logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
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
Toolbox:
  • 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.
$A=64$
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
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...