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Suppose a population A has 100 observations 101, 102,..., 200, and another population B has 100 observations 151,152,...,250. If V_{A}\; \; and\; \; V_{B} represent the variances of the two populations, respectively, then V_{A}/ V_{B}  is

  • Option 1)

    1

  • Option 2)

    9/4

  • Option 3)

    4/9

  • Option 4)

    2/3

 

Answers (1)

best_answer

As we learnt in 

Variance -

In case of discrete data 

\dpi{100} \sigma ^{2}= \left ( \frac{\sum x_{i}^{2}}{n} \right )-\left ( \frac{\sum x_{i}}{n} \right )^{2}

 V_{A} = \sqrt{\frac{\sum{(x-\bar{x_{A}}})^{2}}{n}},\, \, \, \, \, \, \, \, V_{B} = \sqrt{\frac{\sum{(x-\bar{x_{B}}})^{2}}{n}},

Now  x A={101,102,...200}

\bar{x}_{A}=150.5

Similarly  , B={1.51,152,....250}

\bar{x}_{B}=200.5

Then  \sqrt{\frac{\sum{(x-\bar{x_{A}}})^{2}}{n}}=\sqrt{\frac{((-49.5)^{2}+(-48.5)^{2}+......)+(49.5^{2})+(48.5)^{2}+.........)}{100}}                    

V_{A}=\sqrt{\frac{2\left [ \left ( 49.5 \right )^{2}+\left ( 48.5 \right )^{2}+----\left ( 0.5 \right )^{2} \right ]}{100}} ----- (1)

V_{B}=\sqrt{\frac{\left [ \left ( -49.5 \right )^{2}+\left ( -48.5 \right )^{2}+--- \right ]+\left [ \left ( 49.5 \right )^{2}+\left (48.5 \right )^{2}+--- \right ]}{100}}

=\sqrt{\frac{2\left [ \left ( 49.5 \right )^{2}+\left ( 48.5 \right )^{2}+---\left ( 0.5 \right ) \right ]}{100}^{2}} ----(2)

from(1) and (2), V_{A}=V_{B}

=\sqrt{\frac{\left ( a^{2}+a^{2}+--- \right )n\:times+\left ( 1-a \right )^{2}+\left (( -a \right )^{2}+---n\: times)}{2n}}

=\sqrt{\frac{2na^{2}}{2n}}=\left | a \right | --- (1)

but \frac{a}{q}\:\ std \:deviation=2 - - - -- - -\left ( 2 \right )

from (1) and (2)

\left | a \right |=2


Option 1)

1

Option 2)

9/4

Option 3)

4/9

Option 4)

2/3

Posted by

prateek

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