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Question

Question: Assertion: Variance of \[2{{x}_{1}},2{{x}_{2}},....,2{{x}_{n}}\] is \[4{{\sigma }^{2}}\] Reason: a...

Assertion: Variance of 2x1,2x2,....,2xn2{{x}_{1}},2{{x}_{2}},....,2{{x}_{n}} is 4σ24{{\sigma }^{2}}
Reason: arithmetic mean of 2x1,2x2,....,2xn2{{x}_{1}},2{{x}_{2}},....,2{{x}_{n}} is 4Xˉ4\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

Explanation

Solution

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 2x1,2x2,....,2xn2{{x}_{1}},2{{x}_{2}},....,2{{x}_{n}} and solve the question.

Complete step by step solution:
Xˉ\bar{X} is the AM and σ2{{\sigma }^{2}} is the variance of n observations x1,x2,....,xn{{x}_{1}},{{x}_{2}},....,{{x}_{n}}.
The AM of the series 2x1,2x2,....,2xn2{{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,
=2x1+2x2+....2xnn= \dfrac{2{{x}_{1}}+2{{x}_{2}}+....2{{x}_{n}}}{n}
=2(x1+x2+....xn)n= \dfrac{2({{x}_{1}}+{{x}_{2}}+....{{x}_{n}})}{n}
=2Xˉ= 2\bar{X}
So, we got the AM as 2Xˉ\Rightarrow 2\bar{X}. So, statement two that is the reason is a false statement.
We know that σ2{{\sigma }^{2}} is the variance of n observations x1,x2,....,xn{{x}_{1}},{{x}_{2}},....,{{x}_{n}}. When the terms are doubled that is when 2x1,2x2,....,2xn2{{x}_{1}},2{{x}_{2}},....,2{{x}_{n}} the variance will be as follows.
We know that the formulae for variance is σ2=(Xxˉ)2n\Rightarrow {{\sigma }^{2}}=\dfrac{{{\left( X-\bar{x} \right)}^{2}}}{n}.
When the terms are doubled both the XX and xˉ\bar{x} will be doubled. So, the new variance will be as follows.
σ2=(Xxˉ)2n\Rightarrow {{\sigma }^{2}}=\dfrac{{{\left( X-\bar{x} \right)}^{2}}}{n}
σ2=(2X2xˉ)2n\Rightarrow {{\sigma }^{2}}=\dfrac{{{\left( 2X-2\bar{x} \right)}^{2}}}{n}
=4(Xxˉ)2n= 4\dfrac{{{\left( X-\bar{x} \right)}^{2}}}{n}
=4σ2= 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 σ2=(Xxˉ)2n {{\sigma }^{2}}=\dfrac{{{\left( X-\bar{x} \right)}^{2}}}{n} to solve the question.