JEE Main On 16 April 2018 Question 25
Question: The mean and the standard deviation (s.d.) of five observations are 9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes10, then their s.d. is? [JEE Main 16-4-2018]
Options:
A) 0
B) 4
C) 2
D) 1
Show Answer
Answer:
Correct Answer: C
Solution:
$ \bar{x}=\frac{\sum{xi}}{n}=9 $ $ s.d=\sqrt{\sum{\frac{xi-\bar{x}}{n}=0}} $
$ \Rightarrow $ $ X _{i}=\bar{x}\forall i $
$ \therefore $ each term in the original observation
$ \Rightarrow $ observation $ 1={9,9,9,9,9} $ $ \bar{x}=\frac{\sum{xi}}{n}=\frac{9+9+9+9x _5}{5} $
$ \Rightarrow $ $ 10=\frac{36+X _5}{5} $
$ \Rightarrow $ $ x _5=14 $ $ s.d=\sqrt{\frac{\sum{{{(x _{i}-{x^{-1}})}^{2}}}}{n}} $
$ =\frac{\sqrt{{{(9-10)}^{2}}+{{(9-10)}^{2}}+{{(9-10)}^{2}}+(10-10)}}{5} $
$ =\sqrt{\frac{4+4^{2}}{5}} $ $ =\sqrt{\frac{20}{5}} $
