Find the mean and variance for the following frequency distribution.
Classes
$0-10$
$10-20$
$20-30$
$30-40$
$40-50$
Frequencies
$5$
$8$
$15$
$16$
$6$
Find the mean and variance for the following frequency distribution.
| Classes | $0-10$ | $10-20$ | $20-30$ | $30-40$ | $40-50$ |
| Frequencies | $5$ | $8$ | $15$ | $16$ | $6$ |
| Class | Frequency ${f_i}$ | Mid-point ${x_i}$ | ${y_i} = \frac{{{x_i} - 25}}{{10}}$ | ${y_i}^2$ | ${f_i}{y_i}$ | ${f_i}{y_i}^2$ |
| $0-10$ | $5$ | $5$ | $-2$ | $4$ | $-10$ | $20$ |
| $10-20$ | $8$ | $15$ | $-1$ | $1$ | $-8$ | $8$ |
| $20-30$ | $15$ | $25$ | $0$ | $0$ | $0$ | $0$ |
| $30-40$ | $16$ | $35$ | $1$ | $1$ | $16$ | $16$ |
| $40-50$ | $6$ | $45$ | $2$ | $4$ | $12$ | $24$ |
| $50$ | $10$ | $68$ |
Mean, $\bar x = A + \frac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{N} \times h$
$ = 25 + \frac{{10}}{{50}} \times 10 = 25 + 2 = 27$
Variance, $\left( {{\sigma ^2}} \right) = \frac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^5 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)}^2}} } \right]$
$=\frac{(10)^{2}}{(50)^{2}}\left[50 \times 68-(10)^{2}\right]$
$=\frac{1}{25}[3400-100]=\frac{3300}{25}$
$=132$
Similar Questions
While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
While calculating the mean and variance of 10 readings, a student wrongly used the reading 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
If the variance of the frequency distribution is $3$ then $\alpha$ is ......
$X_i$
$2$
$3$
$4$
$5$
$6$
$7$
$8$
Frequency $f_i$
$3$
$6$
$16$
$\alpha$
$9$
$5$
$6$
If the variance of the frequency distribution is $3$ then $\alpha$ is ......
| $X_i$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
| Frequency $f_i$ | $3$ | $6$ | $16$ | $\alpha$ | $9$ | $5$ | $6$ |
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The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking it was found that, in the observations, $20$ was misread as $5$ . Then, the correct variance is equal to......
The mean and standard deviation of $15$ observations are found to be $8$ and $3$ respectively. On rechecking it was found that, in the observations, $20$ was misread as $5$ . Then, the correct variance is equal to......
- [JEE MAIN 2022]
For the frequency distribution :
Variate $( x )$
$x _{1}$
$x _{1}$
$x _{3} \ldots \ldots x _{15}$
Frequency $(f)$
$f _{1}$
$f _{1}$
$f _{3} \ldots f _{15}$
where $0< x _{1}< x _{2}< x _{3}<\ldots .< x _{15}=10$ and
$\sum \limits_{i=1}^{15} f_{i}>0,$ the standard deviation cannot be
For the frequency distribution :
| Variate $( x )$ | $x _{1}$ | $x _{1}$ | $x _{3} \ldots \ldots x _{15}$ |
| Frequency $(f)$ | $f _{1}$ | $f _{1}$ | $f _{3} \ldots f _{15}$ |
where $0< x _{1}< x _{2}< x _{3}<\ldots .< x _{15}=10$ and
$\sum \limits_{i=1}^{15} f_{i}>0,$ the standard deviation cannot be
- [JEE MAIN 2020]
If each of given $n$ observations is multiplied by a certain positive number $'k'$, then for new set of observations -
If each of given $n$ observations is multiplied by a certain positive number $'k'$, then for new set of observations -