Find the mean and variance for the following frequency distribution.
Classes | $0-30$ | $30-60$ | $60-90$ | $90-120$ | $120-150$ | $50-180$ | $180-210$ |
$f_i$ | $2$ | $3$ | $5$ | $10$ | $3$ | $5$ | $2$ |
Class |
Freq ${f_i}$ |
Mid-point ${x_i}$ |
${y_i} = \frac{{{x_i} - 105}}{{30}}$ | ${y_i}^2$ | ${f_i}{y_i}$ | ${f_i}{y_i}^2$ |
$0-30$ | $2$ | $15$ | $-3$ | $9$ | $-6$ | $18$ |
$30-60$ | $3$ | $45$ | $-2$ | $4$ | $-6$ | $12$ |
$60-90$ | $5$ | $75$ | $-1$ | $1$ | $-5$ | $5$ |
$90-120$ | $10$ | $105$ | $0$ | $0$ | $0$ | $0$ |
$120-150$ | $3$ | $135$ | $1$ | $1$ | $3$ | $3$ |
$150-180$ | $5$ | $165$ | $2$ | $4$ | $10$ | $20$ |
$180-210$ | $2$ | $195$ | $3$ | $9$ | $6$ | $18$ |
$30$ | $2$ | $76$ |
Mean, $ \bar x = A + \frac{{\sum\limits_{i = 1}^7 {{f_i}{y_i}} }}{N} \times h$
$ = 105 + \frac{2}{{30}} \times 30 = 105 + 2 = 107$
Variance, $\left( {{\sigma ^2}} \right) = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^7 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)}^2}} } \right]$
$=\frac{(30)^{2}}{(30)^{2}}\left[30 \times 76-(2)^{2}\right]$
$=2280-4$
$=2276$
The mean and variance of eight observations are $9$ and $9.25,$ respectively. If six of the observations are $6,7,10,12,12$ and $13,$ find the remaining two observations.
Find the mean and variance for the first $n$ natural numbers
In any discrete series (when all values are not same) the relationship between $M.D.$ about mean and $S.D.$ is
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:
Let $v_1 =$ variance of $\{13, 1 6, 1 9, . . . . . , 103\}$ and $v_2 =$ variance of $\{20, 26, 32, . . . . . , 200\}$, then $v_1 : v_2$ is