Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Statistics
0 votes

Let $a,b,c,d,e$ be the observations with mean m and standard deviations $S$. The standard deviation of the observations $a+k,b+k,c+k,d+k,e+k$ is

$\begin{array}{1 1}(A)\; s \\(B)\;ks\\(C)\;s+k\\(D)\;s/k \end{array} $

Can you answer this question?

1 Answer

0 votes
  • The formula to calculate the SD is $SD(\sigma) =\sqrt { \large\frac{\sum \limits_{i=1}^{n} (x_i -\bar{x})^2}{n}}$
Step 1:
Given the observations $a,b,c,d,e$
Given mean=m
$m= \large\frac{a+b+c+d+e}{5}$
Let the new mean be n
The new observations are
$n= \large\frac{a+k+b+k+c+k+d+k+e+k}{5}$
$\quad= \large\frac{a+b+c+d+e+5k}{5}$
$\quad= \large\frac{a+b+c+d+e}{5}+\frac{5k}{5}$
The standard deviation $S = \sqrt {\large\frac{(a-m)^2+(b-m)+(c-m)^2+(d-m)^2+(e-m)^2}{5}}$
new standard deviation will be
$S_1= \sqrt {\large\frac{ (a+k-n)^2+(b+k-n)^2+(c+k-n)^2+(d+k-n)^2 +(e+k-m)^2}{5}}$
We know that
$\qquad= a+k-m-k$
$\qquad= a-m$
Similarly, $b+k-n =b-m$
$e+k-n =e-m$
substituting the values ,
$S_1= \sqrt { \large\frac{(a-m)^2+(b-m)^2+(c-m)^2 +(d-m)^2+(e-m)^2}{5}}$
The new standard deviation is s.
Thereby adding or substracting any constant to the observation does not change the value of standard deviation .
Hence A is the correct answer.
answered Jul 8, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App