Q. The standard deviation of the numbers 31, 32, 33,…., 46, 47 is

A) $\sqrt{\frac{17}{12}}$
B) $\sqrt{\frac{47^{2} - 1}{12}}$
C) $2\sqrt{6}$
D) $4\sqrt{3}$
Explanation

(c)
$\overline{x}$=$\frac{31 + 32 + 33 + \ldots + 47}{47} = \frac{663}{17} = 39$
$\therefore\sum_{i = 1}^{17}{\left( x_{i} - \overline{x} \right)^{2} = (31 - 39)^{2} + (32 - 39)^{2}}$
$+ (33 - 39)^{2} + (34 - 39)^{2} + (35 - 39)^{2}$
+$(36 - 39)^{2} + (37 - 39)^{2} + (38 - 39)^{2}$
$+ (39 - 39)^{2} + (40 - 39)^{2} + (41 - 39)^{2}$
$+ (42 - 39)^{2} + (43 - 39)^{2} + (44 - 39)^{2}$
$+ (45 - 39)^{2}$+$(46 - 39)^{2} + (47 - 39)^{2}$
$= 64 + 49 + 36 + 25 + 16 + 9 + 4 + 1 + 0 + 1$
+4+9+16+25+36+49+64
=408
Hence, standard deviation =$\sqrt{\frac{408}{17} =}\sqrt{24} = 2\sqrt{6}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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