Question

If the standard deviation of the values 2, 4, 6, 8 is 2.33, then the standard deviation of the values 4, 6, 8, 10 will be.
(a) 0
(b) 2.58
(c) 4.66
(d) None of these

Answer 1

Hint: To solve this question we have to first know that how can we calculate the standard deviation, suppose the mean of the given n terms is $\overline{x}$ then standard deviation (s) of that terms will be given by, $s=\sqrt{\dfrac{\sum\limits_{i=1}^{i=n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{n-1}}$where ${{x}_{i}}$ represents the ${{i}^{th}}$ term. So we will first calculate the mean first of the given terms then after using the mentioned formula we will calculate mean.

Complete step by step answer:
We are given the terms as,
4, 6, 8, 10
And we have to calculate the standard deviation of these terms,
But first we need to know first that hoe can we calculate standard deviation,
If there are n given terms is $\overline{x}$ then standard deviation (s) is given as,
$s=\sqrt{\dfrac{\sum\limits_{i=1}^{i=n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{n-1}}$
And here ${{x}_{i}}$ represents the ${{i}^{th}}$ term.
So we will first calculate the mean as,
= $\dfrac{sum\,of\,all\,terms}{total\,number\,of\,terms}$
= \[\dfrac{4+6+8+10}{4}\]
= $\dfrac{28}{4}$
= 7
Now using mentioned formula we will calculate the standard deviation as,
$s=\sqrt{\dfrac{\sum\limits_{i=1}^{i=n}{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}}{n-1}}$
\[\begin{align}
  & s=\sqrt{\dfrac{{{\left( 4-7 \right)}^{2}}+{{\left( 6-7 \right)}^{2}}+{{\left( 8-7 \right)}^{2}}+{{\left( 10-7 \right)}^{2}}}{4-1}} \\
 & s=\sqrt{\dfrac{9+1+1+9}{3}} \\
 & s=\sqrt{\dfrac{20}{3}} \\
 & s=2.58 \\
\end{align}\]
Hence we get option (b) as the correct answer.


Note:
 In this question we are already given some different terms and mentioned their standard deviation, it is just to confuse students so you don’t need to pay much attention to that. And you also have to have some prior knowledge about standard deviation and mean to solve this problem so kindly go through these topics.