CBSE Class 11 Applied Mathematics Analysis of Data Notes Set 02

Revision Notes for Class 11 Mathematics: Chapter 13 Statistics

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Measures of Dispersion

Measure of Dispersion

The dispersion or scatter in a data is measured on the basis of the observations and the type of the measure of central tendency used there. Consider the runs scored by two batsmen in their last ten matches as follows:
Batsman A: 30, 91, 0, 64, 42, 80, 30, 5, 117, 71
Batsman B: 53, 46, 48, 50, 53, 53, 58, 60, 57, 52

Range

From the runs scored by the two batsmen A and B, we get some idea of the variability in the scores on the basis of the minimum and maximum runs in each series. To obtain a single number for this, we find the difference of the maximum and minimum values of each series. This difference is called the ‘range’ of the data.
In case of batsman A, range = 117 − 0 = 117, and for batsman B, range = 60 − 46 = 14. Clearly, range of A > range of B. Therefore the scores are scattered or dispersed in case of A, while for B they are close to each other.
Thus, Range of a series = Maximum value − Minimum value.
The range of data gives us a rough idea of variability or scatter but does not tell about the dispersion of the data from a measure of central tendency. For this purpose, we need some other measure of variability. Clearly, such a measure must depend upon the differences (or deviations) of the values from the central tendency.

Mean Deviation

The deviation of an observation x from a fixed value ‘a’ is the difference x − a. In order to find a suitable measure of dispersion, we require the distance of each value from a central tendency or a fixed number ‘a’. Recall that the absolute value of the difference of two numbers gives the distance between the numbers when represented on a number line. Thus, to find the measure of dispersion from a fixed number ‘a’, we may take the mean of the absolute values of the deviations from the central value. This mean is called the ‘mean deviation’. The mean deviation about ‘a’ is denoted as M.D.(a). Therefore
\( \text{M.D.}(a) = \frac{\text{Sum of absolute values of deviations from } a}{\text{Number of observations}} \)
Mean deviation may be obtained from any measure of central tendency. However, mean deviations about the mean and the median are commonly used in statistical studies.

Mean Deviation for Ungrouped Data

Let n observations be \( x_1, x_2, \ldots, x_n \). The following steps are involved in the calculation of the mean deviation about the mean or median:
Step 1: Calculate the measure of central tendency about which we are to find the mean deviation. Let it be ‘a’.
Step 2: Find the deviation of each \( x_i \) from a, i.e. \( x_1 - a, x_2 - a, \ldots, x_n - a \).
Step 3: Find the absolute values of the deviations, i.e. drop the minus sign (−) if it is there: \( |x_1 - a|, |x_2 - a|, \ldots, |x_n - a| \).
Step 4: Find the mean of the absolute values of the deviations. This mean is the mean deviation about a, i.e. \( \text{M.D.}(a) = \frac{\sum_{i=1}^{n} |x_i - a|}{n} \)
Thus \( \text{M.D.}(\bar{x}) = \frac{1}{n}\sum_{i=1}^{n} |x_i - \bar{x}| \), where \( \bar{x} \) = mean, and \( \text{M.D.}(M) = \frac{1}{n}\sum_{i=1}^{n} |x_i - M| \), where M = median.

 

Example 1. Find the mean deviation about the mean for the following data: 6, 7, 10, 12, 13, 4, 8, 12.
Answer:
Step 1: Mean of the given data is \( \bar{x} = \frac{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}{8} = \frac{72}{8} = 9 \)
Step 2: The deviations of the respective observations from the mean \( \bar{x} \), i.e. \( x_i - \bar{x} \), are 6 − 9, 7 − 9, 10 − 9, 12 − 9, 13 − 9, 4 − 9, 8 − 9, 12 − 9, or −3, −2, 1, 3, 4, −5, −1, 3.
Step 3: The absolute values of the deviations, i.e. \( |x_i - \bar{x}| \), are 3, 2, 1, 3, 4, 5, 1, 3.
Step 4: The required mean deviation about the mean is
\( \text{M.D.}(\bar{x}) = \frac{\sum_{i=1}^{8} |x_i - \bar{x}|}{8} = \frac{3 + 2 + 1 + 3 + 4 + 5 + 1 + 3}{8} = \frac{22}{8} \) = 2.75

 

Example 2. Find the mean deviation about the median for the following data: 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.
Answer:
Here the number of observations is 11, which is odd. Arranging the data in ascending order, we have 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21.
Now, median = \( \left(\frac{11 + 1}{2}\right) \)th or 6th observation = 9.
The absolute values of the respective deviations from the median, i.e. \( |x_i - M| \), are 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12.
Therefore \( \sum_{i=1}^{11} |x_i - M| = 58 \)
and \( \text{M.D.}(M) = \frac{1}{11}\sum_{i=1}^{11} |x_i - M| = \frac{58}{11} \) = 5.27

Mean Deviation for Grouped Data

We know that data can be grouped in two ways: (a) discrete frequency distribution, (b) continuous frequency distribution.

(a) Discrete Frequency Distribution

Let the given data consist of n discrete values \( x_1, x_2, \ldots, x_n \) occurring with frequencies \( f_1, f_2, \ldots, f_n \) respectively. This data can be represented in tabular form as given below, and is called a discrete frequency distribution:
x: \( x_1, x_2, x_3, \ldots, x_n \)
f: \( f_1, f_2, f_3, \ldots, f_n \)
(i) Mean deviation about the mean: First of all we find the mean \( \bar{x} \) of the given data by using the formula
\( \bar{x} = \frac{\sum_{i=1}^{n} x_if_i}{\sum_{i=1}^{n} f_i} = \frac{1}{N}\sum_{i=1}^{n} x_if_i \), where \( N = \sum_{i=1}^{n} f_i \)
Then we find the deviations of the observations \( x_i \) from the mean \( \bar{x} \) and take their absolute values, i.e. \( |x_i - \bar{x}| \) for all i = 1, 2, …, n. After this we find the mean of the absolute values of the deviations, which is the required mean deviation about the mean. Thus
\( \text{M.D.}(\bar{x}) = \frac{\sum_{i=1}^{n} f_i|x_i - \bar{x}|}{\sum_{i=1}^{n} f_i} = \frac{1}{N}\sum_{i=1}^{n} f_i|x_i - \bar{x}| \)
(ii) Mean deviation about the median: To find the mean deviation about the median, we find the median of the given discrete frequency distribution. For this, the observations are arranged in ascending order and the cumulative frequencies are obtained. Then we identify the observation whose cumulative frequency is equal to or just greater than \( \frac{N}{2} \), where N is the sum of the frequencies. This value of the observation lies in the middle of the data; therefore it is the required median. After finding the median, we obtain the mean of the absolute values of the deviations from the median. Thus
\( \text{M.D.}(M) = \frac{1}{N}\sum_{i=1}^{n} f_i|x_i - M| \)

 

Example 3. Find the mean deviation about the mean for the following data:
\( x_i \)25681012
\( f_i \)2810785

Answer:
\( x_i \)\( f_i \)\( f_ix_i \)\( |x_i - \bar{x}| \)\( f_i|x_i - \bar{x}| \)
2245.511
58402.520
610601.515
87560.53.5
108802.520
125604.522.5
4030092

\( N = \sum_{i=1}^{6} f_i = 40 \), \( \sum_{i=1}^{6} f_ix_i = 300 \), \( \sum_{i=1}^{6} f_i|x_i - \bar{x}| = 92 \)
Therefore \( \bar{x} = \frac{1}{N}\sum_{i=1}^{6} f_ix_i = \frac{1}{40} \times 300 \) = 7.5
and \( \text{M.D.}(\bar{x}) = \frac{1}{N}\sum_{i=1}^{6} f_i|x_i - \bar{x}| = \frac{1}{40} \times 92 \) = 2.3

 

Example 4. Find the mean deviation about the median for the following data:
\( x_i \)3691213152122
\( f_i \)34524543

Answer:
The given observations are already in ascending order. We get the cumulative frequencies of the data by adding a row:
\( x_i \)3691213152122
\( f_i \)34524543
c.f.37121418232730

Now, N = 30, which is even, so the median is the mean of the 15th and 16th observations. Both of these observations lie in the cumulative frequency 18, for which the corresponding observation is 13.
Therefore, median M = \( \frac{\text{15th observation} + \text{16th observation}}{2} = \frac{13 + 13}{2} \) = 13
Now, the absolute values of the deviations from the median, i.e. \( |x_i - M| \), are given below:
\( |x_i - M| \)107410289
\( f_i \)34524543
\( f_i|x_i - M| \)30282020103227

We have \( \sum_{i=1}^{8} f_i = 30 \) and \( \sum_{i=1}^{8} f_i|x_i - M| = 149 \)
Therefore \( \text{M.D.}(M) = \frac{1}{N}\sum_{i=1}^{8} f_i|x_i - M| = \frac{1}{30} \times 149 \) = 4.97

(b) Continuous Frequency Distribution

A continuous frequency distribution is a series in which the data are classified into different class intervals without gaps along with their respective frequencies.
(i) Mean deviation about the mean: While calculating the mean of a continuous frequency distribution, we make the assumption that the frequency in each class is centred at its mid-point. Here also we write the mid-point of each given class and proceed further as for a discrete frequency distribution to find the mean deviation.

 

Example 5. Find the mean deviation about the mean for the following data.
Marks obtained10 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80
Number of students23814832

Answer:
Marks obtainedNumber of students \( f_i \)Mid-point \( x_i \)\( f_ix_i \)\( |x_i - \bar{x}| \)\( f_i|x_i - \bar{x}| \)
10 – 20215303060
20 – 30325752060
30 – 408352801080
40 – 50144563000
50 – 608554401080
60 – 703651952060
70 – 802751503060
401800400

Here \( N = \sum_{i=1}^{7} f_i = 40 \), \( \sum_{i=1}^{7} f_ix_i = 1800 \), \( \sum_{i=1}^{7} f_i|x_i - \bar{x}| = 400 \)
Therefore \( \bar{x} = \frac{1}{N}\sum_{i=1}^{7} f_ix_i = \frac{1800}{40} \) = 45
and \( \text{M.D.}(\bar{x}) = \frac{1}{N}\sum_{i=1}^{7} f_i|x_i - \bar{x}| = \frac{1}{40} \times 400 \) = 10

Short cut method for calculating the mean deviation about the mean: We can avoid the tedious calculation of \( \bar{x} \) by the step deviation method. In this method, we take an assumed mean which is in the middle of, or just close to the middle of, the data. Then the deviations of the observations (or mid-points of classes) are taken from the assumed mean. This is nothing but shifting the origin from zero to the assumed mean on the number line.
If there is a common factor of all the deviations, we divide them by this common factor to further simplify the deviations. These are known as step deviations. The process of taking step deviations is a change of scale on the number line, as shown below.

Change of origin and change of scale (assumed mean A = 60, h = 10) 0 10 20 30 40 50 60 70 80 90 100 110 120 Observations x −60 −50 −40 −30 −20 −10 0 10 20 30 40 50 60 Deviations x − 60 −6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6 Step deviations (x − 60)/10 A = 60 becomes the new origin 0

The deviations and step deviations reduce the size of the observations, so that the computation becomes simpler. Let the new variable be denoted by \( d_i = \frac{x_i - a}{h} \), where ‘a’ is the assumed mean and h is the common factor. Then the mean \( \bar{x} \) by the step deviation method is given by
\( \bar{x} = a + \frac{\sum_{i=1}^{n} f_id_i}{N} \times h \)

 

Example 6. Find the mean deviation about the mean for the following data by using the step deviation method.
Marks obtained10 – 2020 – 3030 – 4040 – 5050 – 6060 – 7070 – 80
Number of students23814832

Answer:
Marks obtainedNumber of students \( f_i \)Mid-point \( x_i \)\( d_i = \frac{x_i - 45}{10} \)\( f_id_i \)\( |x_i - \bar{x}| \)\( f_i|x_i - \bar{x}| \)
10 – 20215−3−63060
20 – 30325−2−62060
30 – 40835−1−81080
40 – 5014450000
50 – 60855181080
60 – 70365262060
70 – 80275363060
400400

Taking the assumed mean a = 45 and h = 10,
\( \bar{x} = a + \frac{\sum_{i=1}^{7} f_id_i}{N} \times h = 45 + \frac{0}{40} \times 10 \) = 45
and \( \text{M.D.}(\bar{x}) = \frac{1}{N}\sum_{i=1}^{7} f_i|x_i - \bar{x}| = \frac{400}{40} \) = 10

(ii) Mean deviation about the median: The data is first arranged in ascending order. Then the median of the continuous frequency distribution is obtained by first identifying the class in which the median lies (median class) and then applying the formula
\( \text{Median} = l + \frac{\frac{N}{2} - c}{f} \times h \)
where the median class is the class interval whose cumulative frequency is just greater than or equal to \( \frac{N}{2} \), N is the sum of the frequencies, l, f and h are respectively the lower limit, the frequency and the width of the median class, and c is the cumulative frequency of the class just preceding the median class. After finding the median, the absolute values of the deviations of the mid-points \( x_i \) of each class from the median, i.e. \( |x_i - M| \), are obtained. Then
\( \text{M.D.}(M) = \frac{1}{N}\sum_{i=1}^{n} f_i|x_i - M| \)

 

Example 7. Calculate the mean deviation about the median for the following data:
Class0 – 1010 – 2020 – 3030 – 4040 – 5050 – 60
Frequency67151642

Answer:
ClassFrequency \( f_i \)Cumulative frequency c.f.Mid-point \( x_i \)\( |x_i - M| \)\( f_i|x_i - M| \)
0 – 1066523138
10 – 20713151391
20 – 30152825345
30 – 401644357112
40 – 50448451768
50 – 60250552754
50508

The class interval containing the \( \frac{N}{2} \)th or 25th item is 20 – 30. Therefore, 20 – 30 is the median class.
We know that Median = \( l + \frac{\frac{N}{2} - c}{f} \times h \). Here l = 20, c = 13, f = 15, h = 10 and N = 50.
Therefore, Median = \( 20 + \frac{25 - 13}{15} \times 10 \) = 20 + 8 = 28
Thus, the mean deviation about the median is given by
\( \text{M.D.}(M) = \frac{1}{N}\sum_{i=1}^{6} f_i|x_i - M| = \frac{1}{50} \times 508 \) = 10.16

Limitations of Mean Deviation

In a series where the degree of variability is very high, the median is not a representative central tendency. Thus, the mean deviation about the median calculated for such a series cannot be fully relied upon. Mean deviation is calculated on the basis of the absolute values of the deviations and therefore cannot be subjected to further algebraic treatment. This implies that we must have some other measure of dispersion. Standard deviation is such a measure of dispersion.

Skewness

Skewness

Skewness or asymmetry denotes the tendency of a distribution to depart from symmetry.
A frequency distribution is said to be skewed if the frequencies decrease with markedly greater rapidity on one side of the central maximum than on the other side. This characteristic of a frequency distribution is known as skewness, and the measures of asymmetry are usually called ‘measures of skewness’.
The main object of measuring skewness is to know the direction of the variation from an average and to compare frequency distributions and the shapes of their curves.

Types of Skewness

(i) Positive skewness: When in a frequency distribution the frequencies are piled up at the lower values of the variable and spread out over a greater range of values on the high side, the skewness is known as positive.
(ii) Negative skewness: When in a frequency distribution the frequencies are piled up at the higher values of the variable and spread out over a greater range of values on the low side, the skewness is known as negative.

Measures of Skewness

Measures of skewness may be absolute or relative. An absolute measure tells us the direction and extent of asymmetry in a frequency distribution. A relative measure, often called the coefficient of skewness, permits us to compare two or more frequency distributions.

Karl Pearson’s Coefficient of Skewness

(a) Absolute measure, Sk = Mean − Mode = \( \bar{x} - Z \)
Relative measure, \( J = \frac{\text{Mean} - \text{Mode}}{\text{Standard deviation}} \) or \( \frac{\bar{x} - Z}{\sigma} \)
(b) If the mode is not determined (ill-defined), then
Absolute measure, Sk = 3(Mean − Median) = \( 3(\bar{x} - M) \)
Relative measure, \( J = \frac{3(\text{Mean} - \text{Median})}{\text{Standard deviation}} \) or \( \frac{3(\bar{x} - M)}{\sigma} \)
Here the standard deviation of n observations is \( \sigma = \sqrt{\frac{\Sigma(x - \bar{x})^2}{N}} \).

 

Example 8. From the following data, calculate Karl Pearson’s coefficient of skewness:
Year1950195119521953195419551956195719581959
Price index of wheat838793104106109118124126130

Answer:
S. No.X\( X - \bar{X} \)\( (X - \bar{X})^2 \)
183−25625
287−21441
393−15225
4104−416
5106−24
610911
711810100
812416256
912618324
1013022484
ΣX = 10802476

Arithmetic mean \( \bar{X} = \frac{\Sigma X}{N} = \frac{1080}{10} \) = 108
Median, M = size of \( \left(\frac{N + 1}{2}\right) \)th value = \( \frac{10 + 1}{2} \)th value = 5.5th value
\( M = \frac{\text{5th value} + \text{6th value}}{2} = \frac{106 + 109}{2} = \frac{215}{2} \) = 107.5
S.D., \( \sigma = \sqrt{\frac{\Sigma(X - \bar{X})^2}{N}} = \sqrt{\frac{2476}{10}} = \sqrt{247.6} \) = 15.74 (approx.)
\( J = \frac{3(\text{Mean} - \text{Median})}{\text{Standard deviation}} = \frac{3(108 - 107.5)}{15.74} = \frac{3 \times 0.5}{15.74} \) = 0.0953 (approx.)
Since J is positive and small, the distribution is slightly positively skewed.

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