Question

What is the standard deviation of 1, 2, 3, 4 and 5?

Answer 1

First of all calculate the mean of the given data by using the formula $\bar{x}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n}$ where n is the number of observations and $\bar{x}$ is the notation of mean of these observations. Substitute the value of n = 5 and find the mean. Now, to calculate the standard deviation of the given observations use the formula $\sigma =\sqrt{\dfrac{1}{n}\times \sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\bar{x} \right)}^{2}}}}$ where $\sigma$ is the standard deviation, and substitute the obtained value of mean and n = 5 to get the answer.

Complete step-by-step answer:
Here we have been provided with the observations 1, 2, 3, 4 and 5. We have to find the standard deviation of these observations. To calculate the standard deviation first we need to calculate the mean of these observations.
Now, we know that mean of n observations is given as $\bar{x}=\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n}$ where $\bar{x}$ denotes the mean. Clearly we can see that we have only 5 observations so the value of n is equal to 5. Therefore the mean can be given as:
$\begin{aligned} & \Rightarrow \bar{x}=\dfrac{\sum\limits_{i=1}^{5}{{{x}_{i}}}}{5} \\\ & \Rightarrow \bar{x}=\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}+{{x}_{5}}}{5} \\\ \end{aligned}$
Substituting the values of given observations we get,
$\begin{aligned} & \Rightarrow \bar{x}=\dfrac{1+2+3+4+5}{5} \\\ & \Rightarrow \bar{x}=\dfrac{15}{5} \\\ & \therefore \bar{x}=3.......\left( i \right) \\\ \end{aligned}$
Now, the standard deviation of n observations is given as $\sigma =\sqrt{\dfrac{1}{n}\times \sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\bar{x} \right)}^{2}}}}$ where $\sigma$ denotes the standard deviation, so substituting the value of the mean from equation (i) and the value n = 5 we get,

& \Rightarrow \sigma =\sqrt{\dfrac{1}{n}\times \sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\bar{x} \right)}^{2}}}} \\\ & \Rightarrow \sigma =\sqrt{\dfrac{1}{5}\times \sum\limits_{i=1}^{5}{{{\left( {{x}_{i}}-3 \right)}^{2}}}} \\\ & \Rightarrow \sigma =\sqrt{\dfrac{1}{5}\times \left[ {{\left( {{x}_{1}}-3 \right)}^{2}}+{{\left( {{x}_{2}}-3 \right)}^{2}}+{{\left( {{x}_{3}}-3 \right)}^{2}}+{{\left( {{x}_{4}}-3 \right)}^{2}}+{{\left( {{x}_{5}}-3 \right)}^{2}} \right]} \\\ \end{aligned}$$ Substituting the values of observations and simplifying we get, $$\begin{aligned} & \Rightarrow \sigma =\sqrt{\dfrac{1}{5}\times \left[ {{\left( 1-3 \right)}^{2}}+{{\left( 2-3 \right)}^{2}}+{{\left( 3-3 \right)}^{2}}+{{\left( 4-3 \right)}^{2}}+{{\left( 5-3 \right)}^{2}} \right]} \\\ & \Rightarrow \sigma =\sqrt{\dfrac{1}{5}\times \left[ 4+1+0+1+4 \right]} \\\ & \therefore \sigma =\sqrt{2} \\\ \end{aligned}$$ Hence the standard deviation of the given observations is $\sqrt{2}$. **Note:** Note that if you will square both the sides of the relation of the standard deviation then you will get the variance of the given observations. Variance is given as $${{\sigma }^{2}}=\dfrac{1}{n}\times \sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\bar{x} \right)}^{2}}}$$. Substitute the value of n after counting carefully otherwise you may a calculation mistake. Remember the formulas of mean, standard deviation, variance, mean deviation etc. which are used frequently in statistics.