Statistics of Mathematics

Today our topic of discussion is Statistics of Mathematics.

Statistics of Mathematics

 

Statistics of Mathematics

 

Statistics

Owing to the contribution of information and data, the world has become a global village for the rapid advancement of science and information. Globalization has been made possible due to rapid transformation and expansion of information and data.

So, to keep the continuity of development and for participating and contribute in globalization, it is essential for the students at this stage to have clear knowledge about information and data. In the context, to meet the demands of students in acquiring knowledge, information and data have been discussed from class VI and class-wise contents have been arranged step by step.

In continuation of this, the students of this class will know and learn cumulative frequency, frequency polygon, ogive curve in measuring of central tendency mean, median, mode etc. in short-cut method.

At the end of this chapter, the students will be able to –

‣ explain cumulative frequency, frequency polygon and ogive curve.

‣ explain data by the frequency polygon, and ogive curve.

‣ explain the method of measuring of central tendency.

‣ explain the necessity of short-cut method in the measurement of central tendency.

‣ find the mean, median and mode by the short-cut method.

‣  explain the diagram of frequency polygon and ogive curve.

Presentation of Data: We know that numerical information which are not qualitative are the data of statistics. The data under investigation are the raw materials of statistics. They are in unorganized form and it is not possible to take necessary decision directly from the unorganized data.

It is necessary to organize and tabulate the data. And the tabulation of data is the presentation of the data. In previous class we have learnt how to organize the data in tabulation. We know that it is required to determine the range of data for tabulation.

Then determining the class interval and the number of classes by using tally marks, the frequency distribution table is made. Here, the methods of making frequency distribution table are to be re-discussed through example for convenient understanding.

Example 1. In a winter season, the temperature (in celsius) of the month of January in the district of Srimangal is placed below. Find the frequency distribution table of the temperature.

14 ° , 14 °, 14° , 13° , 12° , 13° , 10° , 10° , 11° , 12° , 11° , 10° , 9° , 8° , 9° , 11° , 10° , 10° , 8° , 9° , 7° , 6°  ,6°  , 6° , 6° , 7° , 7° , 8° , 9° , 9° , 8° , 7°.

Solution: Here the minimum and maximum numerical values of the data of temperature are 6 and 14 respectively. Hence the range = (14-6)+1=9

If the class interval is considered to be 3, the numbers of class will be 9/3 or 3 Considering 3 to be the class interval, if the data are arranged in 3 classes, the frequency table will be:

Statistics of Mathematics

Work: Form two groups of all the students studying in your class. Find the frequency distribution table of the weights (in Kgs) of all the members of the groups.

Cumulative Frequency: In Example 1 considering 3 the class interval and determining the number of classes, the frequency distribution table has been made. The number of classes of the mentioned data is 3.

The limit of the first class is 6 deg – 8 deg The lowest range of the class is 6 deg and the highest range is 8 deg The frequency of this class is 11. Similarly, The limit of the second class is 9 deg – 11 deg and the frequency of this class is 13.

Now if the frequency 11 of first class is added to the frequency 13 of the second class, we get 24. This 24 will be the cumulative frequency of the second class and the cumulative frequency of first class as begins with the class will be 11.

Again, if the cumulative frequency 24 of the second class is added to the frequency of the third class, we get 24+7 = 31 which is the cumulative frequency of the third class. Thus cumulative frequency distribution table is made. In the context of the above discussion, the cumulative frequency distribution of temperature in Example 1 is as follows:

Statistics of Mathematics

Example 2. The marks obtained in English by 40 students in an annual examination are given below. Make a cumulative frequency table of the marks obtained.

70, 40, 35, 60, 55, 58, 45, 60, 65, 80, 70, 46, 50, 60, 65, 70, 58, 60, 48, 70, 36, 85, 60, 50, 46, 65, 55, 61, 72, 85, 90, 68, 65, 50, 40, 56, 60, 65, 46, 76

Solution:

Range of the data = (highest numerical value lowest numerical value) + 1 = (9035)+1 55+1=56

56 Let the class interval be 5 then the number of classes

11.2 or 12 [taking the immediate next integer when class interval is fractional]

Hence the cumulative frequency distribution table at a class interval of 5 will be as follows:

Statistics of Mathematics

Variable: We know that the numerical information is the data of statistics. The numbers used in data are variable. Such as, in example 1 the numbers indicating temperatures are variable. Similarly, in example 2, the secured marks used in the data are the variables.

Discrete and Indiscrete variable: The variables used in statistics are of two types. Such as, discrete and indiscrete vari ables. The variables whose values are only integers, are discrete variables.

The marks obtained in example 2 are discrete variables. Similarly, only integers are used in population indicated data. That is why, the variables of data used for population are discrete variables.

A nd the variables whose numerical values can be any real number are indiscrete variables. Such as, in example 1, the temperature indicated data which can be any real number. Besides, any real number can be used for the data related to age, height, weight etc.

That is why, the variables used for those are indiscrete variables. The number between two indiscrete variables can be the value of those variables. Some times it becomes necessary to make class interval indiscrete.

To make the class interval indiscrete, the actual higher limit of a class and the lower limit of the next class are determined by fixing mid-point of a higher limit of any class and the lower limit of the next class. Such as, in example 1 the actual higher-lower limits of the first class are 8.5° and 5.5° respectively and that of the second class are 11.5° 8.5° etc.

Work: Form a group of maximum 40 students of your class. Form frequency distribution table and cumulative frequency table of the group with the weights/heights of the members.

Diagram of Data: We have seen that the collected data under investigation are the raw materials of the statistics. If the frequency distribution and cumulative frequency distribution table are made with them, it becomes clear to comprehend and to draw a conclusion.

If that tabulated data are presented through diagram, they become easier to understand as well as attractive. That is why, presentation of statistical data in tabulation and diagram is widely and frequently used method.

In class VIII, different types of diagram in the form of line graph and histogram have been discussed elaborately and the students have been taught how to draw them. Here, how frequency polygon, pie-chart, ogive curve are drawn from frequency distribution and cumulative frequency table will be discussed.

Frequency Polygon: In class VIII, we have learnt how to draw the histogram of discrete data. Here how to draw frequency polygon from histogram of indiscrete data will be put for discussion through example.

Example 3. The frequency distribution table of the weights (in Kg) of 60 students of class X of a school is given below:

Statistics of Mathematics

1) Draw the histogram of frequency distribution.

2) Draw frequency polygon of the histogram.

Solution: The class interval of the data in the table is discrete. If the class interval are made indiscrete, the table will be:

Statistics of Mathematics

1) Histogram has been drawn taking each square of graph paper as 5 unit of class interval along with x-axis and frequency along with y-axis. The class interval along with x-axis has started from 45.5. The broken segments have been used to show the presence of previous squares starting from from origin to 45.5.

Statistics of Mathematics

 

2) The mid-points of the opposite sides parallel to the base of rectangle of the histogram have been fixed for drawing frequency polygon from histogram. The mid-points have been joined by line segments to draw the frequency polygon (shown in the adjacent figure). The mid-points of the first and the last rectangles have been joined with x-axis representing the class interval by the end points of line segments to show the frequency polygon attractive.

Statistics of Mathematics

 

Example 4. Draw polygon of the following frequency distribution table.

Statistics of Mathematics

Solution: Histogram of frequency distribution is drawn taking each square of graph paper as 10 units of class interval along with x-axis and each square of graph paper as 5 units of frequency along with y-axis.

The mid-points of the sides opposite to the base of rectangle of histogram are identified which are the mid-points of the class. Now the fixed mid-points are joined. The end-points of the first and the last classes are joined to z-axis representing the class interval to draw frequency polygon.

Statistics of Mathematics

Example 5. The frequency distribution table of the marks obtained by 50 students of class X in science are given. Draw the frequency polygon of the data (without using histogram):

Statistics of Mathematics

Solution: Here the given data are discrete. In this case, it is convenient to draw frequency polygon directly by finding the mid-point of class interval.

The Mid-point of the first class interval (31-40) is 31 +40 2 35.5

Statistics of Mathematics

The polygon is drawn by taking each squares of graph paper as 1 units of mid- points of class interval along with z-axis and taking each square of graph paper as 2 units of frequency along with y-axis.

 

Statistics of Mathematics

 

Cumulative Frequency Diagram or Ogive curve: Cumulative frequency diagram or Ogive curve is drawn by taking the upper limit of class interval along with x-axis and cumulative frequency along with y-axis after classification of a data.

Example 6. The frequency distribution table of the marks obtained by 50 students out of 60 students is as follows:

Statistics of Mathematics

Draw the Ogive curve of this frequency distribution.

Solution: The cumulative frequency table of frequency distribution of the given data is :

 

Statistics of Mathematics

 

Ogive curve of cumulative frequency of data is drawn taking each square of graph paper as two units of upper limit of class interval along with x-axis and cumulative frequency along with y-axis.

 

Statistics of Mathematics

 

Central Tendency: Central tendency and its measurement have been discussed in class VII and VIII. We have seen if the data under investigation are arranged in order of values, the data cluster round near any central value.

A gain if the disorganized data are placed in frequency distribution table, the frequency is found to be abundant in a middle class i.e. frequency is maximum in middle class. In fact, the tendency of data to be clustered around the central value is number and it represents the data.

The central tendency is measured by this number. Generally, the measurement of central tendency is of three types (1) Arithmetic means (2) Median (3) Mode:

Arithmetic Mean: We know if the sum of data is divided by the numbers of the data, we get the arithmetic mean. But this method is complex, time consuming and there is every possibility of committing mistake for large numbers of data. In such cases, the data are tabulated through classification and the arithmetic mean is determined by short-cut method.

Example 7. The frequency distribution table of the marks obtained by the students of a class is as follows. Find the arithmetic mean of the marks.

Statistics of Mathematics

Solution: Here class interval is given and that is why it is not possible to know the individual marks of the students. In such case, it becomes necessary to know the mid-value of the class.

Class upper value +class lower value

Mid-value of the class= 2

If the class mid-value is x_{i} (i=1…k) the mid-value related table will be as follows:

Statistics of Mathematics

 

The required arithmetic mean

 

Statistics of Mathematics

 

Arithmetic mean of classified data (short-cut method): The short-cut method is easy for determining arithmetic mean of classified data. The steps to determine mean by short-cut method are:

  1. To find the mid-value of classes.
  2. To take convenient approximated mean (a) from the mid-values.
  3. To determine steps deviation, the difference between class mid-values and approximate mean are divided by the class interval i.e. mid value – approximate mean steps deviation u class interval
  1. To multiply the steps deviation by the corresponding class frequency.
  2. To determine the mean of the deviation and to add this mean with approximate mean to find the required mean.

Short-cut method: The formula used for determining the mean of the data by this method is

 

Statistics of Mathematics

 

where, overline x = required mean, a = approximate mean, f_{i} = class frequency of ith class, u_{i}*f_{i} =* the product of step deviation with class intervals of ith class and h = class interval, k = number of class and n= total number of frequency.

Example 8. The production cost (in hundred taka) of a commodity at different stages is shown in the following table. Find the mean of the expenditure by short- cut method.

Statistics of Mathematics

Solution: To determine mean in the light of followed steps in short-cut method, the table will be:

Statistics of Mathematics

Mean overline x =a+ sum i=1 ^ n f i u i nh = 20 – 37/188 * 4 = 20 – 0.79 = 19.21

∴ Mean production cost is Tk. 19 hundred.

Weighted mean: In many cases the numerical values x_{1}, x_{2} ,…,x n of statistical data under investigation may be influenced by different reasons/ importance/ weight. In such case, the values of the data x_{1}, x_{2} ,…,x n along with their reasons/ importance/ weight w_{1}, w_{2} ,…,w n are considered to find the arithmetic mean. If the values of n numbers of data are 1.2, and their weights are w₁, w2,…, wn, the weighted mean will be

 

Statistics of Mathematics

Statistics of Mathematics

 

Example 9. The rate of passing in degree Honours class and the number of students of some department of a University are presented in the table below. Find the mean rate of passing in degree Honours class of those departments of the university.

Statistics of Mathematics

Solution: Here, the rate of passing and the number of students are given. The weight of rate of passing is the number of students. If the variables of rate of passing are z and numerical variable of students is w, the table for determining the arithmetic mean of given weight will be as follows:

Statistics of Mathematics

∴ Mean rate of passing 77.14

Work: Collect the rate of passing students and their numbers in S.S.C. examination of some schools in your Upazilla and find mean rate of passing.

Median:

We have already learnt in class VIII the value of the data which divide the data when arranged in ascending order into two equal parts are median of the data.

We have also learnt if the numbers of data are n and n is an odd number, the median will be the value of (n + 1)/2 th term. But if n is an even number, the median will be numerical mean of the value of n/2 and (n/2 + 1) th terms. Here we present through example how mean is determined with or without the help formulae.

Example 10. The frequency distribution table of 51 students is placed below. Find the median.

Statistics of Mathematics

Solution: Cumulative frequency distribution table for finding mean is as follows

Statistics of Mathematics

Here, n = 51 is an odd number.

∴ Median = the value of (51 + 1)/2 th term = the value of26th term = 165

Required median 165 cm.

Note: The value of the terms from 23th to 38th is 165.

Example 11. The frequency distribution table of marks obtained in mathematics of 60 students is as follows. Find the median :

Statistics of Mathematics

Solution: Cumulative frequency distribution table for determining median is:

Statistics of Mathematics

Here, n = 60 which is an even number.

∴ Median = {(60/2)th term +(60/2+1)th term}= (30th term +31th term)/2

 = (70 + 80)/2 = 75

∴ Required median 75

Work:

1) Make frequency distribution table of the heights (in cm.) of 49 students of your class and find the mean without using any formula.

2) From the above problem, deduct the heights of 9 students and then find the median of heights (in cm.) of 40 students.

Determining Median of Classified Data: If the number of classified data is n, the value of n/2 * th term of classified data is median. And the formula used to determine the median or the value of n/2 * th term is:

Median = L + (n/2 – F_{c}) * h/f_{m} where L is the lower limit of the median class, n is the frequency, F_{c} is the cumulative frequency of previous class to median class, f_{m} is the frequency of median class and h is the class interval.

Example 12.

Determine median from the following frequency distribution table:

Statistics of Mathematics

1) What is a frequency distribution table?

2) Determine median from the frequency distribution table given above.

3) Draw the frequency polygon of the given data.

Solution:

1) Frequency distribution table refers to organize and tabulate a dataset by determining specific class interval and number of classes.

2) The frequency distribution table to determine median is given below:

 

Statistics of Mathematics

 

Here, n = 70 and n/2 = 70/2 or 35

Therefore, median is the value of 35th term. 35th term lies in the class (48-53). Hence the median class is (48-53).

L = 48, F_{c} = 31, f_{m} = 25 and h = 6

So median = 48 + (35 – 31) * 6/25 = 48 + 4 * 6/25 = 48 + 0.96 = 48.96

∴ Required median 48.96.

3) The table useful to draw the frequency polygon is given below: The median of the class prior to the first class would be 26.5 and the median of the class following the last class would be 68.5.

Now, taking convenient unit for median values in along the X axis where, sign indicates 0-26.5 and taking length of side of each sqaure as 2 of frequency along Y axis, the frequency polygon has been drawn.

 

Statistics of Mathematics

 

Work: Make two groups with all the students of your class. (a) Make a frequency distribution table of the time taken by each of you to solve a problem, (b) find the median from the table.

Mode:

In class VIII, we have learned that the number which appears maximum times in a data is the mode of the data. In a data, there may be one or more than one mode. If there is no repetition of a member in a data, data will have no mode. Now we shall discuss how to determine the mode of classified data using formula.

fi Determining Mode of Classified Data: Mode = L+- xh, where fi+f2

L is the lower limit of mode-class i.e. the class where the mode lies,

fi frequency of mode-class frequency of the class previous

f2 frequency of mode class – frequency of next class of mode class and h =class interval.

Example 13. Following is a table.

Statistics of Mathematics

1) What is central tendency?

2) Find mode from the above table.

3) Draw ogive curve of the given data.

Solution:

1) If the disorganized data of statistics are arranged according to the value, the data cluster round near any central value. Moreover, abundance of data is observed in a single class when these data are presented in some frequency distribution table. This tendency of data to cluster around central value is known as central tendency.

2) The table to determine mode is given below:

Statistics of Mathematics

Mode = L + xh

Here, the maximum numbers of repetition of frequency is 12 which lies in the class (61-70).

∴ L = 61, f₁ = 12-84, f212-9=3,h=10

∴ Mode = 4 40 61 + x 10 = 61+10=61+ = 61+5.7 66.7 4+3 7

Required mode 66.7

3) The table to draw the ogive curve is given below:

Statistics of Mathematics

Using convenient scaling along X axis where (sign indicates 0-30 and taking per square unit as 5 units of cumulative frequency along Y axis points along higher limits of classes have been marked. Then these marked points have been joined and thus the Ogive curve has been drawn.

 

Statistics of Mathematics

 

Example 14. Find mode from the frequency distribution table given below.

Statistics of Mathematics

Solution: Here, maximum numbers of frequency are 25 which lie in the class (41-50). So it is evident that mode is in this class. We know that

Mode= L+ xh. Here, L = 41 f_{1} = 25 + 0 = 25 f_{2} = 25 – 20 = 5 [If the fi+f2 frequency is maximum in the first class, the frequency of previous class is zero]

∴ Mode = 41 + 25 25+5 25 x 10 = 41+ x 1041+8.33 = 49.33 30

Required mode 49.33.

Example 15. Find mode from the frequency distribution table given below.

Statistics of Mathematics

Solution: The maximum numbers of Class Frequency frequency are 25 which lie in the class (4150). So it is obvious that this class is the class of mode. We f1 know that, Mode = L+F1/F1+F2 x h

Here, L = 4, f1 = 25 – 20 = 5, f2 = 25 + 0 = 25  [ If the frequency is maximum h =10 in the last class, the frequency of following class is zero]

∴ Mode = 41 + 25+5 5 x 10 = 41 + x 10 = 41 + 30 513 41+1.6742.67

Required mode 42.67 (approximately)

If the frequency is maximum in the first class, the frequency of previous class is zero. If the frequency is maximum in the last class, the frequency of following class is zero.

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