Find the variance and standard deviation for the following data:

${x_i}$ $4$ $8$ $11$ $17$ $20$ $24$ $32$
${f_i}$ $3$ $5$ $9$ $5$ $4$ $3$ $1$

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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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If the data $x_1, x_2, ...., x_{10}$ is such that the mean of first four of these is $11$, the mean of the remaining six is $16$ and the sum of squares of all of these is $2,000$; then the standard deviation of this data is

  • [JEE MAIN 2019]