Question

Find the standard deviation 50, 56, 59, 60, 63, 67, 68.

Answer 1

Hint: We first define the notion of variance and standard deviation. We use the formulas to find the mean and the squares of the entries. Then we put the values in the formula to find the value of the variance. In the end, we need to find the root value.

Complete step-by-step solution
The given observations are discrete points. To find the standard deviation we first need to find the mean of the observations.
There are 7 entries. So, $n=7$.
We define the terms as ${{x}_{i}},i=1(1)7$. Here the terms are ${{x}_{1}}=50,{{x}_{2}}=56,{{x}_{3}}=59,{{x}_{4}}=60,{{x}_{5}}=63,{{x}_{6}}=67,{{x}_{7}}=68$
The mean of the observations is $\overline{x}=\dfrac{1}{n}\sum{{{x}_{i}}}$. We put the values to find the mean
$\begin{align}
  & \overline{x}=\dfrac{1}{n}\sum{{{x}_{i}}} \\
 & =\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}+{{x}_{6}}+{{x}_{7}}}{7} \\
 & =\dfrac{50+56+59+60+63+67+68}{7} \\
 & =\dfrac{423}{7} \\
\end{align}$
Now we find the standard deviation. Slandered deviation is the root mean square value of the differences of the observations from the mean value.
So, the differences in every observation are termed as root mean square deviation.
We take ${{\left( {{x}_{i}}-\overline{x} \right)}^{2}}$ which defines the square of deviation. We take the mean value of these squares.as variance. $v=\dfrac{1}{n}\sum{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}$.
It can be also termed as \[v=\dfrac{1}{n}\sum{{{\left( {{x}_{i}} \right)}^{2}}}-{{\left( \overline{x} \right)}^{2}}\].
The standard deviation is the root value of the variance.
So, \[s.d.=\sqrt{v}=\sqrt{\dfrac{1}{n}\sum{{{\left( {{x}_{i}} \right)}^{2}}}-{{\left( \overline{x} \right)}^{2}}}\].
We now find the value of \[\dfrac{1}{n}\sum{{{\left( {{x}_{i}} \right)}^{2}}}\]. We are taking mean value of the squares of the entries.
\[\begin{align}
  & \dfrac{1}{n}\sum{{{\left( {{x}_{i}} \right)}^{2}}} \\
&=\dfrac{{{x}_{1}}^{2}+{{x}_{2}}^{2}+{{x}_{3}}^{2}+{{x}_{4}}^{2}+{{x}_{5}}^{2}+{{x}_{6}}^{2}+{{x}_{7}}^{2}}{7} \\
 & =\dfrac{{{50}^{2}}+{{56}^{2}}+{{59}^{2}}+{{60}^{2}}+{{63}^{2}}+{{67}^{2}}+{{68}^{2}}}{7} \\
 & =\dfrac{2500+3136+3481+3600+3969+4489+4624}{7} \\
 & =\dfrac{25799}{7} \\
\end{align}\]
We also find the value of \[{{\left( \overline{x} \right)}^{2}}\] which is \[{{\left( \overline{x} \right)}^{2}}={{\left( \dfrac{423}{7} \right)}^{2}}=\dfrac{178929}{49}\].
Now we place all the values to find the standard deviation.
\[\begin{align}
  & s.d.=\sqrt{\dfrac{1}{n}\sum{{{\left( {{x}_{i}} \right)}^{2}}}-{{\left( \overline{x} \right)}^{2}}} \\
 & =\sqrt{\dfrac{25799}{7}-\left( \dfrac{178929}{49} \right)} \\
 & =\sqrt{\dfrac{180593-178929}{49}} \\
 & =\sqrt{\dfrac{1664}{49}}=\dfrac{8\sqrt{26}}{7} \\
\end{align}\]
The standard deviation 50, 56, 59, 60, 63, 67, 68 is \[\dfrac{8\sqrt{26}}{7}\].

Note: The conversion of the formula from $v=\dfrac{1}{n}\sum{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}}$ to \[v=\dfrac{1}{n}\sum{{{\left( {{x}_{i}} \right)}^{2}}}-{{\left( \overline{x} \right)}^{2}}\] is just simplification.
$\begin{align}
  & v=\dfrac{1}{n}\sum{{{\left( {{x}_{i}}-\overline{x} \right)}^{2}}} \\
 & \Rightarrow v=\dfrac{1}{n}\sum{\left( {{x}_{i}}^{2}-2x\overline{x}+{{\overline{x}}^{2}} \right)} \\
 & \Rightarrow v=\dfrac{1}{n}\sum{{{\left( {{x}_{i}} \right)}^{2}}}-2\overline{x}\dfrac{1}{n}\sum{x}+{{\left( \overline{x} \right)}^{2}} \\
 & \Rightarrow v=\dfrac{1}{n}\sum{{{\left( {{x}_{i}} \right)}^{2}}}-2{{\left( \overline{x} \right)}^{2}}+{{\left( \overline{x} \right)}^{2}} \\
 & \Rightarrow v=\dfrac{1}{n}\sum{{{\left( {{x}_{i}} \right)}^{2}}}-{{\left( \overline{x} \right)}^{2}} \\
\end{align}$
That’s why we used the square values of the entries.