Unit – 3

Statistics

3.1 Basic Statistics

STATISTICS is a branch of science dealing with the collection of data, organizing, summarizing, presenting and analyzing data and drawing valid conclusions and thereafter making reasonable decisions on the basis of such analysis.

Collection of data

The collection of data constitutes the starting point of any statistical investigation. Data may be collected for each and every unit of the whole lot (population), for it would ensure greater accuracy. But complete enumeration is prohibitively expensive and time consuming. Ash search out of a very large number of items a few of them (sample) are selected and conclusions drawn on the basis of this sample are taken to hold for the population.

Classification of data

The data collected in the course of an enquiry is not in an easily assimilable form. As such its proper classification is necessary for making intelligent interferences. The classification is done by dividing the raw data into a convenient number of groups according to the values of the variable and finding the frequency of the variable in each group.

Let us, for example, consider the raw data relating to marks obtained in Mechanics by a group of 64 students

This data can conveniently be grouped and shown in a tabular form as follows

It would be seen from the above table that there is one student getting marks between 50 -54, to students getting marks between 55-59, 9 students getting marks between 60-64 and so on. Thus the 64 figure have been put into only ten groups, called as classes. The width of the class is called the class interval and the number in that interval is called the frequency. The midpoint of the mid value of the class is called the class mark. The above table showing the classes and the corresponding frequencies is called a frequency table. Thus a set of row data are summarised by distributing it into a number of classes along with their frequencies is known as a frequency distribution. while forming the frequency distribution the number of classes should not ordinarily exceed 20 and should not in general be less than 10. As far as possible the class intervals should be of equal width.

Cumulative frequency

in some investigations we require the number of items less than a certain value. We add up The frequencies of the classes up to that value and call this number as the cumulative frequency. In the above table the third column shows the cumulative frequency, i.e. the number of students getting less than 54 marks, less than 59 marks and so on.

Graphical representation

A convenient way of representing a sample frequency distribution is by means of graphs. It gives to the eyes the general run of the observations and at the same time makes the raw data readily intelligible. We give below the important types of graphs in use:

1)     Histogram. A histogram is drawn by erecting rectangles over the class intervals such that the areas of the rectangles are proportional to the class frequencies. If the class intervals are of equal size the height of the rectangles will be proportional to the class frequencies themselves. (Figure 25.1)

2)     Frequency polygon. A frequency polygon for an ungrouped data can be obtained by joining points plotted with the variable values as the abscissa And The frequencies as the ordinates. For a grouped distribution the abscissa of the points will be the mid values of the class intervals. In case of intervals are equal the frequency polygon can be obtained by joining the middle points of the upper sides of the rectangles of the histogram by straight lines (shown by dotted lines in figure 25.1). if the class intervals become very very small the frequency polygon takes the form of a smooth curve called the frequency curve.

3)     Cumulative frequency curve Ogive. Very often it is desired to show in a diagrammatic form not the relative frequencies in the various intervals but the cumulative frequencies above or below a given value. For example we may wish to read of from a diagram the number of proportions of people whose income is not less than any given amount or proportion of people whose height does not exceed any stated value. Diagrams of this type are known as cumulative frequency curves or ogives. these are of two kinds more than or less than and typical day look somewhat like a long-drawn S (figure 25.2).

Example. Draw the histogram frequency polygon frequency curve and the ogives 'less than' and 'more than' from the following distribution of marks obtained by 49 students.

Solution. In figure 25.1 the rectangles show the histogram, the dotted polygon represent the frequency polygon and the smooth curve in the frequency curve.

The ogives 'less than' and 'more than' are shown in figure 25.2.

3.2 Measures of Central tendency: Moments, skewness and Kurtosis

Average or measures of Central tendency

An average is a value which is representative of a set of data. Average value may also be termed as measures of Central tendency. There are five types of averages in common.

(i)                Arithmetic average or mean

(ii)              Median

(iii)           Mode

(iv)            Geometric mean

(v)              Harmonic mean

Arithmetic mean

If are n numbers then their arithmetic mean (A.M) is defined by

If the number occurs times X and so on then

This is known as direct method.

Example 1. Find the mean of 20, 22, 25, 28, 30.

Solution.

Example 2. Find the mean of the following:

Solution. fx = 8×5 + 10×8 + 15×8 + 20×4 = 40+80+120+80=320

 f = 5+8+8+4=25

A.M.=

(b) Short cut method

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

Example 3. Find the arithmetic mean for the following distribution

Solution. Let assumed mean (a) = 25

(C) Step diffusion method

Let a be the assumed mean, i the width of the class interval and

Example 4. Find the arithmetic mean of the data given in example 3 by step deviation method.

Solution. Let a =25

Median

Median is defined as the measure of the central atom when they are arranged in ascending or descending order of magnitude .

When the total number of the items is odd and equal to say n of item gives the median.

When the total number of The frequencies is even, say n, then there are two middle items and so the mean of the values of th items is the median.

Example 5. Find the median of 6, 8, 9, 10, 11, 12, 13.

Solution. Total number of items =7

The middle item

Median= value of the 4th item = 10

For grouped data, median

Where l is the lower limit of the median class, f is the frequency of the class, i is the width of the class interval, F is the total of all the the preceding frequencies of the median class and N is total frequency of the data.

Example 6. Find the value of median from the following data

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

Median = size of

327.5th item lies in 10-15 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

Mode

Mode is defined to be the size of the variable which occurs most frequently.

Example 7. Find the mode of the following items

0,1,6,7,2,3,7,6,6,2,6,0,5,6,0.

Solution. 6 occurs 5 times and no other item occurs 5 or more than 5 times, hence the mode is 6.

For grouped data,

Where l is the lower limit of the modal class, f is the frequency of the modal class, i is the width of the class, is the frequency before the model class and frequency of the modal class.

Empirical formula

Mean – Mode =3 [Mean – Median]

Example 8. Find the mode from the following data

Solution.

Geometric mean

If be n values of variates x, then the geometric mean

Example 10. Calculate the harmonic mean of 4,8,16.

Solution.

Average deviation on mean deviation

It is the mean of the absolute values of the definitions of given set of numbers from their arithmetic mean.

If be a set of numbers with frequencies respectively. Let x be the arithmetic mean of the numbers

Mean deviation =

Example 11. Find the mean deviation of the following frequency distribution

Solution. Let a = 15

Average deviation=

MOMENTS

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 about any point a is defined by

Relation between moments about mean and moment about any point:

where and

In particular

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

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

MOMENT GENERATING FUNCTION

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

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:

Skewness denotes the opposite of symmetry. It is lack of symmetry. In a symmetrical series, the mode, the median, and the arithmetic average are identical.

Coefficient of skewness

KURTOSIS: It measures the degree of peakedness of a distribution and is given by Measure of kurtosis.

Negative skewness               Positive skewness              A: Mesokurtic B: Leptokurtic

                                                                                                    C: Playkurtic

If , the curve is normal or mesokurtic.

If , the curve is peaked or leptokurtic.

If , the curve is flat topped or platykurtic

Example. The first four moments about the working mean 28.5 of distribution are 0.2 94, 7.1 44, 42.409 and 454.98. Calculate the moments about the mean. Also evaluate and comment upon the skewness and kurtosis of the distribution.

Solution. The first four moments about the arbitrary origin 28.5 are

, which indicates considerable skewness of the distribution.

, which shows that the distribution is leptokurtic.

Example. Calculate the median, quartiles and the quartile coefficient of skewness from the following data:

Solution. Here total frequency

The cumulative frequency table is

Now, N/2 =230/2= 115th item which lies in 110 – 120 group.

Median or

Also, is 57.5th or 58th item which lies in 90-100 group.

Similarly 3N/4 = 172.5 i.e. is 173rd item which lies in 120-130 group.

Hence quartile coefficient of skewness =

3.3 Probability distributions: Binomial, Poisson and Normal & Evaluation of statistical parameters for these three         distributions

A probability distribution is an arithmetical function which defines completely possible values &possibilities that a random variable can take in a given range. This range will be bounded between the minimum and maximum possible values. But exactly where the possible value is possible to be plotted on the probability distribution depends on a number of influences. These factors include the distribution's mean, SD, Skewness, and kurtosis.

Binomial Distribution:

BINOMIAL DISTRIBUTION

To find the probability of the happening of an event once, twice, thrice,…r times ….exactly in n trails.

Let the probability of the happening of an event A in one trial be p and its probability of not happening be 1 – p – q.

We assume that there are n trials and the happening of the event A is r times and its not happening is n – r times.

This may be shown as follows

AA……A                                 

r times                                       n – r times                          (1)

A indicates its happening its failure and P (A) =p and P (

We see that (1) has the probability

pp…p qq….q=

r times              n-r times                                                       (2)

Clearly (1) is merely one order of arranging r A’S.

The probability of (1) =Number of different arrangements of r A’s and (n-r)’s

The number of different arrangements of r A’s and (n-r)’s

Probability of the happening of an event r times =

If r = 0, probability of happening of an event 0 times

If r = 1, probability of happening of an event 1 times

If r = 2, probability of happening of an event 2 times

If r = 3, probability of happening of an event 3 times and so on.

These terms are clearly the successive terms in the expansion of

Hence it is called Binomial Distribution.

Example. If on an average one ship in every ten is wrecked. Find the probability that out of 5 ships expected to arrive, 4 at least we will arrive safely.

Solution. Out of 10 ships one ship is wrecked.

I.e. nine ships out of 10 ships are safe, P (safety) =

P (at least 4 ships out of 5 are safe) = P (4 or 5) = P (4) + P(5)

Example. The overall percentage of failures in a certain examination is 20. if 6 candidates appear in the examination what is the probability that at least five pass the examination?

Solution. Probability of failures = 20%

Probability of (P) =

Probability of at least 5 pass = P(5 or 6)

Example. The probability that a man aged 60 will live to be 70 is 0.65. what is the probability that out of 10 men, now 60, at least seven will live to be 70?

Solution. The probability that a man aged 60 will live to be 70

Number of men= n = 10

Probability that at least 7 men will live to 70 = (7 or 8 or 9 or 10)

 = P (7)+ P(8)+ P(9) + P(10) =

Example. assuming that 20% of the population of a city are literate so that the chance of an individual being literate is and assuming that hundred investigators each take 10 individuals to see whether they are illiterate, how many investigators would you expect to report 3 or less were literate.

Solution.

Required number of investigators = 0.879126118× 100 =87.9126118

 = 88 approximate

Mean or binomial distribution

Since,

STANDARD DEVIATION OF BINOMIAL DISTRIBUTION

We know that           (1)

r is the deviation of items (successes) from 0.

Putting these values in (1) we have

Hence for the binomial distribution, Mean

Example. A die is tossed thrice. A success is getting 1 or 6 on a TOSS. Find the mean and variance of the number of successes.

Solution.

RECURRENCE RELATION FOR THE BINOMIAL DISTRIBUTION

By Binomial Distribution

On dividing (2) by (1) , we get

Poisson Distribution:

Poisson distribution is a particular limiting form of the Binomial distribution when p (or q) is very small and n is large enough.

Poisson distribution is

where m is the mean of the distribution.

Proof. In Binomial Distribution

Taking limits when n tends to infinity

MEAN OF POISSON DISTRIBUTION

STANDARD DEVIATION OF POISSON DISTRIBUTION

Hence mean and variance of a Poisson distribution are equal to m. Similarly we can obtain,

MEAN DEVIATION

Show that in a Poisson distribution with unit mean, and the mean deviation about the mean is 2/e times the standard deviation.

Solution. But mean = 1 i.e. m =1 and S.D. =

Mean Deviation =

MOMENT GENERATING FUNCTION OF POISSON DISTRIBUTION

Solution.

Let be the moment generating function then

CUMULANTS

The cumulant generating function is given by

Now cumulant =coefficient of in K (t) = m

i.e. , where r = 1,2,3,…

Mean =

RECURRENCE FORMULA FOR POISSON DISTRIBUTION

SOLUTION. By Poisson distribution

On dividing (2) by (1) we get

Example. Assume that the probability of an individual coal miner being killed in a mine accident during a year is . Use appropriate statistical distribution to calculate the probability that in a mine employing 200 miners, there will be at least one fatal accident in a year.

Solution.

Example. Suppose 3% of bolts made by a machine are defective, the defects occuring at random during production. If bolts are packaged 50 per box, find

(a)  Exact probability and

(b) Poisson approximation to it, that a given box will contain 5 defectives.

Solution.

(a)  Hence the probability for 5 defectives bolts in a lot of 50.

(b) To get Poisson approximation m = np =

Required Poisson approximation=

Example. In a certain factory producing cycle tyres, there is a smallchance of 1 in 500 tyres to be defective. The tyres are supplied in lots of 10. Using Poisson distribution, calculate the approximate number of lots containing no defective, one defective and two defective tyres, respectively, in a consignment of 10,000 lots.

Solution.

Normal Distribution

Normal distribution is a continuous distribution. It is derived as the limiting form of the Binomial distribution for large values of n and p and q are not very small.

The normal distribution is given by the equation

    (1)

Where = mean, = standard deviation, =3.14159…e=2.71828…

On substitution in (1) we get      (2)

Here mean = 0, standard deviation = 1

(2) is known as standard form of normal distribution.

MEAN FOR NORMAL DISTRIBUTION

Mean         [Putting

STANDARD DEVIATION FOR NORMAL DISTRIBUTION

Put,

MEDIAN OF THE NORMAL DISTRIBUTION

If a is the median then it divides the total area into two equal halves so that

Where,

Suppose mean, then

Thus,

Similarly, when mean, we have a =

Thus, median = mean =

MEA DEVIATION ABOUT THE MEAN

Mean deviation

MODE OF THE NORMAL DISTRIBUTION

We know that mode is the value of the variate x for which f (x) is maximum. Thus by differential calculus f (x) is maximum if and

Where,

Thus mode is and model ordinate =

NORMAL CURVE

Let us show binomial distribution graphically. The probabilities of heads in 1 tosses are

. it is shown in the given figure.

If the variates (head here) are treated as if they were continuous, the required probability curve will be a normal curve as shown in the above figure by dotted lines.

Properties of the normal curve

The curve is drawn, if mean (origin of x) and standard deviation are given. The value of

can be calculated from the fact that the area of the curve must be equal to the total number of observations.

(a)

AREA UNDER THE NORMAL CURVE

By taking the standard normal curve is formed.

The total area under this curve is 1. The area under the curve is divided into two equal parts by z = 0. Left hand side area and right hand side area to z = 0 is 0.5. The area between the ordinate z = 0.

Example. On a final examination in mathematics, the mean was 72, and the standard deviation was 15. Determine the standard scores of students receiving graders.

(a)  60

(b) 93

(c)  72

Solution. (a)

(b)

(c)

Example. Find the area under the normal curve in each of the cases

(a)  Z = 0 and z = 1.2

(b) Z = -0.68 and z = 0

(c)  Z = -0.46 and z = -2.21

(d) Z = 0.81 and z = 1.94

(e) To the left of z = -0.6

(f)    Right of z = -1.28

Solution. (a) Area between Z = 0 and z = 1.2 =0.3849

(b)Area between z = 0 and z = -0.68 = 0.2518

(c)Required area = (Area between z = 0 and z = 2.21) + (Area between z = 0 and z =-0.46)\

= (Area between z = 0 and z = 2.21)+ (Area between z = 0 and z = 0.46)

=0.4865 + 0.1772 = 0.6637

(d)Required area = (Area between z = 0 and z = 1.+-(Area between z = 0 and z = 0.81)

 = 0.4738-0.2910=0.1828

(e) Required area = 0.5-(Area between z = 0 and z = 0.6)

                                 = 0.5-0.2257=0.2743

(f)Required area = (Area between z = 0 and z = -1.28)+0.5

                               = 0.3997+0.5

                               = 0.8997

Example. The mean inside diameter of a sample of 200 washers produced by a machine is 0.0502 cm and the standard deviation is 0.005 cm. The purpose for which these washers are intended allows a maximum tolerance in the diameter of 0.496 to 0.508 cm, otherwise the washers are considered defective. Determine the percentage of defective washers produced by the machine assuming the diameters are normally distributed.

Solution.

Area for non – defective washers = Area between z = -1.2

And z = +1.2

=2 Area between z = 0 and z = 1.2

=2 (0.3849)-0.7698=76.98%

Percentage of defective washers = 100-76.98=23.02%

Example. A manufacturer knows from experience that the resistance of resistors he produces is normal with mean and standard deviation . What percentage of resistors will have resistance between 98 ohms and 102 ohms?

Solution. ,

Area between

= (Area between z = 0 and z = +1)

= 2 (Area between z = 0 and z = +1)=2 0.3413 = 0.6826

Percentage of resistors having resistance between 98 ohms and 102 ohms = 68.26

Example. In a normal distribution, 31% of the items are 45 and 8% are over 64. Find the mean and standard deviation of the distribution.

Solution. Let be the mean and the S.D.

If x = 45,

If x = 64,

Area between 0 and

[From the table, for the area 0.19, z = 0.496)

Area between z = 0 and z =

(from the table for area 0.42, z = 1.405)

Solving (1) and (2) we get

3.4 Correlation and regression – Rank correlation

Correlation

So far we have confined our attention to the analysis of observations on a single variable. There are however, many phenomenon where the changes in one variable are related to the changes in the other variable. For instance the yield of a crop varies with the amount of rainfall, the price of a commodity increases with the reduction in its supply and so on. Such a data connecting two variables is called bivariate population.

To obtain a measure of relationship between the two variables, we plot their corresponding values on the graph taking one of the variable along the x axis and the other along the y axis. (Figure 25.6).

Let the origin be shifted to , where re the means of X’s and y's that the new coordinates are given by

Now the points (X,Y) are so distributed over the four quadrants of XY plane that the product XY is positive in the first and third quadrant but negative in the second and fourth quadrants. the algebraic sum of the products can be taken as describing the trend of the dots in all the quadrants.

(i)                If XY is positive, the trend of the dots is through the first and third quadrants.

(ii)              If XY is negative the trend of two dots is in the second and fourth quadrants and

(iii)           If XY is zero, the points indicate no trend i.e. the points are evenly distributed over the quadrants.

The XY or better still XY i.e. the average of n products may be taken as a measure of correlation. If we put X and Y in their units, i.e. taking , as the unit for x and for y, then

Is the measure of correlation.

Coefficient of correlation

The numerical measure of correlation is called the coefficient of correlation and is defined by the relation

Where, X = deviation from the mean = = devaluation from the mean

= Standard deviation of x series, = standard deviation of y series and n = number of the values of the two variables

Methods of calculation

(a)  Direct method. Substituting the value of in the above formula we get

Another form of the formula (1) which is quite handy for calculation is

(b) Step deviation method. The direct method becomes very lengthy and tedious if the means of the two series are not integers. In such cases, use is made of assumed means. If are step deviations from the assumed means, then

(c)  Coefficient of correlation for grouped data. When x and y series are both given as frequency distributions these can be represented by a two way table known as the correlation table. the coefficient of correlation for such a bivariate frequency distribution is calculated by the formula

Where = derivation of the central values from the assumed mean of x series

derivation of the central values from the assumed mean of y series

is the frequency corresponding to the pair (x, y)

is the total number of frequency.

Example. Psychological test of the intelligence and of Engineering ability were applied to 10 students. Here is a record of ungrouped data showing intelligence ratio (I.R) and Engineering ratio (E.R). Calculate the coefficient of correlation.

Solution. We construct the following table

From this table, mean of x, i.e. and mean of y, i.e.

Substituting these value in the formula (1)p.744 we have

Example. The correlation table given below shows that the ages of husband and wife of 53 married couples living together on the census night of 1991. Calculate the coefficient of correlation between the age of the husband and that of the wife.

Solution.

With the help of the above correlation table, we have

Lines of Regression

It frequently happens that the dots of the scatter diagram generally tends to cluster along a well- defined direction which suggests a linear relationship between the variables x and y. Such a line of best fit for the given distribution of dots is called the line of regression (figure 25.6). in fact there are two such lines, one giving the best possible mean values of y for each specified value pf x and the other giving the best possible mean values of x for given value of y. the former is known as the line of regression of y on x and the latter as the line of regression of x on y.

Consider first the line of regression of y on x. Let the straight line satisfying the general trend of n dots in a scatter diagram be

             (1)

We have to determine the constant a and b so that (1) gives for each value of x, the best estimate for the average value of y in accordance with the principle of least squares therefore, the normal equation for a and b are

  i.e.

This shows that i.e. the mean of x and y lie on (1).

Shifting the origin to (3) takes the form of

Cor. The correlation coefficient r is the geometric mean between the two regression coefficients

For

Example. The two regression equations of the variable x and y are x = 19.13 and y = 11.64 – 0.50 x. Find (i) mean of x’s (ii) mean of y’s and (iii) the correlation coefficient between x and y.

Solution. Since the mean of x’s and the mean of y’s lie on the two regression lines, we have

Multiplying (ii) by 0.87 and subtracting from (i) we have

Regression coefficient of y and x is -0.50 and that of x and y is -0.87.

Now since the coefficient of correlation is the geometric mean between the two regression coefficients.

[-ve sign is taken since both the regression coefficients are –ve]

Example. If is the angle between the two regression lines show that

Explain the significance when .

Solution. The equations to the line of regression of y on x and x on y are

Their slopes are

Thus,

When  r = 0,i.e. when the variable are independent, the two lines of regression are perpendicular to each other.

When . Thus the line of regression coincide i.e. there is perfect correlation between the two variables.

Example. While calculating correlation coefficient between two variables x and y from 25 pairs of observations, the following results were obtained : n = 25, Later it was discovered at the time of checking that the pairs of values x -8,6 and y = 12, 8 were copied down as x = 6,8 and y = 14,6. Obtain the correct value of correlation coefficients.

Solution. To get the correct results, we subtract the incorrect values and add the corresponding correct values.

The correct results would be

RANK CORRELATION

A group of n individuals may be arranged in order to merit with respect to some characteristics. The same group would give different orders for different characteristics. Considering the orders corresponding to two characteristics A and B, the correction between these n pairs of rank is called the rank correlation in the characteristics A and B for that group of individuals.

Let be the ranks of the ith individuals in A and B respectively. Assuming that no two individuals are bracketed equal in either case, each of the variables taking the values 1,2,3,…,n we have

If X, Y be the deviations of x, y from their means, then

Now let,

Hence the correlation coefficient between these variables is

This is called the rank correlation coefficient and is denoted by

Example. Ten participants in a contest are ranked by two judges as follows:

Calculate the rank correlation coefficient

Solution. If

Hence,

Example. Three judges A,B,C give the following ranks. Find which pair of judges has common approach

Solution. Here n = 10

Since is maximum, the pair of judge A and C have the nearest common approach.

3.5 Curve fitting by the method of least squares- fitting of straight lines

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

On simplification equation (2) and (3) becomes

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

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

(b)To fit the parabola

The normal equations are

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

Example. Find the best values of a and b so that y = a + bx fits the data given in the table

Solution.

 y = a + bx

Normal equations, y= na+ bx  (2)

On putting the values of

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 squares, find the straight line that best fits the following data :

Solution. Let the equation of the straight line best fit be y = a + bx.     (1)

Normal equations are

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

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

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

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

Solution. Let the equation of second degree polynomial be

Normal equations are

On putting the values of x, y, xy,

have

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

The required polynomial of second degree is

3.6 Second degree parabolas and more general 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 least square method:

Solution. Taking

Taking

The equation is transformed to

Normal equations are

On solving these equations we get

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

Solution. We shift the origin to (2.5, 0) antique 0.5 as the new unit. This amounts to changing the variable x to X, by the relation X = 2x – 5.

Let the parabola of fit be y = a + bX The values of X etc. Are calculated as below: