Question

Given N = 10, \[\sum{{{x}_{i}}=60}\] and \[\sum{{{x}_{i}}^{2}=1000}\] . The standard deviation is

Answer 1

Hint: Now we know that the formula for standard deviation is $\sigma =\sqrt{\dfrac{\sum\limits_{1}^{n}{{{\left( {{x}_{i}}-\mu \right)}^{2}}}}{N}}$ Now we will try to write the equation in terms of ${{\sigma }^{2}}$ and then open the brackets of numerator by formula ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ Now we know that \[\dfrac{\sum\limits_{1}^{n}{\left( {{x}_{i}} \right)}}{N}=\mu \] and \[\sum\limits_{1}^{n}{1=N}\] hence using this we will get a simplified equation for standard deviation. Now out of the given data we will find mean by the formula \[\dfrac{\sum\limits_{1}^{n}{\left( {{x}_{i}} \right)}}{N}=\mu \] and then substitute the values in the obtained equation. Hence we can easily find the standard deviation of the given data.

Complete step-by-step answer:
Now we are given with N = 10, \[\sum{{{x}_{i}}=60}\] and \[\sum{{{x}_{i}}^{2}=1000}\]
Now we know that standard deviation is given by $\sigma =\sqrt{\dfrac{\sum\limits_{1}^{n}{{{\left( {{x}_{i}}-\mu \right)}^{2}}}}{N}}$ where $\mu $ is the mean of data.

Hence we can say that ${{\sigma }^{2}}=\dfrac{\sum\limits_{1}^{n}{{{\left( {{x}_{i}}-\mu \right)}^{2}}}}{N}$
Now let us simplify the equation above.
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$
Using this in the numerator we get.
\[{{\sigma }^{2}}=\dfrac{\sum\limits_{1}^{n}{\left( {{x}_{i}}^{2}+{{\mu }^{2}}-2{{x}_{i}}\mu \right)}}{N}\]
Now let us open the summation. Hence we get
\[{{\sigma }^{2}}=\dfrac{\sum\limits_{1}^{n}{\left( {{x}_{i}}^{2} \right)}+\sum\limits_{1}^{n}{\left( {{\mu }^{2}} \right)}-\sum\limits_{1}^{n}{\left( 2{{x}_{i}}\mu \right)}}{N}\]
Now we know that $\mu $ is the mean and is constant. Hence we can take it out of summation.
\[\begin{align}
  & {{\sigma }^{2}}=\dfrac{\sum\limits_{1}^{n}{\left( {{x}_{i}}^{2} \right)}+{{\mu }^{2}}\sum\limits_{1}^{n}{1}-2\mu \sum\limits_{1}^{n}{\left( {{x}_{i}} \right)}}{N} \\
 & \Rightarrow {{\sigma }^{2}}=\dfrac{\sum\limits_{1}^{n}{\left( {{x}_{i}}^{2} \right)}}{N}+\dfrac{{{\mu }^{2}}\sum\limits_{1}^{n}{1}}{N}-2\mu \dfrac{\sum\limits_{1}^{n}{\left( {{x}_{i}} \right)}}{N} \\
\end{align}\]
Now we know that \[\dfrac{\sum\limits_{1}^{n}{\left( {{x}_{i}} \right)}}{N}=\mu \] and \[\sum\limits_{1}^{n}{1=N}\] . Hence we get
\[\begin{align}
  & {{\sigma }^{2}}=\dfrac{\sum\limits_{1}^{n}{\left( {{x}_{i}}^{2} \right)}}{N}+\dfrac{{{\mu }^{2}}N}{N}-2\mu \mu \\
 & \Rightarrow {{\sigma }^{2}}=\dfrac{\sum\limits_{1}^{n}{\left( {{x}_{i}}^{2} \right)}}{N}-{{\mu }^{2}} \\
\end{align}\]
\[{{\sigma }^{2}}=\dfrac{\sum\limits_{1}^{n}{\left( {{x}_{i}}^{2} \right)}}{N}-{{\mu }^{2}}.................\left( 1 \right)\]
Now we have the reduced simplified formula for standard deviation.
Now we know that N = 10 and $\sum{{{x}_{i}}=60}$
Hence we can say that mean $\mu =\dfrac{\sum{{{x}_{i}}}}{N}=\dfrac{60}{10}=6$
Now substituting the values of $\mu =6$ , N = 10 and \[\sum{{{x}_{i}}^{2}=1000}\] in equation (1) we get.
\[\begin{align}
  & {{\sigma }^{2}}=\dfrac{1000}{10}-{{\left( 6 \right)}^{2}} \\
 & \Rightarrow {{\sigma }^{2}}=100-36 \\
 & \Rightarrow {{\sigma }^{2}}=64 \\
 & \Rightarrow \sigma =8 \\
\end{align}\]
Hence we get the standard deviation is equal to 8.

Note: The formula for standard deviation is ${{\sigma }^{2}}=\dfrac{\sum\limits_{1}^{n}{{{\left( {{x}_{i}}-\mu \right)}^{2}}}}{N}$ . after some algebraic manipulation we arrive at an equivalent formula \[{{\sigma }^{2}}=\dfrac{\sum\limits_{1}^{n}{\left( {{x}_{i}}^{2} \right)}}{N}-{{\mu }^{2}}\] . We can directly use this formula for faster calculation as this is a general alternate formula for standard deviation. Though it is easy to derive the equation too.