Unit-3

Statistics

3.1 Measure 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-

Example-1: Find the mean of 26, 15, 29, 36, 35, 30, 14, 21, 25 .

Sol.

Example-2: Find the mean of the following dataset.

Sol.

We have the following table-

Then Mean will be-

Short cut method

Let a be the assumed mean, d the deviation of the variate x from a. Then

Example1. Find the arithmetic mean for the following distribution:

Solution. Let assumed mean (a) = 25

Class	Midvalue	Frequency
40— 50
Total

(c) Step deviation method

Let be the assumed mean, the width ofthe class interval and

Example 2. Find the arithmetic mean of the data given in example by step deviation method

Solution. Let

Class	Mid‐value	frequency	a
Total

Median-

Median is the mid-value 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.

Note-The data should be arranged in ascending r descending order

2. If the total number of values in the data set is even then the median is the mean of the item.

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: Find the median of the following dataset-

Sol.

So that median class is 20-30.

Now putting the values in the formula-

So that the median is 28.57

Example: Find the value of Median from the following data:

No. of days for which absent (less than)
No. of students

Solution. The given cumulative frequency distribution will first be converted into ordinary frequency as under

Class Interval	Cumulative frequency	Ordinary frequency
0-5 5-10 15-20 20-25 25-35 30-35 35-40 40-45	29 465 582 634 644 650 653 655	29=29 224-29=195 465-224= 582-465=117 634-582=52 644-634=10 650-644=6 653-650=3 655-653=2

Median size of or 327.Item

327. Item lies in 10‐15 which is the median class.

Where stands for the lower limit of the median class,

stands for the total frequency,

stands for the cumulative frequency just preceding the median class, stands for class interval

stands for frequency for the median class.

Median

Mode-

A value in the data which is most frequent is known as mode.

Example: Find the mode of the following data points-

Sol. Here 6 has the highest frequency so that the mode is 6.

Mode for grouped data-

Here,

Example: Find the mode of the following dataset-

Sol.

Here the highest frequency is 9. So that the modal class is 40-50,

Put the values in the given data-

Hence the mode is 42.86

Example: Find the mode from the following data:

Solution.

Age	Frequency	Cumulative frequency
0-6 6-12 12-18 24-30 30-36 36-42	6 11 25 35 12 6	6 17 42 77 95 107 113

Mode

GEOMETRIC MEAN

, , be values of variates , then the geometric mean

Example 7. Find the geometric mean of 4, 8, 16.

Solution..

HARMONIC MEAN

The harmonic mean of a series of values is defined as the reciprocal of the arithmetic mean of their reciprocals. Thus be the harmonic mean, then

Example 8: Calculate the harmonic mean of 4, 8, 16.

Solution:

Note-

1.

2.

3. Mean – Mode = [Mean - Median]

Key takeaways-

 

2.

3.2 Measures of dispersion and Standard Deviation

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

Range-

This is one of the simplest measures of dispersion. The difference between the maximum and minimum value of the dataset is known as the range.

Range = Max. value – Min. value

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 =

Quartile deviation-

Example- Find the quartile deviation of the following data-

Sol.

Here N/4 = 28/4 = 7 so that the 7’th observation falls in class 10 – 20.

And

3N/4 = 21, and 21’st observation falls in the interval 30 – 40 which is the third quartile.

The quartiles can be calculated as below-

And

Hence the quartile deviation is-

Mean deviation-

The mean deviation can be defined as-

Here A is assumed mean.

Example: Find the mean deviation from the mean of the following data-

Sol.

Then mean deviation from mean-

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-1: Compute the variance and standard deviation.

Sol.

Class	Mid-value (x)	Frequency (f)
0-10	5	3	1470.924
10-20	15	5	737.250
20-30	25	7	32.1441
30-40	35	9	555.606
40-50	45	4	1275.504
Sum			4071.428

Then standard deviation,

Example-2: Calculate the standard deviation of the following frequency distribution-

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.

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-

Coefficient of Range =

3.3 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.

Solution:

We already calculated

Now, 

A.M    

A.M    

Coefficient of Variation

Key takeaways-

3.4 Moments, Skewness, and Kurtosis

The rth moment of a variable x about the mean x is usually denoted by is given by

The rth moment of a variable x aboutany point a is defined by

The relation between moments about mean and moment about any point:

where and

In particular

Note. 1. The sum ofthe coefficients ofthe various terms on the right‐hand side is zero.

2. The dimension of each term on the right‐hand side is the same as that ofterms on the left.

MOMENT GENERATING FUNCTION

The moment generating function ofthe variate about is defined as the expected value of and is denoted by .

where , ‘ is the moment of order about

Hence coefficient of or

again )

Thus, the moment generating function about the point moment generating function about the origin.

Skewness-

The word skewness means lack of symmetry-

The examples of the symmetric curve, positively skewed, and negatively skewed curves are given as follows-

1. Symmetric curve-

2. Positively skewed-

3. Negatively skewed-

To measure the skewness, we use Karl Pearson’s coefficient of skewness.

Then the formula is as follows-

Note- the value of Karl Pearson’s coefficient of skewness lies between -1 to +1.

Kurtosis-

It is the measurement of the degree of peakedness of a distribution

Kurtosis is measured as-

Calculation of kurtosis-

The second and fourth central moments are used to measure kurtosis.

We use Karl Pearson’s formula to calculate kurtosis-

Now, three conditions arise-

1. If , then the curve is mesokurtic.

2. If , then the curve is platykurtic

3. if  , then the curve is said to be leptokurtic.

Example: If the coefficient of skewness is 0.64. The standard deviation is 13 and the mean is 59.2, then find the mode and median.

Sol.

We know that-

So that-

And we also know that-

Example: Calculate Karl Pearson’s coefficient of skewness of marks obtained by 150 students.

Sol. The mode is not well defined so that first we calculate mean and median-

Class	f	x	CF		fd
0-10	10	5	10	-3	-30	90
10-20	40	15	50	-2	-80	160
20-30	20	25	70	-1	-20	20
30-40	0	35	70	0	0	0
40-50	10	45	80	1	10	10
50-60	40	55	120	2	80	160
60-70	16	65	136	3	48	144
70-80	14	75	150	4	56	244

Now,

And

Standard deviation-

Then-

Key takeaways-

If

, then the curve is mesokurtic.

 If

, then the curve is platykurtic

 if 

, then the curve is said to be leptokurtic.

3.5 Curve fitting of a straight line

Method of Least Squares:

Let     … (1)  

be the straight line to be fitted to the given data points .

Let be the theoretical value for .

Then 

For S to be minimum

or

 [To generalize , is written as y]

or

On Simplification equation (2) and (3) becomes

The equations (3) and (4) are known as Normal equations.

On solving equations (3) and (4), we get the values of a and b.

(b) To fit the parabola:

The normal equations are

On solving three equations, we get the values of a, b and c.

Note:

1. The normal equation (4) has been obtained by puttingon both sides of                                 

equation (1). Equation (5) is obtained by multiplying on both sides of (1).

2. The normal equation (7), (8), (9) are obtained by multiply by and             on both sides of equation (6).

Example: Find the best values of a and b so that fit the data given in the table.

Solution: 

0	1	0	0
1	2.9	2.9	1
2	4.8	9.6	4
3	6.7	20.1	9
4	8.6	13.4	16

Normal equations     …. (2)

  …. (3)

On putting the values of in (2) and (3), we have

    …. (4)

   …. (5)

On solving (4) and (5), we get

On Substituting the values of a and b in (1), we get

Example: By the method of least squats, find the straight line that best fits the following data:

	1	2	3	4	5
	14	27	40	55	68

Solution:  Let the equation of the straight line best fit be   …. (1)

1	14	14	1
2	27	54	4
3	40	120	9
4	55	220	16
5	68	340	25

Normal equations are

Putting the values of in (2) and (3), we have

On solving (4) and (5), we get

On Substituting the values of a and b in (1), we get

Example: Find the least-squares approximation of the second degree for the discrete data.             

	-2	-1	0	1	2
	15	1	1	3	19

Solution:

Let the equation of second-degree polynomial be

-2	15	-30	4	60	-8	16
-1	1	-1	1	1	-1	1
0	1	0	0	0	0	0
1	3	3	1	3	1	1
2	19	38	4	76	8	16

Normal equations are     

On putting the values of in equations (2), (3), (4), we have

On solving (5), (6), (7), we get

The required polynomial of the second degree is

3.6 Fitting Parabola and related curves

Change of Scale: If the data is of equal interval in large numbers then we change the scale as .

Example:Fit a second-degree parabola to the following data by the least-squares method.

	1929	1930	1931	1932	1933	1934	1935	1936	1937
	352	356	357	358	360	361	361	360	359

Solution: Taking

Taking 

The equation is transformed into

1929	-4	352	-5	20	16	-80	-64	256
1930	-3	360	-1	3	9	-9	-27	81
1931	-2	357	0	0	4	0	-8	16
1932	-1	358	1	-1	1	1	-1	1
1933	0	360	3	0	0	0	0	0
1934	1	361	4	4	1	4	1	1
1935	2	361	4	8	4	16	8	16
1936	3	360	3	9	9	27	27	81
1937	4	359	2	8	16	32	64	256
Total

Normal equations are

On solving these equations, we get

Example: Fit a second-degree parabola to the following data:

Solution: Let and so that the parabola of fit becomes

  …. (i)

The normal equations are

Saving these as simultaneous equations we get

(i) becomes

Or 

Hence

Example: Fit a second degree parabola to the following data:

	1.5	2	2.5	3	3.5	4
	1.3	1.6	2	2.7	3.4	4.1

Solution: We shift the origin to (2.5, 0) and take 0.5 as the new unit. This amounts to changing the variable to X, by the relation

Let the parabola of fit be . The values of etc., ae calculated below:

1.0	-3	1.1	-3.3	9	9.9	-27	81
1.5	-2	1.3	-2.6	4	5.2	-8	16
2.0	-1	1.6	-1.6	1	1.6	-1	1
2.5	0	2	0	0	0	0	0
3.0	1	2.7	2.7	1	2.7	1	1
3.5	2	3.4	6.8	4	13.6	8	16
4.0	3	4.1	12.3	9	36.9	27	81
Total	0	16.2	14.2	28	69.9	0	196

The normal equations are

Solving these as simultaneous equations, we get

Replacing X by in the above equation, we get

Which simplifies by . This is the required parabola of best fit.

3.7 Correlation and regression

When two variables are related in such a way that a change in the value of one variable affects the value of the other variable, then these two variables are said to be correlated and there is a correlation between two variables.

Example- Height and weight of the persons of a group.

The correlation is said to be a perfect correlation if two variables vary in such a way that their ratio is constant always.

Scatter diagram-

Karl Pearson’s coefficient of correlation-

Here- and

Note-

1. Correlation coefficient always lies between -1 and +1.

2. Correlation coefficient is independent of the change of origin and scale.

3. If the two variables are independent then the correlation coefficient between them is zero.

Example: Find the correlation coefficient between age and weight of the following data-

Sol.

X	y					( ) )
30	56	-10	100	-4	16	40
44	55	4	16	-5	25	-20
45	60	5	25	0	0	0
43	64	3	9	4	16	12
34	62	-6	36	2	4	-12
44	63	4	16	3	9	12
Sum= 240	360	0	202	0	70	32

Karl Pearson’s coefficient of correlation-

Here the correlation coefficient is 0.27.which is the positive correlation (weak positive correlation), this indicates that as age increases, the weight also increases.

Short-cut method to calculate correlation coefficient-

Here,

Example: Find the correlation coefficient between the values X and Y of the dataset given below by using the short-cut method-

Sol.

X	Y
10	90	-20	400	20	400	-400
20	85	-10	100	15	225	-150
30	80	0	0	10	100	0
40	60	10	100	-10	100	-100
50	45	20	400	-25	625	-500
Sum = 150	360	0	1000	10	1450	-1150

Short-cut method to calculate correlation coefficient-

Spearman’s rank correlation-

When the ranks are given instead of the scores, then we use Spearman’s rank correlation to find out the correlation between the variables.

Spearman’s rank correlation coefficient can be defined as-

Example: Compute the Spearman’s rank correlation coefficient of the dataset given below-

Sol.

Person	Rank in test-1	Rank in test-2	d =
A	9	1	8	64
B	10	2	8	64
C	6	3	3	9
D	5	4	1	1
E	7	5	2	4
F	2	6	-4	16
G	4	7	-3	9
H	8	8	0	0
I	1	9	-8	64
J	3	10	-7	49
Sum				280

Regression-

Regression is the measure of the average relationship between the independent and dependent variable

Regression can be used for two or more than two variables.

There are two types of variables in regression analysis.

1. Independent variable

2. Dependent variable

The variable which is used for prediction is called the independent variable.

It is known as a predictor or regressor.

The variable whose value is predicted by an independent variable is called the dependent variable or regressed or explained variable.

The scatter diagram shows the relationship between the independent and dependent variable, then the scatter diagram will be more or less concentrated around a curve, which is called the curve of regression.

When we find the curve as a straight line then it is known as the line of regression and the regression is called linear regression.

Note- regression line is the best fit line that expresses the average relation between variables.

Equation of the line of regression-

Let

y = a + bx …………..  (1)

is the equation of the line of y on x.

Let be the estimated value of for the given value of .

So that, According to the principle of least squares, we have the determined ‘a’ and ‘b’ so that the sum of squares of deviations of observed values of y from expected values of y,

That means-

Or

   …….. (2)

Is the minimum.

Form the concept of maxima and minima, we partially differentiate U with respect to ‘a’ and ‘b’ and equate to zero.

Which means

And

These equations (3) and (4) are known as the normal equation for a straight line.

Now divide equation (3) by n, we get-

This indicates that the regression line of y on x passes through the point
.

We know that-

The variance of variable x can be expressed as-

Dividing equation (4) by n, we get-

From equation (6), (7), and (8)-

Multiply (5) by, we get-

Subtracting equation (10) from equation (9), we get-

Since ‘b’ is the slope of the line of regression y on x and the line of regression passes through the point (), so that the equation of the line of regression of y on x is-

This is known as the regression line of y on x.

Note-

are the coefficients of regression.

2.

Example: Two variables X and Y are given in the dataset below, find the two lines of regression.

Sol.

The two lines of regression can be expressed as-

And

x	y			xy
65	66	4225	4356	4290
66	68	4356	4624	4488
67	65	4489	4225	4355
67	69	4489	4761	4623
68	74	4624	5476	5032
69	73	4761	5329	5037
70	72	4900	5184	5040
71	70	5041	4900	4970
Sum = 543	557	36885	38855	37835

Now-

And

The standard deviation of x-

Similarly-

Correlation coefficient-

Put these values in the regression line equation, we get

Regression line y on x-

Regression line x on y-

A regression line can also be found by the following method-

Example: Find the regression line of y on x for the given dataset.

Sol.

Let y = a + bx is the line of regression of y on x, where ‘a’ and ‘b’ are given as-

We will make the following table-

x	y	Xy
4.3	12.6	54.18	18.49
4.5	12.1	54.45	20.25
5.9	11.6	68.44	34.81
5.6	11.8	66.08	31.36
6.1	11.4	69.54	37.21
5.2	11.8	61.36	27.04
3.8	13.2	50.16	14.44
2.1	14.1	29.61	4.41
Sum = 37.5	98.6	453.82	188.01

Using the above equations we get-

On solving these both equations, we get-

a = 15.49 and b = -0.675

So that the regression line is –

y = 15.49 – 0.675x 

Note – Standard error of predictions can be found by the formula given below-

Difference between regression and correlation-

1. Correlation is the linear relationship between two variables while regression is the average relationship between two or more variables.

2. There are only limited applications of correlation as it gives the strength of linear relationship while the regression is to predict the value of the dependent variable for the given values of independent variables.

3. Correlation does not consider dependent and independent variables while regression considers one dependent variable and other independent variables.

Key takeaways-

2.    

3.    

4.    

3.8 Reliability of regression estimates

Besides, to study the reliability of regression estimates we require to know the standard error.

The standard error of regression estimate of y on is

The Standard error of Regression estimate of on is

1. Discuss the Reliability of Regression Estimates:

Solution:

For A,

	45	38	59	64	72
	2025	1444	3481	4096	5184

For B,

	60	48	82	93	45
	2025	1444	3481	4096	5184

Now,

	45	38	59	64	72
	60	48	82	93	45
	2700	1824	4838	5952	3240

The standard error of Regression of estimates of y on x is

…..(Standard error of Regression of estimates of y on x is )

Key takeaways-

2.

References

1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.

2. N.P. Bali and Manish Goyal, A textbook of Engineering Mathematics, Laxmi Publications.

3. Higher engineering mathematic, Dr. B.S. Grewal, Khanna publishers