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Consider a finite sequence of random values $X=[x_1,x_2,\dots x_n]$. Let $\mu_x$ be the mean and $\sigma_x$ be the standard deviation of $X$. Let another finite sequence $Y$ of equal length be derived from this as $y_i=a*x_i+b$, where $a$ and $b$ are positive constants. Let $\mu_y$ be the mean and $\sigma_y$ be the standard deviation of this sequence.

Which one of the following statements is INCORRECT?

  1. Index position of mode of $X$ in $X$ is the same as the index position of mode of $Y$ in $Y$
  2. Index position of median of $X$ in $X$ is the same as the index position of median of $Y$ in $Y$
  3. $\mu_y=a \mu_x + b$
  4. $\sigma_y=a \sigma_x + b$
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4 Answers

Best answer
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44 votes
Answer - $D$.

Mean, median and mode are linear functions over a random vaiable.

So, multiplying by constants or adding constants wont change their relative position.

Standard deviation is not a linear function over a random variable.
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Taking example here also give ansing easy

Say

$X= \left [ 2,3,4,4,5 \right ]$

and value of $a=2 , b=100$

So, according to question $Y= \left [ 104,106,108,108,110 \right ]$

Now come to option

A) correct, because for X mode is 4 and for Y mode is 108 which has same index position

B) correct , because for X median is 4and for Y median is 108 which has same index position

C) both gives mean =107.2

D) will be incorrect in option elimination

and also by calculation

$\sigma _{y}=0$

while $a\sigma _{x}+b=2.28$

https://www.khanacademy.org/math/statistics-probability/summarizing-quantitative-data/mean-median-basics/a/mean-median-and-mode-review

https://en.wikipedia.org/wiki/Standard_deviation

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Let Variance of X = p

$\sigma _{x}$ = $\sqrt{p}$

Variance of Y = $a^{2}p$ 

 ($\because$ When we add a constant to each element of the list, the standard deviation (or variance) remains unchanged. This is because, the mean also gets added by the same constant and hence the deviation from the mean remains the same for each element. when we multiply a random variable with a constant, since we square the values while calculating variance, the value of a(constant) will also gets squared)

$\sigma _{y}$ = $a\sqrt{p}$ = a * $\sigma _{x}$

$\therefore $ $\sigma _{y}$ =  $a \sigma _{x}$

Answer D is incorrect

 

Answer:

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