Find the mean and variance for the data
${x_i}$ | $92$ | $93$ | $97$ | $98$ | $102$ | $104$ | $109$ |
${f_i}$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
The data is obtained in tabular form as follows.
${x_i}$ | ${f_i}$ | ${f_i}{x_i}$ | ${{x_i} - \bar x}$ | ${\left( {{x_i} - \bar x} \right)^2}$ | ${f_i}{\left( {{x_i} - \bar x} \right)^2}$ |
$92$ | $3$ | $276$ | $-8$ | $64$ | $192$ |
$93$ | $2$ | $186$ | $-7$ | $49$ | $98$ |
$97$ | $3$ | $291$ | $-3$ | $9$ | $27$ |
$98$ | $2$ | $196$ | $-2$ | $4$ | $8$ |
$102$ | $6$ | $612$ | $2$ | $4$ | $24$ |
$104$ | $3$ | $312$ | $4$ | $16$ | $48$ |
$109$ | $3$ | $327$ | $9$ | $81$ | $243$ |
$22$ | $2200$ | $640$ |
Here, $N = 22,\sum\limits_{i = 1}^7 {{f_i}{x_i}} = 2200$
$\therefore \bar x = \frac{1}{n}\sum\limits_{i = 1}^7 {{f_i}{x_i}} = \frac{1}{{22}} \times 2200 = 100$
Variance $\left( {{\sigma ^2}} \right) = \frac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{{\left( {{x_i} - \bar x} \right)}^2} = } \frac{1}{{22}} \times 640 = 29.09$
Find the mean and variance for the following frequency distribution.
Classes | $0-30$ | $30-60$ | $60-90$ | $90-120$ | $120-150$ | $50-180$ | $180-210$ |
$f_i$ | $2$ | $3$ | $5$ | $10$ | $3$ | $5$ | $2$ |
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
The variance of $20$ observations is $5 .$ If each observation is multiplied by $2,$ find the new variance of the resulting observations.
If each of given $n$ observations is multiplied by a certain positive number $'k'$, then for new set of observations -
If the mean deviation about the mean of the numbers $1,2,3, \ldots ., n$, where $n$ is odd, is $\frac{5(n+1)}{n}$, then $n$ is equal to