Consider three observations a, b and c, such that b = a + c. If the standard deviation of a+2, b+2, c+2 is d, then which of the following is true? [16 Mar 2021 Shift 1]

Correct Answer is : b2=3(a2+c2)−9d2b² = 3(a² + c²) - 9 d²b2=3(a2+c2)−9d2

Correct Answer is : b2=3(a2+c2)−9d2b² = 3(a² + c²) - 9 d²b2=3(a2+c2)−9d2

Test Index

Statistics Part 1

Section: Mathematics

Question : 42 of 100

Marks: +1, -0

a, b

c

b = a + c

a+2, b+2, c+2

d

Solution:

Given, three observations

= a, b, c

b = a + c

Standard deviation

= \sigma

\sigma = \sqrt{ \; \frac{ \sum (x_i - \mu)^2 }{ N } }

where,

\mu

is the mean of observation

x_i

and

N

is the total number of observations.

When

x_1 = a+2, x_2 = b+2, x_3 = c+2

\mu = \; \frac{a+2+b+2+c+2}{3} = \; \frac{2b}{3} + 2

x_1 - \mu = (a+2) - \left( \frac{2b}{3} + 2 \right) = a - \frac{2b}{3}

x_2 - \mu = (b+2) - \left( \frac{2b}{3} + 2 \right) = b - \frac{2b}{3}

x_3 - \mu = (c+2) - \left( \frac{2b}{3} + 2 \right) = c - \frac{2b}{3}

d = \sqrt{ \; \frac{ \left(a - \frac{2b}{3}\right)^2 + \left(b - \frac{2b}{3}\right)^2 + \left(c - \frac{2b}{3}\right)^2 }{3} }

3 d^2 = a^2 + b^2 + c^2 + \; \frac{12 b^2}{9} - \; \frac{4}{3} (a b + b^2 + b c)

\Rightarrow 3 d^2 = a^2 + b^2 + c^2 + \; \frac{4}{3} b^2 - \; \frac{4}{3} [ b(a+c) + b^2 ]

\Rightarrow 3 d^2 = a^2 + b^2 + c^2 + \; \frac{4}{3} b^2 - \; \frac{4}{3} [ b \cdot b + b^2 ]

\Rightarrow 3 d^2 = a^2 + b^2 + c^2 + \; \frac{4}{3} b^2 - \; \frac{4}{3} \cdot 2 b^2

\Rightarrow 3 d^2 = a^2 + c^2 - \; \frac{b^2}{3}

\Rightarrow b^2 = 3 a^2 + 3 c^2 - 9 d^2

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