Chapter 13: Statistics
13.1 Measures of Dispersion
In earlier classes, we have studied measures of central tendency: Mean, Median, and Mode. These measures give us a single value around which the data is clustered. However, they do not tell us how the data is scattered around the central value. The scattering of data is called dispersion.
Some important measures of dispersion are:
- Range
- Mean Deviation
- Variance
- Standard Deviation
13.2 Range
The range is the simplest measure of dispersion. It is the difference between the maximum and minimum values of a series. \[ \text{Range} = x_{\text{max}} - x_{\text{min}} \]
13.3 Mean Deviation
Mean deviation is the arithmetic mean of the absolute deviations of the observations from an appropriate measure of central tendency (usually mean or median).
For Ungrouped Data
Let \( x_1, x_2, \dots, x_n \) be \( n \) observations. Let \( a \) be the measure of central tendency (Mean \( \bar{x} \) or Median \( M \)). Then, Mean Deviation about \( a \) is: \[ M.D.(a) = \frac{1}{n} \sum_{i=1}^{n} |x_i - a| \]
For Grouped Data
Let the observations \( x_1, x_2, \dots, x_n \) have frequencies \( f_1, f_2, \dots, f_n \) respectively. Let \( N = \sum f_i \). Then: \[ M.D.(a) = \frac{1}{N} \sum_{i=1}^{n} f_i |x_i - a| \]
13.4 Variance and Standard Deviation
The square of the standard deviation is called variance. It overcomes the drawback of taking absolute values in Mean Deviation by squaring the deviations before averaging them.
Variance (\( \sigma^2 \))
For ungrouped data: \[ \sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 \] A commonly used alternative formula is: \[ \sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2 \]
Standard Deviation (\( \sigma \))
The standard deviation is the positive square root of the variance. \[ \sigma = \sqrt{ \frac{1}{n} \sum_{i=1}^{n} (x_i - \bar{x})^2 } \]
Coefficient of Variation (C.V.)
To compare the variability or dispersion of two or more series, we use the Coefficient of Variation. \[ C.V. = \frac{\sigma}{\bar{x}} \times 100 \] A series with lesser C.V. is more consistent (less dispersed).
Competency-Based Questions
Question 1:
The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation if the wrong item is omitted.
Answer 1:
Given: Number of observations \( n = 20 \)
Incorrect Mean \( \bar{x} = 10 \)
Incorrect Standard Deviation \( \sigma = 2 \)
Incorrect observation \( = 8 \)
First, let’s find the correct sum of observations.
\[ \text{Incorrect sum of observations } \sum x = n \bar{x} = 20 \times 10 = 200 \]
Since observation 8 is omitted, the new number of observations \( n’ = 19 \).
\[ \text{Correct sum of observations } \sum x’ = 200 - 8 = 192 \]
\[ \text{Correct Mean } \bar{x}’ = \frac{192}{19} \approx 10.105 \]
Now calculate the correct variance. We know:
\[ \sigma^2 = \frac{1}{n} \sum x^2 - (\bar{x})^2 \]
\[ 2^2 = \frac{1}{20} \sum x^2 - (10)^2 \]
\[ 4 = \frac{1}{20} \sum x^2 - 100 \]
\[ 104 = \frac{1}{20} \sum x^2 \Rightarrow \text{Incorrect } \sum x^2 = 2080 \]
When the item 8 is removed, the correct sum of squares:
\[ \text{Correct } \sum x’^2 = 2080 - 8^2 = 2080 - 64 = 2016 \]
Now find the new variance:
\[ \text{New Variance } \sigma’^2 = \frac{\text{Correct } \sum x’^2}{n’} - (\bar{x}’)^2 \]
\[ \sigma’^2 = \frac{2016}{19} - \left(\frac{192}{19}\right)^2 = \frac{2016}{19} - \frac{36864}{361} \]
\[ \sigma’^2 = \frac{2016 \times 19 - 36864}{361} = \frac{38304 - 36864}{361} = \frac{1440}{361} \approx 3.988 \]
The new standard deviation is:
\[ \sigma’ = \sqrt{ \frac{1440}{361} } = \frac{\sqrt{1440}}{19} = \frac{12\sqrt{10}}{19} \approx \frac{37.947}{19} \approx 1.997 \]
Conclusion: The correct mean is 10.105 and the correct standard deviation is 1.997.
Question 2:
Determine the range and mean deviation about the mean for the simple data set: 6, 7, 10, 12, 13, 4, 8, 12.
Answer 2:
Range:
Maximum value = 13, Minimum value = 4.
Range = 13 - 4 = 9.
Mean Deviation about mean:
Data points: 4, 6, 7, 8, 10, 12, 12, 13.
Number of observations \( n = 8 \).
Mean \( \bar{x} = \frac{4 + 6 + 7 + 8 + 10 + 12 + 12 + 13}{8} = \frac{72}{8} = 9 \).
Deviations \( |x_i - \bar{x}| \):
\( |4 - 9| = 5 \)
\( |6 - 9| = 3 \)
\( |7 - 9| = 2 \)
\( |8 - 9| = 1 \)
\( |10 - 9| = 1 \)
\( |12 - 9| = 3 \)
\( |12 - 9| = 3 \)
\( |13 - 9| = 4 \)
Sum of absolute deviations = 5 + 3 + 2 + 1 + 1 + 3 + 3 + 4 = 22.
Mean Deviation = \( \frac{22}{8} = 2.75 \).
Conclusion: Range = 9, Mean Deviation about Mean = 2.75.