Let V _ 1 = variance of \ 13,16,19 . . . . . . . . . . 103\ ~and V _ 2 = variance of \ 20,26,32 . . . . . . . . . . 200\ . Then V _ 1 : V _ 2 is

Correct answer: D. 1: 4 V _ 1 = variance of \ 13,16,19 . . . . . . . . . . 103\ = variance of \ 3,6, 9 . . . . . . . . . . 93\ = 9 of variance of \ 1,2, 3 . . . . . . . . . . 31\ V _ 2 = variance of \ 20,26,32 . . . . . . . . . . 200\ = variance of \ 6,12,18 . . . . . . . . . . 186\ = 36 of variance of \ 1,2, 3 . . . . . . . . . . 31\ V _ 1 V _ 2 = (1)/(4)

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Statistics

NTA Abhyas 2020 · Mathematics · Statistics

Let V _ 1 = variance of \ 13,16,19 . . . . . . . . . . 103\ ~and V...

Let

{V}_{1} =

variance of \left\{13,16,19 . . . . . . . . . . 103\right\}

~and

{V}_{2} =

variance of \left\{20,26,32 . . . . . . . . . . 200\right\}

. Then

{V}_{1}: {V}_{2}

Options

A $1: 2$
B $1: 1$
C $4: 9$
D $1: 4$

Correct answer

D. $1: 4$

Step-by-step solution

{V}_{1} =

variance of \left\{13,16,19 . . . . . . . . . . 103\right\}

<br />= variance of \left\{3,6, 9 . . . . . . . . . . 93\right\}

= 9

of variance of \left\{1,2, 3 . . . . . . . . . . 31\right\}

<br />

{V}_{2} =

variance of \left\{20,26,32 . . . . . . . . . . 200\right\}

= variance of \left\{6,12,18 . . . . . . . . . . 186\right\}

<br />

= 36

of variance of \left\{1,2, 3 . . . . . . . . . . 31\right\}

\dfrac{{V}_{1}}{{V}_{2}} = \dfrac{1}{4}

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