If displaystyle∑9i =1(xi - 5 ) = 9 and displaystyle∑ 9i =1(xi - 5 )2 = 45 , then the standard deviation of the 9 items x1 , x2 , .......,x9 is :

Question

If displaystyle∑9i =1(xi - 5 ) = 9 and displaystyle∑ 9i =1(xi - 5 )2 = 45 , then the standard deviation of the 9 items x1 , x2 , .......,x9 is : (A) 9 (B) 4 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

Standard deviation of xi-5 is σ=√( displaystyle∑i=19(xi-5)2/9)-(( displaystyle∑i=19(xi-5)/9))2 ⇒ σ=√5-1=2 As, standard deviation remains constant if observations are added/subtracted by a fixed quantity. So, σ of xi is 2

Q. If $i = 1 \sum 9 (x_{i} - 5) = 9$ and $i = 1 \sum 9 (x_{i} - 5)^{2} = 45,$ then the standard deviation of the 9 items $x_{1}, x_{2}, ......., x_{9}$ is :

9

4

2

3

Standard deviation of $x_{i} - 5$ is
$σ = \frac{i = 1 \sum 9 ( x _{i} - 5 ) ^{2}}{9} - ⎝ ⎛ \frac{i = 1 \sum 9 ( x _{i} - 5 )}{9} ⎠ ⎞^{2}$
$\Rightarrow σ = 5 - 1 = 2$
As, standard deviation remains constant if observations are added/subtracted by a fixed quantity.
So, $σ$ of $x_{i}$ is $2$

Q. If i=1∑9​(xi​−5)=9 and i=1∑9​(xi​−5)2=45, then the standard deviation of the 9 items x1​,x2​,.......,x9​ is :

9

4

2

3

Standard deviation of xi​−5 is σ=9i=1∑9​(xi​−5)2​−⎝⎛​9i=1∑9​(xi​−5)​⎠⎞​2​ ⇒σ=5−1​=2 As, standard deviation remains constant if observations are added/subtracted by a fixed quantity. So, σ of xi​ is 2

Q. If $i = 1 \sum 9 (x_{i} - 5) = 9$ and $i = 1 \sum 9 (x_{i} - 5)^{2} = 45,$ then the standard deviation of the 9 items $x_{1}, x_{2}, ......., x_{9}$ is :