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

Hint: 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 \[i = 1,2,......100\]
\[ \Rightarrow {x_1} - \overline x = 101 - \dfrac{{301}}{2}\] \[ \Rightarrow \dfrac{{202 - 301}}{2}\] \[ \Rightarrow \dfrac{{ - 99}}{2}\]
\[ \Rightarrow {x_2} - \overline x = 102 - \dfrac{{301}}{2}\] \[ \Rightarrow \dfrac{{204 - 301}}{2}\] \[ \Rightarrow \dfrac{{ - 97}}{2}\]
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\[ \Rightarrow {x_{100}} - \overline x = 200 - \dfrac{{301}}{2}\] \[ \Rightarrow \dfrac{{400 - 301}}{2}\] \[ \Rightarrow \dfrac{{99}}{2}\]
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
\[{y_1} - \overline y = 151 - \dfrac{{401}}{2}\] \[ \Rightarrow \dfrac{{302 - 401}}{2}\] \[ \Rightarrow \dfrac{{ - 99}}{2}\]
\[{y_2} - \overline y = 152 - \dfrac{{401}}{2}\] \[ \Rightarrow \dfrac{{304 - 401}}{2}\] \[ \Rightarrow \dfrac{{97}}{2}\]
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\[{y_{100}} - \overline y = 250 - \dfrac{{401}}{2}\] \[ \Rightarrow \dfrac{{500 - 401}}{2}\] \[ \Rightarrow \dfrac{{99}}{2}\]
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.