Find the mean and variance for the first $10$ multiples of $3$
The first $10$ multiples of $3$ are
$3,6,9,12,15,18,21,24,27,30$
Here, number of observations, $n=10$
Mean, $\bar x = \frac{{\sum\limits_{i = 1}^{10} {{x_i}} }}{{10}} = \frac{{165}}{{10}} = 16.5$
The following table is obtained.
${x_i}$ | $\left( {{x_i} - \bar x} \right)$ | ${\left( {{x_i} - \bar x} \right)^2}$ |
$3$ | $-13.5$ | $182.25$ |
$6$ | $-10.5$ | $110.25$ |
$9$ | $-7.5$ | $56.25$ |
$12$ | $-4.5$ | $20.25$ |
$15$ | $-1.5$ | $2.25$ |
$18$ | $1.5$ | $2.25$ |
$21$ | $4.5$ | $20.25$ |
$24$ | $7.5$ | $56.25$ |
$27$ | $10.5$ | $110.25$ |
$30$ | $13.5$ | $182.25$ |
Variance $\left( {{\sigma ^2}} \right) = \frac{1}{n}\sum\limits_{i = 1}^{10} {{{\left( {{x_1} - \bar x} \right)}^2} = } \frac{1}{{10}} \times 742.5 = 74.25$
$742.5$
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 data is obtained in tabular form as follows.
${x_i}$ | $60$ | $61$ | $62$ | $63$ | $64$ | $65$ | $66$ | $67$ | $68$ |
${f_i}$ | $2$ | $1$ | $12$ | $29$ | $25$ | $12$ | $10$ | $4$ | $5$ |
If $x_1, x_2,.....x_n$ are $n$ observations such that $\sum\limits_{i = 1}^n {x_i^2} = 400$ and $\sum\limits_{i = 1}^n {{x_i}} = 100$ , then possible value of $n$ among the following is
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 :
The mean and variance of $7$ observations are $8$ and $16$ respectively. If two observations are $6$ and $8 ,$ then the variance of the remaining $5$ observations is: