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Let  x_{1},x_{2},......................,x_{n} be n observations, and let  \bar{x}  be their arithematic mean and  \sigma ^{2} be their variance .

Statement 1: Variance of 2x_{1},2x_{2},..............,2x_{n} \: is \: 4\sigma ^{2}

Statement 2: Arithmetic mean of 2x_{1},2x_{2},..............,2x_{n} \: is \: 4 \bar{x}

Option: 1

Statement 1 is false, statement 2 is true


Option: 2

Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1


Option: 3

Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1


Option: 4

Statement 1 is true, statement 2 is false


Answers (1)

As learnt

ARITHMETIC Mean -

For the values x1, x2, ....xn of the variant x the arithmetic mean is given by 

\bar{x}= \frac{x_{1}+x_{2}+x_{3}+\cdots +x_{n}}{n}

in case of discrete data.

-

 

 and

 

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}

-

 

 

If all the observations are doubles, mean is also doubled but variance is four times the old variance. 

 

:\: \: x_{1},x_{2},x_{3}.............x_{n},A.M.=\bar{x},Variance = \sigma ^{2}

Statement 2 : A.M.\:of \: 2x_{1},2x_{2},.............2x_{n}

= \frac{2\left ( x_{1}+x_{2}+.............+x_{n} \right )}{n}= 2\bar{x}

Given A.M.=4\bar{x}\therefore Statement\: \: 2\: \: is\: \: false.

 

Posted by

Ramraj Saini

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