Standard Deviation

Standard Deviation

The standard deviation, which is shown by Greek letter s (read as sigma) is extremely useful in judging the representativeness of the mean. The concept of standard deviation, which was introduced by Karl Pearson, has a practical significance because it is free from all defects, which exists in a range, quartile deviation or average deviation.

Standard deviation is calculated as the square root of average of squared deviations taken from actual mean. In other words, it is the square root of the variance. It is also called root mean square deviation.

Standard Deviation for an entire population \(= \sigma = \sqrt{\cfrac{\sum(x – \mu)^2}{N}}\)

Where:

\(\sigma =\) population standard deviation;
\(x =\) each value in the population;
\(\mu =\) the mean of the values; and
\(N =\) the number of values (the population).

Standard Deviation for a sample of a population \(= s = \sqrt{\cfrac{\sum(x – \bar{x})^2}{N – 1}}\)

Where:

\(s =\) sample standard deviation;
\(x =\) each value in the sample;
\(\bar{x} =\) the mean of the values; and
\(N =\) the number of values (the sample size).

Examples:

a) Four friends were comparing their scores on a recent essay. The scores were as follows: 6, 2, 3, 1. Calculate the standard deviation of their scores.

b) A sample of four students was taken to see how many pencils they were carrying. The responses were: 2, 2, 5, 7. Calculate the standard deviation of their responses.

Solution:

a) Step 1: Find the mean.
\(\mu = \cfrac{6 + 2 + 3 + 1}{4} = \cfrac{12}{4} = 3\)

Step 2: Subtract the mean from each score \((x – \mu)\).
\(6 – 3 = 3\)
\(2 – 3 = -1\)
\(3 – 3 = 0\)
\(1 – 3 = -2\)

Step 3: Square each deviation \((x – \mu)^2\).
\((3)^2 = 9\)
\((-1)^2 = 1\)
\((0)^2 = 0\)
\((-2)^2 = 4\)

Step 4: Add the squared deviations \((\sum(x – \mu)^2)\).
\(9 + 1 + 0 + 4 = 14\)

Step 5: Divide the sum by the number of scores \(\left(\cfrac{\sum(x – \mu)^2}{N}\right)\).
\(\cfrac{14}{4} = 3.5\)

Step 6: Take the square root of the result from Step 5 \(\left(\sqrt{\cfrac{\sum(x – \mu)^2}{N}}\right)\).
\(\sqrt{3.5} \approx 1.87\)

The standard deviation is approximately \(1.87\).

b) Step 1: Find the mean.
\(s = \cfrac{2 + 2 + 5 + 7}{4} = \cfrac{16}{4} = 4\)

Step 2: Subtract the mean from each score \((x – \bar{x})\).
\(2 – 4 = -2\)
\(2 – 4 = -2\)
\(5 – 4 = 1\)
\(7 – 4 = 3\)

Step 3: Square each deviation \((x – \bar{x})^2\).
\((-2)^2 = 4\)
\((-2)^2 = 4\)
\((1)^2 = 1\)
\((3)^2 = 9\)

Step 4: Add the squared deviations \(\sum(x – \bar{x})^2\).
\(4 + 4 + 1 + 9 = 18\)

Step 5: Divide the score by one less than the number of values \(\left(\cfrac{\sum(x – \bar{x})^2}{N – 1}\right)\).
\(\cfrac{18}{4 – 1} = \cfrac{18}{3} = 6\)

Step 6: Take the square root of the result from Step 5 \(\left(\sqrt{\cfrac{\sum(x – \bar{x})^2}{N – 1}}\right)\).
\(\sqrt{6} \approx 2.45\)

The standard deviation is approximately \(2.45\).

Merits of Standard Deviation

  1. Standard deviation is the best measure of dispersion because it takes into account all the items and is capable of future algebraic treatment and statistical analysis.
  2. It is possible to calculate standard deviation for two or more series.
  3. This measure is most suitable for making comparisons among two or more series about variability.
  4. It is least affected by sampling fluctuations.
  5. It can tactfully avoid the complication of considering the negative algebraic sign while calculating deviations.

Demerits of Standard Deviation

  1. It involves complicated and laborious numerical calculations especially when the information is large enough.
  2. It is neither easy to take up, nor simple to calculate.
  3. It is considerably affected by the extreme values of the given variable.

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