Question

Two samples of sizes 60 and 90 have 52 and 48 as the respective arithmetic means and 9 and 12 as the respective standard deviations. Find standard deviation of the combined sample of size 150.

Answer 1

Hint: We use the formula for arithmetic mean of the combined sample as ${{\mu }_{c}}=\dfrac{{{n}_{1}}{{\mu }_{1}}+{{n}_{2}}{{\mu }_{2}}}{{{n}_{1}}+{{n}_{2}}}$ and then we find the standard deviation of the combined sample using the formula ${{\sigma }_{c}}=\sqrt{\dfrac{{{n}_{1}}\left( {{\sigma }_{1}}^{2}+{{d}_{1}}^{2} \right)+{{n}_{2}}\left( {{\sigma }_{2}}^{2}+{{d}_{2}}^{2} \right)}{{{n}_{1}}+{{n}_{2}}}}$ where ${{n}_{1}},{{n}_{2}}$ are sample sizes, ${{\mu }_{1}},{{\mu }_{2}}$ are arithmetic means and ${{\sigma }_{1}},{{\sigma }_{2}}$ are standard deviations of the samples respectively.

Complete step-by-step answer:

We know from statistics that that if there are $n$ number of data points in the population say ${{x}_{1}},{{x}_{2}},...,{{x}_{n}}$ are given then the mean of the data points is denoted as $\mu $ and is given by

\[\mu =\dfrac{{{x}_{1}}+{{x}_{2}}+...={{x}_{n}}}{n}=\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}}\]

The mean of population signifies where the data points are centred around and that is why mean is also called a measure of central tendency along with median and mode. The standard deviation abbreviated as SD and denoted by the Greek alphabet $\sigma $ is given by the formula

\[\sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\mu \right)}^{2}}}}{n}}\]

The standard deviation signifies how the data is spread across the population or how deviated the data points are from the mean. If standard deviation is lower that means the data points are close mean and higher value of standard deviation means the data points are farther form mean. \[\]

We also know that if there are two sample sizes say ${{n}_{1}},{{n}_{2}}$ have arithmetic means ${{\mu }_{1}},{{\mu }_{2}}$ respectively and standard deviations ${{\sigma }_{1}},{{\sigma }_{2}}$ respectively then the arithmetic mean of combined sample ${{\mu }_{c}}$ with size${{n}_{1}}+{{n}_{2}}$ is given by

\[{{\mu }_{c}}=\dfrac{{{n}_{1}}{{\mu }_{1}}+{{n}_{2}}{{\mu }_{2}}}{{{n}_{1}}+{{n}_{2}}}\]

The standard deviation of combined sample ${{\sigma }_{c}}$ is given by

\[{{\sigma }_{c}}=\sqrt{\dfrac{{{n}_{1}}\left( {{\sigma }_{1}}^{2}+{{d}_{1}}^{2} \right)+{{n}_{2}}\left( {{\sigma }_{2}}^{2}+{{d}_{2}}^{2} \right)}{{{n}_{1}}+{{n}_{2}}}}\]

Here we have denoted the differences of sample means from the combined mean as

\[{{d}_{1}}={{\mu }_{1}}-{{\mu }_{c}},{{d}_{2}}={{\mu }_{2}}-{{\mu }_{c}}\]

We are given in the question that two samples of sizes 60 and 90 have 52 and 48 as the respective arithmetic means and 9 and 12 as the respective standard deviations. So we have

\[{{n}_{1}}=60,{{n}_{2}}=90,{{\mu }_{1}}=52,{{\mu }_{2}}=48,{{\sigma }_{1}}=9,{{\sigma }_{2}}=12\]

Let us use the formula for arithmetic mean for combined sample and have,

\[{{\mu }_{c}}=\dfrac{60\times 52+90\times 48}{60+90}=\dfrac{7440}{150}=49.6\]

Let us find the differences of sample means from the combined mean as

\[ {{d}_{1}}={{\mu }_{1}}-{{\mu }_{c}}=52-49.6=2.4 \]

\[ {{d}_{2}}={{\mu }_{2}}-{{\mu }_{c}}=48-49.6=-1.6 \]

Let us use the formula for standard deviation for combined sample and have,

\[   {{\sigma }_{c}}=\sqrt{\dfrac{60\left( {{9}^{2}}+{{2.4}^{2}} \right)+90\left( {{12}^{2}}+{{\left( -1.6 \right)}^{2}} \right)}{60+90}} \]

\[ \Rightarrow {{\sigma }_{c}}=\sqrt{\dfrac{60\left( 86.76 \right)+90\left( 146.56 \right)}{150}} \]

\[ \Rightarrow {{\sigma }_{c}}=\sqrt{\dfrac{18396}{150}}=\sqrt{122.64}=11.074 \]

Note: We rejected the negative sign on square root because standard deviation is a positive quantity. The square of standard deviation is called variance and is denoted by ${{\sigma }^{2}}$. The ratio of mean to standard deviation is called coefficient of variation. The alternative formula for standard deviation for combined sample is $\sqrt{\dfrac{\left( {{n}_{1}}-1 \right){{\sigma }_{1}}^{2}+\left( {{n}_{2}}-1 \right){{\sigma }_{2}}^{2}+\dfrac{{{n}_{1}}{{n}_{2}}}{{{n}_{1}}+{{n}_{2}}}{{\left( {{\mu }_{1}}-{{\mu }_{2}} \right)}^{2}}}{{{n}_{1}}+{{n}_{2}}}}$.