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+3 votes
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Which of the following assertions are CORRECT?

P: Adding 7 to each entry in a list adds 7 to the mean of the list 

Q: Adding 7 to each entry in a list adds 7 to the standard deviation of the list 

R: Doubling each entry in a list doubles the mean of the list

S: Doubling each entry in a list leaves the standard deviation of the list unchanged


(A) P, Q
(B) Q, R
(C) P, R
(D) R, S

asked in Numerical Ability by Veteran (12.9k points)   | 362 views

2 Answers

+13 votes
Best answer

Suppose we double each entry of a list


Initial Mean $(M_I)=\frac{ \sum_{i=1}^{n} x_i}{n}$

New Mean $(M_N)= \frac{\sum_{i=1}^{n}2\times x_i}{n}$

$=\frac{2}{n} \sum_{i=1}^{n}x_i$

So, when each entry in the list is doubled, mean also gets doubled.




Standard Deviation $\sigma_I = \sqrt {\sum_{i=1}^{n} \left(M_I - x_i\right)^2}$

New Standard Deviation $\sigma_N = \sqrt {\sum_{i=1}^{n} \left(M_N - 2 \times x_i\right)^2}$

$= \sqrt {\sum_{i=1}^{n} \left(2 \times \left(M_I - x_i\right)\right)^2}$

$= 2  \sigma_I$

So, when each entry is doubled, standard deviation also gets doubled. 


 

When we add a constant to each element of the list, it gets added to the mean as well. This can be seen from the formula of mean. 
 

 


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.



So, here P and R are correct. 

answered by Veteran (281k points)  
edited by
Mean = (x1+x2+x3+...+xn)/n
The answer has "the division by n" part missing.
Thanks :) Corrected now..
+1 vote
C. P,R are correct.
answered by Loyal (3.4k points)  


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