Calculate the mean, variance and standard deviation for the following distribution:
Class | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ | $80-90$ | $90-100$ |
$f_i$ | $3$ | $7$ | $12$ | $15$ | $8$ | $3$ | $2$ |
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 $\left( \sigma \right) = \sqrt {201} = 14.18$
Find the standard deviation for the following data:
${x_i}$ | $3$ | $8$ | $13$ | $18$ | $25$ |
${f_i}$ | $7$ | $10$ | $15$ | $10$ | $6$ |
Let $S$ be the set of all values of $a_1$ for which the mean deviation about the mean of $100$ consecutive positive integers $a _1, a _2, a _3, \ldots ., a _{100}$ is $25$. Then $S$ is
Let $ \bar x , M$ and $\sigma^2$ be respectively the mean, mode and variance of $n$ observations $x_1 , x_2,...,x_n$ and $d_i\, = - x_i - a, i\, = 1, 2, .... , n$, where $a$ is any number.
Statement $I$: Variance of $d_1, d_2,.....d_n$ is $\sigma^2$.
Statement $II$ : Mean and mode of $d_1 , d_2, .... d_n$ are $-\bar x -a$ and $- M - a$, respectively
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