Question

What are the mean and standard deviation of \left\\{ 34,\text{ 98, 20, -1200, -90} \right\\}?

Answer 1

We find the mean of the given data using the formula $Mean\left( \mu \right)=\dfrac{{{x}_{1}}+{{x}_{2}}+\ldots {{x}_{n}}}{n},$ where n is the number of terms in the data set and ${{x}_{1}},{{x}_{2}},\ldots {{x}_{n}}$ are the numbers in the data set. Then we calculate the standard deviation with the help of the mean by using the formula $\sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\mu \right)}^{2}}}}{n}}.$ Here ${{x}_{i}}$ stands for the ith data and $\mu$ is the mean.

Complete step by step solution:
In order to solve this question, let us first compute the mean of the given set of data. This can be calculated by using the formula
$\Rightarrow \mu =\dfrac{{{x}_{1}}+{{x}_{2}}+\ldots {{x}_{n}}}{n}$
Substituting the data 34, 98, 20, -1200 and -90 in the above formula with n as 5,
$\Rightarrow \mu =\dfrac{34+98+20-1200-90}{5}$
Adding all the values in the numerator,
$\Rightarrow \mu =\dfrac{-1138}{5}$
Dividing the value -1138 by 5, we get the mean as
$\Rightarrow \mu =-227.6$
We now calculate the value for the standard deviation of the above data and this can be calculated using the formula,
$\Rightarrow \sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\mu \right)}^{2}}}}{n}}$
Here, $\sigma$ is the standard deviation, $\mu$ is the mean, n is the number of terms in the data set and ${{x}_{i}}$ is the data. Substituting the values,
$\Rightarrow \sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{5}{{{\left( {{x}_{i}}-\left( -227.6 \right) \right)}^{2}}}}{n}}$
Expanding the summation,
$\Rightarrow \sigma =\sqrt{\dfrac{{{\left( {{x}_{1}}+227.6 \right)}^{2}}+{{\left( {{x}_{2}}+227.6 \right)}^{2}}+{{\left( {{x}_{3}}+227.6 \right)}^{2}}+{{\left( {{x}_{4}}+227.6 \right)}^{2}}+{{\left( {{x}_{5}}+227.6 \right)}^{2}}}{n}}$
We know the data values are 34, 98, 20, -1200 and -90 and substituting these for ${{x}_{1}},{{x}_{2}}\ldots {{x}_{5}},$
$\Rightarrow \sigma =\sqrt{\dfrac{{{\left( 34+227.6 \right)}^{2}}+{{\left( 98+227.6 \right)}^{2}}+{{\left( 20+227.6 \right)}^{2}}+{{\left( -1200+227.6 \right)}^{2}}+{{\left( -90+227.6 \right)}^{2}}}{5}}$ Adding the terms in the brackets,
$\Rightarrow \sigma =\sqrt{\dfrac{{{\left( 261.6 \right)}^{2}}+{{\left( 325.6 \right)}^{2}}+{{\left( 247.6 \right)}^{2}}+{{\left( -972.4 \right)}^{2}}+{{\left( 137.6 \right)}^{2}}}{5}}$
Squaring all the terms in the numerator,
$\Rightarrow \sigma =\sqrt{\dfrac{68434.56+106015.36+61305.76+945561.76+18933.76}{5}}$
Adding all the terms in the numerator and dividing by 5,
$\Rightarrow \sigma =\sqrt{\dfrac{1200251.2}{5}}=\sqrt{240050.24}$
Taking the square of the above value, we get
$\Rightarrow \sigma =489.9492$
Hence, the mean for the above data is -227.6 and the standard deviation for the same set of data is 489.9492.

Note: We need to know the calculation of mean and standard deviation. These form the basis for many statistical calculations in mathematics. We need to be careful while calculating the mean and standard deviation. The sign of all the terms must be taken into consideration. Neglecting the sign could lead to wrong results.