Ask Questions, Get Answers

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

Find mean and standard deviation using short - cut method.

$\begin{array}{1 1}(A)\;9.09\\(B)\;1.69\\(C)\;4.89\\(D)\;5.125\end{array} $

Can you answer this question?

1 Answer

0 votes
  • The formula used for this problem is $Mean=A +\large\frac{\sum f_id_i}{\sum f_i}$
  • Standard deviation $\sigma = \sqrt { \large\frac{f_id_i^2}{\sum f_i} -\bigg( \large\frac{\sum f_id_i}{\sum f_i} \bigg)^2}$
Step 1:
Mean $= A+\large\frac{\sum f_id_i}{\sum f_i}$
Where by $A= \bigg(\large\frac{n+1}{2} \bigg)$ th observation
$\qquad= \bigg( \large\frac{9+1}{2} \bigg) $th observation
$\qquad= 5th$ obsevation.
Step 2:
Maen $=64+ \large\frac{0}{100}$
$\qquad= 64$
Step 3:
Standard deviation $\sigma = \sqrt { \large\frac{f_id_i^2}{\sum f_i} -\bigg( \large\frac{\sum f_id_i}{\sum f_i} \bigg)^2}$
$\qquad= \sqrt { \large\frac{286}{100} -\bigg(\frac{ 0}{100}\bigg)^2}$
$\qquad= \sqrt {2.86}$
$\qquad= 1.69$
Hence B is the correct answer.
answered Jun 27, 2014 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App