Q. The standard deviation for the scores 1, 2, 3, 4, 5, 6 and 7 is 2. Then, the standard deviation of 12, 23, 34, 45, 56 , 67 and 78 is
A) 2
B) 4
C) 22
D) 11
Explanation
(c)
Here, $n = 7,\$sum=315
$\therefore\ \ \ \ Mean = \frac{315}{7} = 45$
Now, standard deviation
$= \sqrt{\frac{\begin{array}{r}
(12 - 45)^{2} + (23 - 45)^{2} + \\
(34 - 45)^{2} + (45 - 45)^{2} \\
+ (56 - 45)^{2} + (67 - 45)^{2} \\
+ (78 - 45)^{2}
\end{array}}{7}}$
$= \sqrt{\frac{2(1089 + 484 + 121)}{7} = \sqrt{\frac{3388}{7}}}$
$\sqrt{484} = 22$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.