Question

What is the standard deviation of the set of observations 32, 28, 29, 30, 31?
A. 1.6
B. $ \sqrt{2} $
C. 2
D. None of these.

Answer 1

Hint: Recall that the $ standard\ deviation=\sigma =\sqrt{\dfrac{{{(deviations)}^{2}}}{n}} $ .
Deviation is the difference between the observation and the mean of the set of observations.
There are 5 observations. Their average can be calculated by adding them and dividing the sum by 5.

Complete step-by-step answer:
The sum of the 5 observations is $ 32+28+29+30+31=150 $ .
The mean of these observations = $ \dfrac{150}{5}=30 $ .
The deviations from the mean are $ (32-30),\ (28-30),\ (29-30),\ (30-30),\ (31-30) $ respectively.
i.e. $ 2,-2,-1,0,1 $ respectively.
Using the definition of standard deviation $ (\sigma ) $ , we have:
  $ \sigma =\sqrt{\dfrac{{{2}^{2}}+{{(-2)}^{2}}+{{(-1)}^{2}}+{{0}^{2}}+{{1}^{2}}}{5}} $
On simplifying squares, we get:
⇒ $ \sigma =\sqrt{\dfrac{4+4+1+0+1}{5}} $
⇒ $ \sigma =\sqrt{\dfrac{10}{5}} $
⇒ $ \sigma =\sqrt{2} $
The standard deviation $ (\sigma ) $ of the given observations is $ \sqrt{2} $ .
So, the correct answer is “ $ \sqrt{2} $ ”.

Note: The standard deviation of an arithmetic progression can also be calculated as: $ \sigma =|d|\sqrt{\dfrac{(n-1)(n+1)}{12}} $ .
In the given case, the numbers 28, 29, 30, 31 and 32 are in arithmetic progression with a common difference of 1 and $ n=5 $ .
Standard deviation = $ \sigma =|1|\sqrt{\dfrac{(5-1)(5+1)}{12}}=\sqrt{\dfrac{4\times 6}{12}}=\sqrt{2} $ .
The mean of a set of observation is calculated as: Mean = $ \bar{x}=\dfrac{{{x}_{1}}+{{x}_{2}}+\ ...\ +{{x}_{n}}}{n} $ .
The quantities $ ({{x}_{1}}-\bar{x}),\ ({{x}_{2}}-\bar{x}),\ ...,\ ({{x}_{n}}-\bar{x}) $ are the deviations of the observations.
Variance: $ {{\sigma }^{2}}=\dfrac{{{({{x}_{1}}-\bar{x})}^{2}}+{{({{x}_{2}}-\bar{x})}^{2}}+...+{{({{x}_{n}}-\bar{x})}^{2}}}{n}=\dfrac{\sum{{{({{x}_{i}}-\bar{x})}^{2}}}}{n}=\dfrac{\sum{x_{i}^{2}}}{n}-{{\bar{x}}^{2}} $ .
Standard Deviation: $ \sigma =\sqrt{{{\sigma }^{2}}}=\sqrt{Variance} $ .
SD is the best measure of spread of an approximately normal distribution.
The square of Standard Deviation is known as Variance.
Variance is the mean or average of the squares of the deviations or differences in the values from the mean.