Question

How do you find the standard deviation of $3.4,6.1,7.3,4.5,2.3,4.4,9.4,3.4,2.3,7.2?$

Answer 1

We find the mean of the values given. Then we find the size of the population. And apply these values in the formula for the standard deviation.

Complete step-by-step answer:
Consider the given population $3.4,6.1,7.3,4.5,2.3,4.4,9.4,3.4,2.3,7.2$
First, we are going to find the mean of the population.
Mean of the population is the average of the given values. So, the formula for finding the mean is given by, $\mu =\dfrac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n}$ when $n$ observations are given.
We can interpret this as the sum of observations given divided by the number of observations.
Here the observations are $3.4,6.1,7.3,4.5,2.3,4.4,9.4,3.4,2.3,7.2$
The number of observations is $10.$
Now the sum of the observations is given by,
$\Rightarrow 3.4+6.1+7.3+4.5+2.3+4.4+9.4+3.4+2.3+7.2=50.3$
Now, the mean is $\mu =\dfrac{50.3}{10}=5.03$
Now, we have to find the standard deviation using the formula $\sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\mu \right)}^{2}}}}{N}}$ where $\mu$ is the mean and $N$ is the size of the population.
Let us find the value of $\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\mu \right)}^{2}}}$
Let us find the differences separately square them. And then we add them. After that, we divide the value with $N$ and find the square root.
Let us start from finding the differences,
$\Rightarrow {{x}_{1}}-\mu =3.4-5.03=-1.63$
$\Rightarrow {{x}_{2}}-\mu =6.1-5.03=1.07$
$\Rightarrow {{x}_{3}}-\mu =7.3-5.03=2.27$
$\Rightarrow {{x}_{4}}-\mu =4.5-5.03=-0.53$
$\Rightarrow {{x}_{5}}-\mu =2.3-5.03=-2.73$
$\Rightarrow {{x}_{6}}-\mu =4.4-5.03=-0.63$
$\Rightarrow {{x}_{7}}-\mu =9.4-5.03=4.37$
$\Rightarrow {{x}_{8}}-\mu =3.4-5.03=-1.63$
$\Rightarrow {{x}_{9}}-\mu =2.3-5.03=-2.73$
$\Rightarrow {{x}_{10}}-\mu =7.2-5.03=2.17$
Now we square these differences,
$\Rightarrow {{\left( -1.63 \right)}^{2}}=2.6569$
$\Rightarrow {{\left( 1.07 \right)}^{2}}=1.1449$
$\Rightarrow {{\left( 2.27 \right)}^{2}}=5.1529$
$\Rightarrow {{\left( -0.53 \right)}^{2}}=0.2809$
$\Rightarrow {{\left( -2.73 \right)}^{2}}=7.4529$
$\Rightarrow {{\left( -0.63 \right)}^{2}}=0.3969$
$\Rightarrow {{\left( 4.37 \right)}^{2}}=19.0969$
$\Rightarrow {{\left( -1.63 \right)}^{2}}=2.6569$
$\Rightarrow {{\left( -2.73 \right)}^{2}}=7.4529$
$\Rightarrow {{\left( 2.17 \right)}^{2}}=4.7089$
The next step is to add these squared values,
$\Rightarrow 2.6569+1.1449+5.1529+0.2809+7.4529+0.3969+19.0969+2.6569+7.4529+4.7089=51.001$
Now we divide this sum by the population size,
The population size is $N=10$
So, we get, $\dfrac{51.001}{10}=5.1001$
This is the time to find the square root of the quotient we found,
We will get, $\sqrt{5.1001}=2.2583401$
Hence, the standard deviation is $\sigma =2.26.$

Note: The standard deviation is a measure of the amount of variation or dispersion of a set of values. According to the formula, the standard deviation is the square root of the sum of squares of the differences of the values and the mean divided by the population size. Remember that the square of the standard deviation is the variance of the population.