If variables $x$ and $u$ are related by $u = \frac{x - a}{h}$,then the correct relationship between $\sigma_x$ and $\sigma_u$ is:
- A$\sigma_x = |h| \sigma_u$
- B$\sigma_x = h + \sigma_u$
- C$\sigma_u = |h| \sigma_x$
- D$\sigma_u = h + \sigma_x$
Explore More
Similar Questions
The variance of the following data is:
$x_i$ $6$ $10$ $14$ $18$ $24$ $28$ $30$ $f_i$ $2$ $4$ $7$ $12$ $8$ $4$ $3$
| $x_i$ | $6$ | $10$ | $14$ | $18$ | $24$ | $28$ | $30$ |
|---|---|---|---|---|---|---|---|
| $f_i$ | $2$ | $4$ | $7$ | $12$ | $8$ | $4$ | $3$ |
If $x_1, x_2, ..., x_n$ are $n$ observations such that $\sum_{i=1}^n x_i^2 = 400$ and $\sum_{i=1}^n x_i = 100$,then which of the following is a possible value of $n$?
Given that the total of $16$ values is $528$ and the sum of the squares of deviations from $33$ is $9158$. The variance is:
The variance of the following frequency distribution is
Class Interval Frequency $0 - 6$ $10$ $6 - 12$ $8$ $12 - 18$ $6$ $18 - 24$ $4$ $24 - 30$ $2$
| Class Interval | Frequency |
|---|---|
| $0 - 6$ | $10$ |
| $6 - 12$ | $8$ |
| $12 - 18$ | $6$ |
| $18 - 24$ | $4$ |
| $24 - 30$ | $2$ |
Let $X = \{x \in N : 1 \leq x \leq 17\}$ and $Y = \{ax + b : x \in X \text{ and } a, b \in R, a > 0\}$. If the mean and variance of the elements of $Y$ are $17$ and $216$ respectively,then $a + b$ is equal to:
DifficultJEE MAIN 2020
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo