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નીચે આપેલ માહિતી માટે વિચરણ અને પ્રમાણિત વિચલન શોધો :
| ${x_i}$ | $4$ | $8$ | $11$ | $17$ | $20$ | $24$ | $32$ |
| ${f_i}$ | $3$ | $5$ | $9$ | $5$ | $4$ | $3$ | $1$ |
$6.77$
$6.77$
$6.77$
$6.77$
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
ધારોકે માહિતી
| $X$ | $1$ | $3$ | $5$ | $7$ | $9$ |
| આવૃતિ $(f)$ | $4$ | $24$ | $28$ | $\alpha$ | $8$ |
નો મધ્યક $5$ છે.જો માહિતીના મધ્યક સાપેક્ષ સરેરાશ વિચલન અને વિચરણ અનુક્રમે $m$ અને $\sigma^2$ હોય, તો $\frac{3 \alpha}{m+\sigma^2}=……..$
આપેલ પ્રત્યેક માહિતી માટે મધ્યક અને વિચરણ શોધો :
| ${x_i}$ | $92$ | $93$ | $97$ | $98$ | $102$ | $104$ | $109$ |
| ${f_i}$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
