Question

How do you find the standard deviation of $5.6,5.2,4.6,4.9,5.7,6.4?$

Answer 1

Hint: We know that the standard deviation of a given data is the square root of the sum of squares of the differences of observations from the arithmetic mean of the observations divided by the number of observations.

Complete step by step solution:
Let us consider the given data $5.6,5.2,4.6,4.9,5.7,6.4.$
We are asked to find the standard deviation of the given data.
We know that the standard deviation of $n$ observations ${{x}_{i}}, i=1,2,...,n$ with arithmetic mean $\bar{x}$ can be found by the formula $\sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\bar{x} \right)}^{2}}}}{n}}.$
We need to find the values required to apply in the formula so that we can find the standard deviation.
Here, we know that the number of observations is $n=6.$
Now, we need to find the numerator.
As a basic task, we are going to find the arithmetic mean of the given observations by dividing the sum of the observations by the number of the observations.
So, we will get $\bar{x}=\dfrac{5.6+5.2+4.6+4.9+5.7+6.4}{6}=\dfrac{32.4}{6}=5.4.$
So, as a first task, we need to find the difference of the observations from the arithmetic mean of the observations.
We will get the differences,
$5.6-5.4=0.2.$
$5.2-5.4=-0.2$
$4.6-5.4=-0.8$
$4.9-5.4=-0.5$
$5.7-5.4=0.3$
$6.4-5.4=1$
Now, we need to find the squares of the differences.
We will get,
${{\left( 0.2 \right)}^{2}}=0.04$
${{\left( -0.2 \right)}^{2}}=0.04$
${{\left( -0.8 \right)}^{2}}=0.64$
${{\left( -0.5 \right)}^{2}}=0.25$
${{\left( 0.3 \right)}^{2}}=0.09$
${{1}^{2}}=1$
Now we need to find the sum of the squares.
We will get $0.04+0.04+0.64+0.25+0.09+1=2.06.$
Therefore, the standard deviation is $\sigma =\sqrt{\dfrac{2.076}{6}}=\sqrt{0.343}=0.5859.$
Hence the standard deviation of the given data is $0.5859.$

Note: We know that the square of the standard deviation is the variance of the observations. So, we can find the variance of the given observations by squaring the obtained standard deviation. Therefore, the variance is ${{\sigma }^{2}}=0.343.$