Find the mean and variance for the data
${x_i}$
$92$
$93$
$97$
$98$
$102$
$104$
$109$
${f_i}$
$3$
$2$
$3$
$2$
$6$
$3$
$3$
Find the mean and variance for the data
| ${x_i}$ | $92$ | $93$ | $97$ | $98$ | $102$ | $104$ | $109$ |
| ${f_i}$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
The data is obtained in tabular form as follows.
| ${x_i}$ | ${f_i}$ | ${f_i}{x_i}$ | ${{x_i} - \bar x}$ | ${\left( {{x_i} - \bar x} \right)^2}$ | ${f_i}{\left( {{x_i} - \bar x} \right)^2}$ |
| $92$ | $3$ | $276$ | $-8$ | $64$ | $192$ |
| $93$ | $2$ | $186$ | $-7$ | $49$ | $98$ |
| $97$ | $3$ | $291$ | $-3$ | $9$ | $27$ |
| $98$ | $2$ | $196$ | $-2$ | $4$ | $8$ |
| $102$ | $6$ | $612$ | $2$ | $4$ | $24$ |
| $104$ | $3$ | $312$ | $4$ | $16$ | $48$ |
| $109$ | $3$ | $327$ | $9$ | $81$ | $243$ |
| $22$ | $2200$ | $640$ |
Here, $N = 22,\sum\limits_{i = 1}^7 {{f_i}{x_i}} = 2200$
$\therefore \bar x = \frac{1}{n}\sum\limits_{i = 1}^7 {{f_i}{x_i}} = \frac{1}{{22}} \times 2200 = 100$
Variance $\left( {{\sigma ^2}} \right) = \frac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{{\left( {{x_i} - \bar x} \right)}^2} = } \frac{1}{{22}} \times 640 = 29.09$
Similar Questions
The variance $\sigma^2$ of the data is $ . . . . . .$
$x_i$
$0$
$1$
$5$
$6$
$10$
$12$
$17$
$f_i$
$3$
$2$
$3$
$2$
$6$
$3$
$3$
The variance $\sigma^2$ of the data is $ . . . . . .$
| $x_i$ | $0$ | $1$ | $5$ | $6$ | $10$ | $12$ | $17$ |
| $f_i$ | $3$ | $2$ | $3$ | $2$ | $6$ | $3$ | $3$ |
- [JEE MAIN 2024]
If the standard deviation of the numbers $-1, 0, 1, k$ is $\sqrt 5$ where $k > 0,$ then $k$ is equal to
If the standard deviation of the numbers $-1, 0, 1, k$ is $\sqrt 5$ where $k > 0,$ then $k$ is equal to
- [JEE MAIN 2019]
The diameters of circles (in mm) drawn in a design are given below:
Diameters
$33-36$
$37-40$
$41-44$
$45-48$
$49-52$
No. of circles
$15$
$17$
$21$
$22$
$25$
Calculate the standard deviation and mean diameter of the circles.
[ Hint : First make the data continuous by making the classes as $32.5-36.5,36.5-40.5,$ $40.5-44.5,44.5-48.5,48.5-52.5 $ and then proceed.]
The diameters of circles (in mm) drawn in a design are given below:
| Diameters | $33-36$ | $37-40$ | $41-44$ | $45-48$ | $49-52$ |
| No. of circles | $15$ | $17$ | $21$ | $22$ | $25$ |
Calculate the standard deviation and mean diameter of the circles.
[ Hint : First make the data continuous by making the classes as $32.5-36.5,36.5-40.5,$ $40.5-44.5,44.5-48.5,48.5-52.5 $ and then proceed.]
If the variance of the following frequency distribution is $50$ then $x$ is equal to:
Class
$10-20$
$20-30$
$30-40$
Frequency
$2$
$x$
$2$
If the variance of the following frequency distribution is $50$ then $x$ is equal to:
| Class | $10-20$ | $20-30$ | $30-40$ |
| Frequency | $2$ | $x$ | $2$ |
- [JEE MAIN 2020]
The standard deviation of $25$ numbers is $40$. If each of the numbers is increased by $5$, then the new standard deviation will be
The standard deviation of $25$ numbers is $40$. If each of the numbers is increased by $5$, then the new standard deviation will be