Find the variance and standard deviation for the following data:
${x_i}$ | $4$ | $8$ | $11$ | $17$ | $20$ | $24$ | $32$ |
${f_i}$ | $3$ | $5$ | $9$ | $5$ | $4$ | $3$ | $1$ |
Presenting the data in tabular form (Table), we get
${x_i}$ | ${f_i}$ | ${f_i}{x_i}$ | ${{x_i} - \bar x}$ | ${\left( {{x_i} - \bar x} \right)^2}$ | ${f_i}{\left( {{x_i} - \bar x} \right)^2}$ |
$4$ | $3$ | $12$ | $-10$ | $100$ | $300$ |
$8$ | $5$ | $40$ | $-6$ | $36$ | $180$ |
$11$ | $9$ | $99$ | $-3$ | $9$ | $81$ |
$17$ | $5$ | $85$ | $3$ | $9$ | $45$ |
$20$ | $4$ | $80$ | $6$ | $36$ | $144$ |
$24$ | $3$ | $72$ | $10$ | $100$ | $300$ |
$32$ | $1$ | $32$ | $18$ | $324$ | $324$ |
$30$ | $420$ | $1374$ |
$N = 30,\sum\limits_{i = 1}^7 {{f_i}{x_i}} = 420,\sum\limits_{i = 1}^7 {{f_i}{{\left( {{x_i} - \bar x} \right)}^2} = 1374} $
Therefore $\bar x = \frac{{\sum\limits_{i = 1}^7 {{f_i}{x_i}} }}{N} = \frac{1}{{30}} \times 420 = 14$
Hence Variance $\left( {{\sigma ^2}} \right) = \frac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{{\left( {{x_i} - \bar x} \right)}^2}} $
$\left( {{\sigma ^2}} \right) = \frac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{{\left( {{x_i} - \bar x} \right)}^2}} $
and Standard deviation $\left( \sigma \right) = \sqrt {45.8} = 6.77$
Find the mean and variance for the first $n$ natural numbers
Find the mean and variance for the data
${x_i}$ | $6$ | $10$ | $14$ | $18$ | $24$ | $28$ | $30$ |
${f_i}$ | $2$ | $4$ | $7$ | $12$ | $8$ | $4$ | $3$ |
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 $\mathrm{X}$ be a random variable with distribution.
$\mathrm{x}$ | $-2$ | $-1$ | $3$ | $4$ | $6$ |
$\mathrm{P}(\mathrm{X}=\mathrm{x})$ | $\frac{1}{5}$ | $\mathrm{a}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\mathrm{~b}$ |
If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :
The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking it was found that, in the observations, $20$ was misread as $5$ . Then, the correct variance is equal to......