Calculate the mean,variance,and standard deviation for the following distribution:
Class $30-40$ $40-50$ $50-60$ $60-70$ $70-80$ $80-90$ $90-100$ $f_i$ $3$ $7$ $12$ $15$ $8$ $3$ $2$
| 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) From the given data,we construct the following table:
Mean $\bar{x} = \frac{\sum f_ix_i}{N} = \frac{3100}{50} = 62$.
Variance $(\sigma^2) = \frac{1}{N} \sum f_i(x_i - \bar{x})^2 = \frac{9090}{50} = 181.8$.
Standard deviation $(\sigma) = \sqrt{181.8} \approx 13.48$.
| Class | $f_i$ | $x_i$ | $f_ix_i$ | $(x_i - \bar{x})^2$ | $f_i(x_i - \bar{x})^2$ |
| $30-40$ | $3$ | $35$ | $105$ | $729$ | $2187$ |
| $40-50$ | $7$ | $45$ | $315$ | $289$ | $2023$ |
| $50-60$ | $12$ | $55$ | $660$ | $49$ | $588$ |
| $60-70$ | $15$ | $65$ | $975$ | $9$ | $135$ |
| $70-80$ | $8$ | $75$ | $600$ | $49$ | $392$ |
| $80-90$ | $3$ | $85$ | $255$ | $529$ | $1587$ |
| $90-100$ | $2$ | $95$ | $190$ | $1089$ | $2178$ |
| Total | $N=50$ | - | $3100$ | - | $9090$ |
Mean $\bar{x} = \frac{\sum f_ix_i}{N} = \frac{3100}{50} = 62$.
Variance $(\sigma^2) = \frac{1}{N} \sum f_i(x_i - \bar{x})^2 = \frac{9090}{50} = 181.8$.
Standard deviation $(\sigma) = \sqrt{181.8} \approx 13.48$.
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DifficultJEE MAIN 2024
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