Q. If the standard deviation of $x_{1},x_{2},\ldots x_{n}$ is 3.5, then the standard deviation of $- 2x_{1} - 3,\ - 2x_{2} - 3,\ \ldots, - 2x_{n} - 3$ is
A) $- 7$
B) $- 4$
C) 7
D) 1.75
Explanation
(c)
If $d_{i} = \frac{x_{i} - A}{h}$, then $\sigma_{x} = |h|\sigma_{d}$
Now, $- 2x_{i} - 3 = \frac{x_{i} + \frac{3}{2}}{- \frac{1}{2}}$
Here, $h = - \frac{1}{2}$
$\therefore\ \ \ \sigma_{d} = \frac{1}{|h|}\sigma_{x}$
$= 2 \times 3.5 = 7$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.