Question

Assertion: Variance of \[2{{x}_{1}},2{{x}_{2}},....,2{{x}_{n}}\] is \[4{{\sigma }^{2}}\]
Reason: arithmetic mean of \[2{{x}_{1}},2{{x}_{2}},....,2{{x}_{n}}\] is \[4\bar{X}\]
A) both assertion and reason are correct and reason is the correct explanation for assertion.
B) both assertion and reason are correct but reason is not the correct explanation for assertion
C) assertion is correct but reason is incorrect
D) both assertion and reason are incorrect

Answer 1

Hint: From the question we have been given two statements in the concept of variance and arithmetic mean and are asked to find the relevant option for the answer. For solving this question we will use the concept of statistics in mathematics. We will use the concept of variance and arithmetic mean and find them for this series \[2{{x}_{1}},2{{x}_{2}},....,2{{x}_{n}}\] and solve the question.

Complete step by step solution:
\[\bar{X}\] is the AM and \[{{\sigma }^{2}}\] is the variance of n observations \[{{x}_{1}},{{x}_{2}},....,{{x}_{n}}\].
The AM of the series \[2{{x}_{1}},2{{x}_{2}},....,2{{x}_{n}}\] will be as follows.
AM is the sum of all terms divided by the total number of terms of the series. So, we get,
\[= \dfrac{2{{x}_{1}}+2{{x}_{2}}+....2{{x}_{n}}}{n}\]
\[= \dfrac{2({{x}_{1}}+{{x}_{2}}+....{{x}_{n}})}{n}\]
\[= 2\bar{X}\]
So, we got the AM as \[\Rightarrow 2\bar{X}\]. So, statement two that is the reason is a false statement.
We know that \[{{\sigma }^{2}}\] is the variance of n observations \[{{x}_{1}},{{x}_{2}},....,{{x}_{n}}\]. When the terms are doubled that is when \[2{{x}_{1}},2{{x}_{2}},....,2{{x}_{n}}\] the variance will be as follows.
We know that the formulae for variance is \[\Rightarrow {{\sigma }^{2}}=\dfrac{{{\left( X-\bar{x} \right)}^{2}}}{n}\].
When the terms are doubled both the \[X\] and \[\bar{x}\] will be doubled. So, the new variance will be as follows.
\[\Rightarrow {{\sigma }^{2}}=\dfrac{{{\left( X-\bar{x} \right)}^{2}}}{n}\]
\[\Rightarrow {{\sigma }^{2}}=\dfrac{{{\left( 2X-2\bar{x} \right)}^{2}}}{n}\]
\[= 4\dfrac{{{\left( X-\bar{x} \right)}^{2}}}{n}\]
\[= 4{{\sigma }^{2}}\]
So, the statement is true.
Therefore, the assertion is correct and reason is incorrect.

So, the correct answer is “Option C”.

Note: Students must be very careful in doing the calculations. Students should have good knowledge in the concept of variance and arithmetic mean. We must know the formulae of variance that is \[ {{\sigma }^{2}}=\dfrac{{{\left( X-\bar{x} \right)}^{2}}}{n}\] to solve the question.