Find the mean and variance for the data $6,7,10,12,13,4,8,12$
$6,7,10,12,13,4,8,12$
Mean, $\bar x = \frac{{\sum\limits_{i = 1}^8 {{x_i}} }}{n}$
$=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}=9$
The following table is obtained
${x_i}$ | $\left( {{x_i} - \bar x} \right)$ | ${\left( {{x_i} - \bar x} \right)^2}$ |
$6$ | $-3$ | $9$ |
$7$ | $-2$ | $4$ |
$10$ | $-1$ | $1$ |
$12$ | $3$ | $9$ |
$13$ | $4$ | $16$ |
$4$ | $-5$ | $25$ |
$8$ | $-1$ | $1$ |
$12$ | $3$ | $9$ |
$74$ |
Variance $\left( {{\sigma ^2}} \right) = \frac{1}{n}\sum\limits_{i = 1}^8 {{{\left( {{x_i} - \bar x} \right)}^2} = \frac{1}{8} \times 74} = 9.25$
If the variance of the first $n$ natural numbers is $10$ and the variance of the first m even natural numbers is $16$, then $m + n$ is equal to
Consider $10$ observation $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{10}$. such that $\sum_{i=1}^{10}\left(x_i-\alpha\right)=2$ and $\sum_{i=1}^{10}\left(x_i-\beta\right)^2=40$, where $\alpha, \beta$ are positive integers. Let the mean and the variance of the observations be $\frac{6}{5}$ and $\frac{84}{25}$ respectively. The $\frac{\beta}{\alpha}$ is equal to :
Find the standard deviation for the following data:
${x_i}$ | $3$ | $8$ | $13$ | $18$ | $25$ |
${f_i}$ | $7$ | $10$ | $15$ | $10$ | $6$ |
The first of the two samples in a group has $100$ items with mean $15$ and standard deviation $3 .$ If the whole group has $250$ items with mean $15.6$ and standard deviation $\sqrt{13.44}$, then the standard deviation of the second sample is:
The variance of the first $n$ natural numbers is