Question

Find the standard deviation of 210, 240, 250, 260, 220, 230 and 270.\[\]
A.18\[\]
B.20\[\]
C.22\[\]
D.25\[\]

Answer 1

Hint: We first find the mean of the population by adding the data points ${{x}_{i}}$ and then dividing by the number of population $n=7$ following the formula for mean $\mu =\dfrac{1}{n}\sum\limits_{i=1}^{n}{{{x}_{i}}}$. We find the mean of squared differences of the from the mean of the given population $\mu $ and then take the positive square root $n=7$ following the formula for standard deviation $\sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\mu \right)}^{2}}}}{n}}$.\[\]

Complete step-by-step solution:
We know from statistics 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 formula first calculates the mean of squared differences of all data points from the mean and then takes the square root. The standard deviation signifies how the data is spread across the population or how deviated the data points are from the mean. If the standard deviation is lower that means the data points are close to mean and a higher value of standard deviation means the data points are farther from the mean. \[\]
The given sample population is 210, 240, 250, 260, 220, 230 and 270. So the number of data points is $n=7$. So the data points are
\[{{x}_{1}}=210,{{x}_{2}}=240,{{x}_{3}}=250,{{x}_{4}}=260,{{x}_{5}}=220,{{x}_{6}}=230,~{{x}_{7}}=270\]
The mean of the population is
\[\mu =\dfrac{1}{n}\sum\limits_{i=1}^{7}{{{x}_{i}}=}\dfrac{210+240+250+260+220+230+270}{7}=\dfrac{1680}{7}=240\]
We need to first find how much each data point is deviated by calculating the squared difference from $\mu $ that is ${{\left( {{x}_{i}}-\mu \right)}^{2}}$ for $i=1,2,...,7$. We have
\[\begin{align}
  & {{\left( {{x}_{1}}-\mu \right)}^{2}}={{\left( 210-240 \right)}^{2}}=900 \\
 & {{\left( {{x}_{2}}-\mu \right)}^{2}}={{\left( 240-240 \right)}^{2}}=0 \\
 & {{\left( {{x}_{3}}-\mu \right)}^{2}}={{\left( 250-240 \right)}^{2}}=100 \\
 & {{\left( {{x}_{4}}-\mu \right)}^{2}}={{\left( 260-240 \right)}^{2}}=400 \\
 & {{\left( {{x}_{5}}-\mu \right)}^{2}}={{\left( 220-240 \right)}^{2}}=400 \\
 & {{\left( {{x}_{6}}-\mu \right)}^{2}}={{\left( 230-240 \right)}^{2}}=100 \\
 & {{\left( {{x}_{7}}-\mu \right)}^{2}}={{\left( 270-240 \right)}^{2}}=900 \\
\end{align}\]
We find the mean of squared difference and then take the squareroot to find the standard deviation in accordance with standard deviation formula. We have, \[\sigma =\sqrt{\dfrac{\sum\limits_{i=1}^{n}{{{\left( {{x}_{i}}-\mu \right)}^{2}}}}{n}}=\sqrt{\dfrac{900+0+100+400+400+100+900}{7}}=\sqrt{\dfrac{2800}{7}}=\sqrt{400}=20\] \[\]
So the standard deviation is found to be 20 and the correct option is B.

Note: We note that the standard deviation is always positive quantity and that is why we have rejected negative square roots here. The standard deviation is also in the same unit of data unlike variance which is the square of standard deviation and denoted by ${{\sigma }^{2}}$.