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Find the variance and standard deviation for the following data:
| ${x_i}$ | $4$ | $8$ | $11$ | $17$ | $20$ | $24$ | $32$ |
| ${f_i}$ | $3$ | $5$ | $9$ | $5$ | $4$ | $3$ | $1$ |
$6.77$
$6.77$
$6.77$
$6.77$
Solution
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 $\left( \sigma \right) = \sqrt {45.8} = 6.77$
Similar Questions
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 :
If the variance of the frequency distribution is $3$ then $\alpha$ is ……
| $X_i$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
| Frequency $f_i$ | $3$ | $6$ | $16$ | $\alpha$ | $9$ | $5$ | $6$ |
