Suppose a population A has 100 observations 101,102, . . .,200 and another population B has 10

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13.Statistics

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Suppose a population $A $ has $100$ observations $ 101,102, . . .,200 $ and another population $B $ has $100$ observation $151,152, . . .,250$ .If $V_A$ and $V_B$ represent the variances of the two populations , respectively then $V_A / V_B$ is

A

$1$

B

$\frac{9}{4}$

C

$\frac{4}{9}$

D

$\frac{2}{3}$

(AIEEE-2006)

Solution

Series $A=101,102 \ldots \ldots 200$

Series $\mathrm{B}=151,152 \ldots \ldots .250$

Here series $\mathrm{B}$ can be obtained if we change the origin of $A$ by $50$ units.

And we know the variance does not change by changing the origin.

So, $\quad V_{A}=V_{B}$

$\Rightarrow \quad \frac{V_{A}}{V_{B}}=1$

Standard 11

Mathematics

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