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13.Statistics
hard
Calculate the mean, variance and standard deviation for the following distribution:
| Class | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ | $80-90$ | $90-100$ |
| $f_i$ | $3$ | $7$ | $12$ | $15$ | $8$ | $3$ | $2$ |
A
$14.18$
B
$14.18$
C
$14.18$
D
$14.18$
Solution
From the given data, we construct the following Table
| Class |
Freq $\left( {{f_i}} \right)$ |
Mid-point $\left( {{x_i}} \right)$ |
${f_i}{x_i}$ | ${\left( {{x_i} – \bar x} \right)^2}$ | ${f_i}{\left( {{x_i} – \bar x} \right)^2}$ |
| $30-40$ | $3$ | $35$ | $105$ | $729$ | $2187$ |
| $40-50$ | $7$ | $45$ | $315$ | $289$ | $2023$ |
| $50-60$ | $12$ | $55$ | $660$ | $49$ | $588$ |
| $60-70$ | $15$ | $6$ | $975$ | $9$ | $135$ |
| $70-80$ | $8$ | $75$ | $600$ | $169$ | $1352$ |
| $80-90$ | $3$ | $85$ | $255$ | $529$ | $1587$ |
| $90-100$ | $2$ | $95$ | $190$ | $1089$ | $2178$ |
| $50$ | $3100$ | $10050$ |
Thus Mean $\bar x = \frac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{x_i}} = \frac{{3100}}{{50}} = 62$
Variance $\left( {{\sigma ^2}} \right) = \frac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{{\left( {{x_i} – \bar x} \right)}^2}} $
$ = \frac{1}{{50}} \times 10050 = 201$
and Standerd deviation $\left( \sigma \right) = \sqrt {201} = 14.18$
Standard 11
Mathematics
Similar Questions
Find the mean and variance for the following frequency distribution.
| Classes | $0-30$ | $30-60$ | $60-90$ | $90-120$ | $120-150$ | $50-180$ | $180-210$ |
| $f_i$ | $2$ | $3$ | $5$ | $10$ | $3$ | $5$ | $2$ |
hard
