Find the variance and standard deviation for the following data:
${x_i}$
$4$
$8$
$11$
$17$
$20$
$24$
$32$
${f_i}$
$3$
$5$
$9$
$5$
$4$
$3$
$1$
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 $\left( \sigma \right) = \sqrt {45.8} = 6.77$
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${x_i}$
$6$
$10$
$14$
$18$
$24$
$28$
$30$
${f_i}$
$2$
$4$
$7$
$12$
$8$
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$3$
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$\mathrm{x}$
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$-1$
$3$
$4$
$6$
$\mathrm{P}(\mathrm{X}=\mathrm{x})$
$\frac{1}{5}$
$\mathrm{a}$
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$\frac{1}{5}$
$\mathrm{~b}$
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| $\mathrm{P}(\mathrm{X}=\mathrm{x})$ | $\frac{1}{5}$ | $\mathrm{a}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\mathrm{~b}$ |
If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :
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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......
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