JEE Main On 08 April 2018 Question 21
Question: If $ \sum\limits _{i=1}^{9}{(x _{i}-5)=9} $ and $ \sum\limits _{i=1}^{9}{{{(x _{i}-5)}^{2}}=45}, $ then the standard deviation of the 9 items $ x _1,x _2,…..,x _9 $ is: [JEE Main Online 08-04-2018]
Options:
A) 2
B) 3
C) 9
D) 4
Show Answer
Answer:
Correct Answer: A
Solution:
Given $ \sum\limits _{i=1}^{9}{(x _{i}-5)=9} $ and $ \sum\limits _{i=1}^{9}{{{(x _{i}-5)}^{2}}=45} $
Variance $ {{\sigma }^{2}}=\frac{1}{x}\sum\limits _{{}}^{{}}{{{(x _{i}-5)}^{2}}-{{[ \frac{1}{x}\sum\limits _{{}}^{{}}{(x _{i}-5)} ]}^{2}}} $
$ =\frac{1}{9}\times 45-{{[ \frac{9}{9} ]}^{2}} $
$ n=9 $ $ {{\sigma }^{2}}=5-1=4 $ Standard Derivation $ =\sqrt{4}=2 $
