Unit3
Summarization Measures
A measure of central tendency is a single value that attempts to describe a set of data by identifying the central position within that set of data. As such, measures of central tendency are sometimes called measures of central location.
Terms for statistics
 Population: Any welldefined set of objects about which a statistical inquiry is being made is the population. The information required from each individual member of the population is called its characteristic.
 Variate and attribute: if the characteristic that is being studied is numerical then it is called variate. The qualitative characteristic under study is called attribute
 Discrete variable: A variate which takes only integer values, like 1, 2, 3, …. 10,100, even 0, 2,5 etc. will come under a discrete variate
 Continuous variate: A variate which is capable of taking all the values is called continuous values.
 Data in statistics: when the characteristic under study of a population is collected, it is called data.
 Frequency distribution: if instead of 10 students, we have 100 students then frequency is better choice to handle this data, frequency will give number of times the relevant variate
 Cumulative Frequencies: for each value x of a variate, we can speak of the total number of observations less than or equal to x, within its range.
 Arithmetic mean (A.M):It is mean or the average value.
Average or measures of Central tendency
Professor Bowley defines the average as
“Statistical constants which enable us to comprehend in a single effort the significance of the whole”
An average is a single value that is the best representative for a given data set.
Measures of central tendency show the tendency of some central values around which data tend to cluster.
The following are the various measures of central tendency
1. Arithmetic mean,
2. Median,
3. Mode,
4. Weighted mean,
5. Geometric mean,
6. Harmonic mean.
The arithmetic mean or mean
The arithmetic mean is a value which is the sum of all observation divided by a total number of observations of the given data set.
If there are n numbers in a dataset then the arithmetic mean will be
If the numbers along with frequencies are given then mean can be defined as
This is known as direct method.
Example: Find the mean of 20, 22, 25, 28, 30.
Sol:
Example: Find the mean of 26, 15, 29, 36, 35, 30, 14, 21, 25 .
Sol.
Example: Find the mean of the following:
Numbers  8  10  15  20 
Frequency  5  8  8  4 
Sol:
fx = 8×5 + 10×8 + 15×8 + 20×4 = 40+80+120+80=320
f = 5+8+8+4=25
A.M.=
Example: Find the mean of the following dataset.
x  20  30  40 
f  5  6  4 
Sol.
We have the following table
X  F  Fx 
20  5  100 
30  6  180 
40  7  160 
 Sum = 15  Sum = 440 
Then Mean will be
(b) Short cut method
Let a be the assumed mean, d the derivation of the variate x from a. Then
Example: Find the arithmetic mean for the following distribution
Class  010  1020  2030  3040  4050 
Frequency  7  8  20  10  5 
Sol:
Let assumed mean (a) = 25
Class  Midvalue (x)  Frequency (f)  Fd  
010  5  7  20  140 
1020  15  8  10  80 
2030  25  20  0  0 
3040  35  10  + 10  +100 
4050  45  5  + 20  +100 
Total 
 50 
 20 
(C) Step deviation method
Let a be the assumed mean, i the width of the class interval and
Example:
Find the arithmetic mean of the data given in above example 3 by step deviation method.
Solution. Let a =25
Class  Midvalue x  Frequency F  f.D  
010  5  7  2  14 
1020  15  8  1  8 
2030  25  20  0  0 
3040  35  10  +1  +10 
4050  45  5  +2  +10 
Total 
 50 
 2 
Median
Median is defined as the measure of the central atom when they are arranged in ascending or descending order of magnitude.
Median is the midvalue of the given data when it is arranged in ascending or descending order.
1. If the total number of values in the data set is odd then the median is the value of item.
2. If the total number of values in the data set is even then the median is the mean of the item.
Note The data should be arranged in ascending or descending order
Example:
Find the median of 6, 8, 9, 10, 11, 12, 13.
Sol:
Total number of items =7
The middle item
Median= value of the 4th item = 10
Example: Find the median of the data given below
7, 8, 9, 3, 4, 10
Sol.
Arrange the data in ascending order
3, 4, 7, 8, 9, 10
So there total 6 (even) observations, then
=
Median for grouped data
Here,
Example 6. Find the value of median from the following data
Number of days for which absent (less than)  5  10  15  20  25  30  35  40  45 
Number of students  29  224  465  582  634  644  650  653  655 
Solution. The given cumulative frequency distribution will first be converted into ordinary frequency as under:
Class interval  Cumulative frequency  Ordinary frequency 
05  29  29=29 
510  224  22429=105 
1015  465  465224=241 
1520  582  582465=117 
2025  634  634582=52 
2530  644  644634=10 
3035  650  650644=6 
3540  653  653650=3 
4045  655  655653=2 
Median = size of
327.5th item lies in 1015 which is the median class
Where l stands for lower limit of median class.
N stands for the total frequency
C stands for cumulative frequency just preceding the median class
i stands for class interval
f stands for frequency for the median class
Example: Find the median of the following dataset
Sol.
Class interval  Frequency  Cumulative frequency 
0  10  3  3 
10 – 20  5  8 
20 – 30  7  15 
30 – 40  9  24 
40 – 50  4  28 
So that median class is 2030.
Now putting the values in the formula
So that the median is 28.57
Mode
Mode is defined to be the size of the variable which occurs most frequently.
Example:
Find the mode of the following items
0,1,6,7,2,3,7,6,6,2,6,0,5,6,0.
Sol:
6 occurs 5 times and no other item occurs 5 or more than 5 times, hence the mode is 6.
Mode for grouped data
Here,
Empirical formula
Mean – Mode =3 [Mean – Median]
Example: Find the mode from the following data
Age  06  612  1218  1824  2430  3036  3642 
Frequency  6  11  25  35  18  12  6 
Solution.
Age  Frequency  Cumulative frequency 
06  6  6 
612  11  17 
1218  42  
1824  35 = f  77 
2430  95  
3036  12  107 
3642  6  113 
Example: Find the mode of the following dataset
Sol.
Class interval  Frequency 
0  10  3 
10 – 20  5 
20 – 30  7 
30 – 40  9 
40 – 50  4 
Here the highest frequency is 9. So that the modal class is 4050,
Put the values in the given data
Hence the mode is 42.86
Example: A survey was conducted in a housing complex, to find out numbers of person in various age groups, who are using the ATM cards of banks. The result of the survey are as shown below.
Age group  Number of persons 
Below 20  10 
Below 40  25 
Below 60  50 
Below 80  70 
Find the mode of above distribution
Solution:
Class interval  Frequencies 
020  10 
2040  2510=15 
4060  5025=25 
6080  7050=20 
The maximum frequency 25 is for the class 4060
Class 4060 is modal class.
By using formula
We get
Mode = 53.3333
Example:
Calculate the value of modal worker family’s monthly income from the following data:  
Less than cumulative frequency distribution Of income per month (in ‘000 Rs)  
 

Sol:  
 This table should be converted into an ordinary frequency Table to determine the modal class.  

 

The value of the mode lies in 25–30 class interval. By inspection also, it can be seeming that this is a modal class. Now L =25, D1 (30–18) = 12, D2 = (3020) =10, h = 5 Using the formula, you can obtain the value of the mode as:  

 
 Thus, the modal worker family’s monthly income is Rs 27.273. 
Geometric mean
If be n values of variates x, then the geometric mean
Example: Calculate the harmonic mean of 4,8,16.
Sol:
Note
1.
2.
3. Mean – Mode = [Mean  Median]
Key takeaways
 A measure of central tendency is a single value that attempts to describe a set of data by identifying the central position within that set of data.
 An average is a single value that is the best representative for a given data set.
 2.
 3. Mean – Mode = [Mean  Median]
Partition values
The values divide the distribution into certain number of equal parts are called partition values.
Data should be arranged in ascending order descending order.
Quartile, deciles and percentile are the partition values.
Note
 Quartile divides the data into four equal parts.
 Deciles and percentiles divide the distribution into ten and hundred equal parts, respectively
Quartile
There are three quartiles, i.e. Q1, Q2 and Q3 which divide the total data into four equal parts when it has been orderly arranged. Q1, Q2 and Q3 are termed as first quartile, second quartile and third quartile or lower quartile, middle quartile and upper quartile, respectively. The first quartile, Q1, separates the first onefourth of the data from the upper three fourths and is equal to the 25th percentile. The second quartile, Q2, divides the data into two equal parts (like median) and is equal to the 50th percentile. The third quartile, Q3, separates the first threequarters of the data from the last quarter and is equal to 75th percentile.
For ungrouped data, we find the quartiles as follows
For grouped data, we find the quartiles as follows
i’th quartile can be find as
l = lower class limit of i'th quartile class,
h = width of the ith quartile class,
N = total frequency,
C = cumulative frequency of pre ith quartile class, and
f = frequencies of ith quartile class.
Deciles
Deciles divide whole distribution in to ten equal parts. There are nine deciles.
For ungrouped data, we find the deciles as follows
For grouped data, we find the deciles as follows
i’th decile can be find as
Percentile
Percentiles divide whole distribution in to 100 equal parts. There are ninety nine percentiles.
For ungrouped data, we find the percentile as follows
For grouped data, we find the percentile as follows
i’th percentile can be find as
Example: Calculate the first and third quartile of the following data
Class interval  f 
010  3 
1020  5 
2030  7 
3040  9 
4050  4 
Sol.
Class interval  f  CF 
010  3  3 
1020  5  8 
2030  7  15 
3040  9  24 
4050  4  28 
 N = 28 

Here N/4 = 28/4 = 7
The 7th observation falls in the class 1020. So, this is the first quartile class. 3N/4 = 21th observation falls in class 3040, so it is the third quartile class.
For first quartile l = 10, f = 5, C = 3, N = 28
We know that
For third quartile l = 30, f = 9, C = 15
Concept of Ogive
For drawing less than cumulative frequency curve (or less than ogive), first of all the cumulative frequencies are plotted against the values (upper limits of the class intervals) up to which they correspond and then we simply join the points by line segments, curve thus obtained is known as less than ogive. Similarly, more than frequency curve (more than ogive) can be obtained by plotting more than cumulative frequencies against lower limits of the class intervals. As we have already mentioned within brackets that less than cumulative frequency curve and more than cumulative frequency curve are also called less than ogive and more than ogive respectively.
We can define as follows
Less Than Ogive: If we plot the points with the upper limits of the classes as abscissae and the cumulative frequencies corresponding to the values less then the upper limits as ordinates and join the points so plotted by line segments, the curve thus obtained is nothing but known as “less than cumulative frequency curve” or “less than ogive”.
More Than Ogive: If we plot the points with the lower limits of the classes as
Abscissae and the cumulative frequencies corresponding to the values more than the lower limits as ordinates and join the points so plotted by line segments, the curve thus obtained is nothing but known as “more than cumulative frequency curve” or “more than ogive”.
Note
Median may also the obtained by drawing dotted vertical line through the point of inter section of both the ogives, when drawn on a single figure.
Example: By drawing ogive, find the median of the following data
Class  010  1020  2030  3040  4050  5060  6070  7080  8090 
Frequency  3  6  10  13  20  18  15  9  6 
Sol:
Key takeaways
 The values divide the distribution into certain number of equal parts are called partition values.
 Quartile divides the data into four equal parts.
The highest peak of the histogram represents the location of the mode of the data set. The mode is the data value that occurs the most often in a data set. For a symmetric histogram, the values of the mean, median, and mode are all the same and are all located at the centre of the distribution.
Draw a histogram for the following information.
Height (feet):  Frequency  Relative Frequency 
02  0  0 
24  1  1 
45  4  8 
56  8  16 
68  2  2 
When drawing a histogram, the yaxis is labelled 'relative frequency' or 'frequency density'. You must work out the relative frequency before you can draw a histogram. To do this, first you must choose a standard width of the groups. Some of the heights are grouped into 2s (02, 24, 68) and some into 1s (45, 56).
Most are 2s, so we shall call the standard width 2. To make the areas match, we must double the values for frequency which have a class division of 1 (since 1 is half of 2). Therefore, the figures in the 45 and the 56 columns must be doubled. If any of the class divisions were 4 (for example if there was a 812 group), these figures would be halved. This is because the area of this 'bar' will be twice the standard width of 2 unless we half the frequency.
Area of bar = frequency x standard width
The arithmetic mean denoted x, of a set of n numbers x1, x2, …, xn is defined as the sum of the numbers divided by n:
The arithmetic mean (usually synonymous with average) represents a point about which the numbers balance. For example, if unit masses are placed on a line at points with coordinates x1, x2, …, xn, then the arithmetic mean is the coordinate of the centre of gravity of the system.
Example: A student obtained the marks 40, 50, 60, 80, and 45 in math, statistics, physics, chemistry and biology respectively. Assuming weights 5, 2, 4, 3, and 1 respectively for the abovementioned subjects, find the weighted arithmetic mean per subject.
Solution:  Mark Obtained  Weight  Wxwx 
Math  4040  55  200200 
Statistics  5050  22  100100 
Physics  6060  44  240240 
Chemistry  8080  33  240240 
Biology  4545  11  4545 
Total 
 ∑w=15∑w=15  ∑wx=825∑wx=825 
Now we will find the weighted arithmetic mean as:
Xw=∑wx∑w=82515=55X
w=∑wx∑w=82515=55 marks/subject.
Example: Suppose that a marketing firm conducts a survey of 1,000 households to determine the average number of TVs each household owns. The data show a large number of households with two or three TVs and a smaller number with one or four. Every household in the sample has at least one TV and no household has more than four.
Solution:
Here’s the sample data for the survey:
Number of TVs per Household  Number of Households 
1  73 
2  378 
3  459 
4  90 
As many of the values in this data set are repeated multiple times, you can easily compute the sample mean as a weighted mean. Follow these steps to calculate the weighted arithmetic mean:
Step 1: Assign a weight to each value in the dataset:
x1=1,w1=73
x2=2,w2=378
x3=3,w3=459
x4=4,w4=90
Step 2: Compute the numerator of the weighted mean formula.
Multiply each sample by its weight and then add the products together:
==w1x1+w2x2+w3x3+w4x4
= (1)(73)+(2)(378)+(3)(459)+(4)(90)
=2566
Step 3: Now, compute the denominator of the weighted mean formula by adding the weights together.
==w1+w2+w3+w4
= 73 + 378 + 459 + 90
=1000
Step 4: Divide the numerator by the denominator
=/ wi
=2566/1000
=2.566
The mean number of TVs per household in this sample is 2.566.
As the name suggests, the measure of dispersion shows the scatterings of the data. It tells the variation of the data from one another and gives a clear idea about the distribution of the data. The measure of dispersion shows the homogeneity or the heterogeneity of the distribution of the observations.
According to Spiegel, the degree to which numerical data tend to spread about an average value is called the variation or dispersion of data.
Classification of Measures of Dispersion
There are two basic kinds of a measure of dispersion
 Absolute measures,
 Relative measures.
Following are the different types of measures of dispersion
According to Spiegel
“The degree to which numerical data tend to spread about an average value is called the variation or dispersion of data”
The different measures of dispersion are
1. Range,
2. Quartile deviation,
3. Mean deviation,
4. Standard deviation
5. Variance.
Significance of measures of dispersion
Measures of variation are pointed out as to how far an average is representative of the entire data. When variation is less, the average closely represents the individual values of the data and when variation is large; the average may not closely represent all the units and be quite unreliable.
Another purpose of measuring variation is to determine the nature and causes of variations in order to control the variation itself. Measurements of dispersion are helpful to control the causes of variation.
Many powerful statistical tools in statistics such as correlation analysis, the testing of hypothesis, the analysis of variance, techniques of quality control, etc. are based on different measures of dispersion.
Range is the simplest measure of dispersion. Range is the difference between the maximum value of the variable and the minimum value of the variable in the distribution.
Example: Find the range of the distribution 4, 22, 14, 12, 16, 8, 13, 17, 21, 6, 5, 26.
Sol.
For the given distribution, the maximum value is 26 and the minimum value is 4, so that the range of the distribution is –
Example Find the range of the data 8, 5, 6, 4, 7, 10, 12, 15, 25, 30
Sol. Here the maximum value is 30 and the minimum value is 4 so that the range is
30 – 4 = 26
Coefficient of range
The coefficient of range can be calculated as follows
Coefficient of Range =
Advantages of range
 It is very simple to calculate
 It has useful applications in areas like order statistics and statistical quality control.
Disadvantages of range
 It utilizes only the maximum and the minimum values of variable in the series and gives no importance to other observations
 It is affected by fluctuations of sampling
 If a single value lower than the minimum or higher than the maximum is added or if the maximum or minimum value is deleted range is seriously affected
The quartiles divide a data set into quarters. The first quartile, (Q1) is the middle number between the smallest number and the median of the data. The second quartile, (Q2) is the median of the data set. The third quartile, (Q3) is the middle number between the median and the largest number.
Quartile deviation or semiinterquartile deviation is
Relative measure of Q.D. Known as Coefficient of Q.D. And is defined as
Example: Find the quartile deviation of the following data.
Class interval  010  1020  2030  3040  4050 
Frequency  3  5  7  9  4 
Sol.
We have N/4 = 28/4 = 7 and 7th observation falls in the class 1020.
This is the first quartile class. Similarly, 3N/4 = 21 and 21st observation falls in the interval 3040. This is the third quartile class.
Class interval  f  CF 
010  3  3 
1020  5  8 
2030  7  15 
3040  9  24 
4050  4  28 
 N = 28 

By using the formula of quartile deviation, we will find 
Therefore
Q = ½ × (Q3 – Q1) = (36.67 – 18) / 2 = 9.335
Example: Find the quartile deviation of the following data
Class  05  510  1015  1520  2025  2530  3035  3540 
Frequency  6  8  12  24  36  32  24  8 
Sol.
We will construct the cumulative frequency table
Class interval  f  CF 
05  6  6 
510  8  14 
1015  12  26 
1520  24  50 
2025  36  86 
2530  32  118 
3035  24  142 
3540  8  150 
 N = 150 

We know that
So that
And
Therefore
Q = ½ × (Q3 – Q1) = (30.52 – 17.40) / 2 = 6.56
Key takeaways
 The measure of dispersion shows the homogeneity or the heterogeneity of the distribution of the observations.
 The degree to which numerical data tend to spread about an average value is called the variation or dispersion of data.
Mean deviation is the average of the sum of the absolute values of deviation from any arbitrary value viz. Mean, median, mode, etc.
The deviation of an observation xi from the assumed mean A is defined as (xi – A).
Therefore,
The mean deviation can be defined as
Mean deviation from mean is defined as
Mean deviation from median is defined as
For frequency distribution
Example: Find the mean deviation from mean of the following data
Sol.
x  F  Fx  x   fx 
1  3  3  3.24  9.72 
2  5  10  2.24  11.20 
3  8  24  1.24  9.92 
4  12  48  0.24  2.88 
5  10  50  0.76  7.60 
6  7  42  1.76  12.32 
7  5  35  2.76  13.80 
Total  50  212  12.24  67.44 
We know that
Example: The students of statistics got the marks as below
16, 24, 13, 18, 15, 10, 23
Find the mean deviation from mean.
Sol.
X  x17  x  
16  1  1 
24  7  7 
13  4  4 
18  1  1 
15  2  2 
10  7  7 
23  6  6 
Sum = 119 
 28 
Then
Hence
Example 11. Find the mean deviation of the following frequency distribution
Class  06  612  1218  1824  2430 
Frequency  8  10  12  9  5 
Solution. Let a = 15
Class  Midvalue x  Frequency f  d = xa  Fd  x14  fx14 
06  3  8  12  96  11  88 
612  9  10  6  60  5  50 
1218  15  12  0  0  1  12 
1824  21  9  +6  54  7  63 
2430  27  5  +12  60  13  65 
Total 
 44 
 42 
 278 
Then mean deviation from mean
Key takeaways
Variance
Variance is the average of the square of deviations of the values taken from mean. Taking a square of the deviation is a better technique to get rid of negative deviations.
Variance is given as
And for a frequency distribution, the formula is
Variance of the combined series
If σ1, σ2 are two standard deviations of two series of sizes n1 and n2 with means ȳ1 and ȳ2. The variance of the two series of sizes n1 + n2 is:
σ 2 = (1/ n1 + n2) ÷ [n1 (σ1 2 + d1 2) + n2 (σ2 2 + d2 2)]
Where, d1 = ȳ 1 −ȳ , d2 = ȳ 2 −ȳ , and ȳ = (n1 ȳ 1 + n2 ȳ 2) ÷ ( n1 + n2).
Coefficient of variation
Coefficient of variation can be calculated as
Note The lower value of C.V, the more constancy of data
Example If student A has a mean 50 with SD 10.Another student B has a mean of 30 with SD = 3.
Which one is the best performer?
Sol. We calculate C.V.
And
Here B has a lower C.V. So that student B is the best performer.
Example: Calculate coefficient variation for the following frequency distribution.
Wages in Rupees earned per day  010  1020  2030  3040  4050  5060 
No. Of Labourers  5  9  15  12  10  3 
Solution:
We already calculated
Now,
A.M
A.M
Coefficient of Variation
Example: Suppose batsman A has mean 50 with SD 10. Batsman B has mean 30 with SD 3. What do you infer about their performance?
Sol.
A has higher mean than B. This means A is a better run maker.
However, B has lower CV (3/30 = 0.1) than A (10/50 = 0.2) and is consequently more consistent.
Note
It is a relative measure of variability. If we are comparing the two data series, the data series having smaller CV will be more consistent.
Standard deviation
It is defined as the positive square root of the arithmetic mean of the square of the deviation of the given values from their arithmetic mean. It is denoted by the symbol .
Where is A.M of the distribution . We have more formulae to calculate the standard deviation.
..
In frequency distribution from, we put where H is generally taken as width of class interval
Shortcut formula to calculate standard deviation
The square of the standard deviation is called known as a variance.
Example: Find the Variance and Standard Deviation of the Following Numbers: 1, 3, 5, 5, 6, 7, 9, 10.
Sol.
The mean = 46/ 8 = 5.75
Step 1: (1 – 5.75), (3 – 5.75), (5 – 5.75), (5 – 5.75), (6 – 5.75), (7 – 5.75), (9 – 5.75), (10 – 5.75)
= 4.75, 2.75, 0.75, 0.75, 0.25, 1.25, 3.25, 4.25
Step 2: Squaring the above values we get, 22.563, 7.563, 0.563, 0.563, 0.063, 1.563, 10.563, 18.063
Step 3: 22.563 + 7.563 + 0.563 + 0.563 + 0.063 + 1.563 + 10.563 + 18.063
= 61.504
Step 4: n = 8, therefore variance (σ2) = 61.504/ 8 = 7.69
Now, Standard deviation (σ) = 2.77
Example: Suppose a series of 100 data points has mean 50 and variance 20. Another series of 200 data points has mean 80 and variance 40. What is the combined variance of the given series?
Sol.
The mean of the combined series
Therefore, d1= 50 – 70 = –20 and d2 = 80 – 70 =10
Variance of the combined series
Example: Find the standard deviation for the following numbers:
10, 27, 40, 60, 33, 30, 10
Sol.
First we prepare the following distribution table
X  
10  100 
27  729 
40  1600 
60  3600 
33  1089 
30  900 
10  100 
Sum = 210  8118 
Then
Mean = 210 / 7 = 30
And standard deviation
Example1: Compute the variance and standard deviation.
Class  Frequency 
010  3 
1020  5 
2030  7 
3040  9 
4050  4 
Sol.
Class  Midvalue (x)  Frequency (f)  
010  5  3  1470.924 
1020  15  5  737.250 
2030  25  7  32.1441 
3040  35  9  555.606 
4050  45  4  1275.504 
Sum 
 4071.428 
Then standard deviation,
Example2: Calculate the standard deviation of the following frequency distribution
Weight  60 – 62  63 – 65  66 – 68  69 – 71  72 – 74 
Item  5  18  42  27  8 
Sol.
Weight  Item (f)  X  d = x – 67  f.d  
60 – 62  5  61  6  30  180 
63 – 65  18  64  3  54  162 
66 – 68  42  67  0  0  0 
69 – 71  27  70  3  81  243 
72 – 74  8  73  6  48  288 
Total 
100 


45 
873 
Example: Calculate S.D for the following distribution.
Wages in rupees earned per day  010  1020  2030  3040  4050  5060 
No. Of Labourers  5  9  15  12  10  3 
Solution:
Wages earned C.I  Mid value  Frequency  
52  5  5  2  10  20 
153  15  9  1  9  9 
25  25  15  0  0  0 
35  35  12  1  12  12 
45  45  10  2  20  40 
55  55  3  3  9  27 
Total   
Using formula,
Key takeaways
 Range = Max. Value – Min. Value
 Coefficient of Range =
References
 Mathematics and Statistics for Business – R. S. Bhardwaj – Excel Books.
 Business Mathematics and Statistics – Subhanjali Chopra – Pearson publication.
 Fundamentals of Business Mathematics and Statistics – ICAI – ICAI.
 Business Mathematics and Statistics – Dr. J K Das, N Das – McGraw Hill Education.
 Mathematical and statistical techniques, Dr. Abhilasha S. Magar & Manohar B. Bhagirath
 IGNOU