 + \frac{5}{{10}} \times 2 = 15$
and Veriance $\left( {{\sigma ^2}} \right) = \frac{1}{n}\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - \bar x} \right)}^2} = \frac{1}{{10}} \times 330 = 33} $
Thus Standard deviation $\left( \sigma \right) = \sqrt {33} = 5.74$
Similar Questions
The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$
Height
Weight
Mean
$162.6\,cm$
$52.36\,kg$
Variance
$127.69\,c{m^2}$
$23.1361\,k{g^2}$
Can we say that the weights show greater variation than the heights?
The following values are calculated in respect of heights and weights of the students of a section of Class $\mathrm{XI}:$
| Height | Weight | |
| Mean | $162.6\,cm$ | $52.36\,kg$ |
| Variance | $127.69\,c{m^2}$ | $23.1361\,k{g^2}$ |
Can we say that the weights show greater variation than the heights?
The mean and standard deviation of some data for the time taken to complete . a test are calculated with the following results:
Number of observations $=25,$ mean $=18.2$ seconds, standard deviation $=3.25 s$
Further, another set of 15 observations $x_{1}, x_{2}, \ldots, x_{15},$ also in seconds, is now available and we have $\sum_{i=1}^{15} x_{i}=279$ and $\sum_{i=1}^{15} x_{i}^{2}=5524 .$ Calculate the standard deviation based on all 40 observations.
The mean and standard deviation of some data for the time taken to complete . a test are calculated with the following results:
Number of observations $=25,$ mean $=18.2$ seconds, standard deviation $=3.25 s$
Further, another set of 15 observations $x_{1}, x_{2}, \ldots, x_{15},$ also in seconds, is now available and we have $\sum_{i=1}^{15} x_{i}=279$ and $\sum_{i=1}^{15} x_{i}^{2}=5524 .$ Calculate the standard deviation based on all 40 observations.
If the variance of observations ${x_1},\,{x_2},\,......{x_n}$ is ${\sigma ^2}$, then the variance of $a{x_1},\,a{x_2}.......,\,a{x_n}$, $\alpha \ne 0$ is
If the variance of observations ${x_1},\,{x_2},\,......{x_n}$ is ${\sigma ^2}$, then the variance of $a{x_1},\,a{x_2}.......,\,a{x_n}$, $\alpha \ne 0$ is
Find the mean and variance for the first $n$ natural numbers
Find the mean and variance for the first $n$ natural numbers
Let the mean and variance of $8$ numbers $x , y , 10$, $12,6,12,4,8$, be $9$ and $9.25$ respectively. If $x > y$, then $3 x-2 y$ is equal to $...........$.
Let the mean and variance of $8$ numbers $x , y , 10$, $12,6,12,4,8$, be $9$ and $9.25$ respectively. If $x > y$, then $3 x-2 y$ is equal to $...........$.
- [JEE MAIN 2023]