Question

Suppose a population A has $100$ observations $101$ , $102$ ,………., $200$ and another population B has $100$ observations $151$ , $152$ , $153$ ,…………., $250$ . If ${V_A}$ and ${V_B}$ represent the variances of the two populations respectively, then $\dfrac{{{V_A}}}{{{V_B}}}$ is
$\left( A \right){\text{ }}1$
$\left( B \right){\text{ }}\dfrac{9}{4}$
$\left( C \right){\text{ }}\dfrac{4}{9}$
$\left( D \right){\text{ }}\dfrac{2}{3}$

Answer 1

In this question first we have to find the mean of both the populations A and B separately. Then after finding the mean find $\sum {{{\left( {{x_i} - \overline x } \right)}^2}}$ for both the populations by using the mean of the populations. Then find the variance of both the populations by substituting the values in the formula $\dfrac{{\sum {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}$ . Then divide the variance of population A by the variance of population B and you will get your answer then on solving it further.

Complete step by step answer:
It is given to us that population A has $100$ observations that is
$A{\text{ }} \Rightarrow {\text{ }}101,102,103,............,200$
It is also given that population B also has $100$ observations that is
${\text{B }} \Rightarrow {\text{ }}151,152,153,............,250$
As we have to find the value of $\dfrac{{{V_A}}}{{{V_B}}}$ . For this first we will find the variance of A.
Variance measures how far a data set is spread out. It is mathematically defined as the average of the squared differences from the mean. Variance is denoted by ${\sigma ^2}$ . So, to calculate the variance first we have to find the mean. Mean is denoted by $\overline x$ . Therefore,
Mean of population A $= {\text{ }}\overline x {\text{ }} = {\text{ }}\dfrac{{101 + 102 + 103 + ......... + 200}}{{100}}$ ------------- (i)
To find the sum of the arithmetic sequence written in the numerator we use the formula $S = \dfrac{n}{2}\left( {a + L} \right)$
where n is the total number of terms in the sequence, a is the first term, L is the last term of the sequence and S is the sum of the Arithmetic sequence. Therefore, equation (i) becomes
$\Rightarrow {\text{ }}\overline x {\text{ }} = {\text{ }}\dfrac{{\dfrac{{100}}{2}\left( {101 + 200} \right)}}{{100}}$
On further solving we get
$\Rightarrow {\text{ }}\overline x {\text{ }} = {\text{ }}\dfrac{{50\left( {301} \right)}}{{100}}$
Zeroes in the numerator and the denominator will cancel out
$\Rightarrow {\text{ }}\overline x {\text{ }} = {\text{ }}\dfrac{{5\left( {301} \right)}}{{10}}$
Ten in the denominator can be cancelled by the number five in the numerator,
$\Rightarrow {\text{ }}\overline x {\text{ }} = {\text{ }}\dfrac{{301}}{2}$
Therefore, the required mean for the population A is $\dfrac{{301}}{2}$ .
Now, the formula to find the variance is ${S^2} = \dfrac{{\sum {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}$ where
${S^2}$ is the sample variance
${x_i}$ is the value of the one observation
$\overline x$ is the mean value of all observations
$n$ is the number of observations
Therefore, we first solve $\left( {{x_i} - \overline x } \right)$ at $i = 1,2,......100$
$\Rightarrow {x_1} - \overline x = 101 - \dfrac{{301}}{2}$ $\Rightarrow \dfrac{{202 - 301}}{2}$ $\Rightarrow \dfrac{{ - 99}}{2}$
$\Rightarrow {x_2} - \overline x = 102 - \dfrac{{301}}{2}$ $\Rightarrow \dfrac{{204 - 301}}{2}$ $\Rightarrow \dfrac{{ - 97}}{2}$
.
.
.
$\Rightarrow {x_{100}} - \overline x = 200 - \dfrac{{301}}{2}$ $\Rightarrow \dfrac{{400 - 301}}{2}$ $\Rightarrow \dfrac{{99}}{2}$
Therefore,
$\Rightarrow {\left( {{x_1} - \overline x } \right)^2} + {\left( {{x_2} - \overline x } \right)^2} + .......... + {\left( {{x_{100}} - \overline x } \right)^2} = 2{\left( {\dfrac{{99}}{2}} \right)^2} + 2{\left( {\dfrac{{97}}{2}} \right)^2} + ......... + 2{\left( {\dfrac{1}{2}} \right)^2}$
Here we multiplied the right hand side by two because ${x_1}$ is $- 99$ and ${x_{100}}$ is $99$ . So they have counted the fifty terms that’s why we take twice of that. By taking out $2$ and $\dfrac{1}{4}$ common from all terms at right hand side we have
$\Rightarrow \dfrac{1}{2}\left( {{1^2} + {3^2} + {5^2} + ....... + {{97}^2} + {{99}^2}} \right)$
Therefore, ${V_A} = \dfrac{{\dfrac{1}{2}\left( {{1^2} + {3^2} + {5^2} + ....... + {{97}^2} + {{99}^2}} \right)}}{{100}}$ -------------- (i)
Similarly we can find mean and variance for B just like we do for A. Therefore, the mean of population B will be
$\overline y = \dfrac{{151 + 152 + ......... + 250}}{{100}}$
Now we will use the formula $S = \dfrac{n}{2}\left( {a + L} \right)$ in the numerator of the above expression
$\overline y = \dfrac{{\dfrac{{100}}{2}\left( {151 + 250} \right)}}{{100}}$
$100$ in both the numerator and the denominator will cancel out each other
$\overline y = \dfrac{1}{2}\left( {151 + 250} \right)$
On adding the terms we get
$\overline y = \dfrac{{401}}{2}$
So, the mean of the population B is $\dfrac{{401}}{2}$ .
Similarly as we done above for A, For B will be
${y_1} - \overline y = 151 - \dfrac{{401}}{2}$ $\Rightarrow \dfrac{{302 - 401}}{2}$ $\Rightarrow \dfrac{{ - 99}}{2}$
${y_2} - \overline y = 152 - \dfrac{{401}}{2}$ $\Rightarrow \dfrac{{304 - 401}}{2}$ $\Rightarrow \dfrac{{97}}{2}$
.
.
.
${y_{100}} - \overline y = 250 - \dfrac{{401}}{2}$ $\Rightarrow \dfrac{{500 - 401}}{2}$ $\Rightarrow \dfrac{{99}}{2}$
Therefore,
$\Rightarrow {\left( {{y_1} - \overline y } \right)^2} + {\left( {{y_2} - \overline y } \right)^2} + .......... + {\left( {{y_{100}} - \overline y } \right)^2} = 2{\left( {\dfrac{{99}}{2}} \right)^2} + 2{\left( {\dfrac{{97}}{2}} \right)^2} + ......... + 2{\left( {\dfrac{1}{2}} \right)^2}$
On further solving the right hand side will be
$\Rightarrow \dfrac{1}{2}\left( {{1^2} + {3^2} + {5^2} + ....... + {{97}^2} + {{99}^2}} \right)$
Therefore,
${V_B} = \dfrac{{\dfrac{1}{2}\left( {{1^2} + {3^2} + {5^2} + ....... + {{97}^2} + {{99}^2}} \right)}}{{100}}$ -------------- (ii)
On dividing equation (i) by (ii) we get
$\dfrac{{(i)}}{{(ii)}} \Rightarrow \dfrac{{{V_A}}}{{{V_B}}} = \dfrac{{\dfrac{{\dfrac{1}{2}\left( {{1^2} + {3^2} + {5^2} + ....... + {{97}^2} + {{99}^2}} \right)}}{{100}}}}{{\dfrac{{\dfrac{1}{2}\left( {{1^2} + {3^2} + {5^2} + ....... + {{97}^2} + {{99}^2}} \right)}}{{100}}}}$
As it is clearly observable that the numerator and the denominator are the same. Therefore, they will cancel each other out. By doing this we have
$\Rightarrow \dfrac{{{V_A}}}{{{V_B}}} = 1$
Hence, the correct options is $\left( A \right){\text{ }}1$.

Note:
Note that we can also find the answer of the given question directly without doing any calculation or we can say it as a shortcut method. Series A: $101$ , $102$ ,………., $200$ and Series B: $151$ , $152$ , $153$ ,…………., $250$ . Series B is obtained by adding a fixed quantity to each item of series A. We know that variance is independent of change of origin; both series have the same variance so ratio of their variances is one.