Textbook: Statistics | Mathematics Class 10 (Maharashtra SSC Board) PDF Download

 Page 1


129
6
Statistics
Statistics is useful in many fields of life: for example, agriculture, economics, 
commerce, medicine, botany, biotechnology, physics, chemistry, education, sociology, 
administration etc. An experiment can have many outcomes. To assess the possibility 
of possible outcomes, one has to carry out the experiment on a large scale and keep 
the record meticulously. Possibilities of different outcomes can be assessed using 
the record. For this purpose, rules are formulated in statistics.
Francis Galton (1822-1911) has done much of fundamental work in 
statistics.He used to prepare questionnaires, distribute 
them among people and request them to fill them up. 
He collected information from a number of people 
and recorded their backgrounds, financial situations, 
likes and dislikes, health etc. on a large scale. By that 
time, it was known that the fingerprints of different 
people are different. He collected finger-prints of a 
large number of people and invented a method of their 
classification. Using statistical methods, he showed that the possibility of 
finger prints of two different people being identical is nearly zero. This result 
made it possible to identify a person from his finger-prints. This method of 
identifying criminals was accepted in the judiciary. He had done much work 
in the field of anthropology of humans and other animals also. 
Let’s recall.
 We usually find a specific property in the numerical data collected in a survey 
that the scores have a tendency to cluster around a particular score. This score is 
a representative number of the group. The number is called the measure of central 
tendency.
 In the previous standards we have studied the measures of central tendency, 
namely the mean, median and mode, for ungrouped data.
Francis Galton
   · Measures of a central tendency- 
   mean, median and mode from grouped frequency table. 
   · Graphical representation of statistical data -
    histogram, frequency polygon, pie diagram
Let’s study.
Page 2


129
6
Statistics
Statistics is useful in many fields of life: for example, agriculture, economics, 
commerce, medicine, botany, biotechnology, physics, chemistry, education, sociology, 
administration etc. An experiment can have many outcomes. To assess the possibility 
of possible outcomes, one has to carry out the experiment on a large scale and keep 
the record meticulously. Possibilities of different outcomes can be assessed using 
the record. For this purpose, rules are formulated in statistics.
Francis Galton (1822-1911) has done much of fundamental work in 
statistics.He used to prepare questionnaires, distribute 
them among people and request them to fill them up. 
He collected information from a number of people 
and recorded their backgrounds, financial situations, 
likes and dislikes, health etc. on a large scale. By that 
time, it was known that the fingerprints of different 
people are different. He collected finger-prints of a 
large number of people and invented a method of their 
classification. Using statistical methods, he showed that the possibility of 
finger prints of two different people being identical is nearly zero. This result 
made it possible to identify a person from his finger-prints. This method of 
identifying criminals was accepted in the judiciary. He had done much work 
in the field of anthropology of humans and other animals also. 
Let’s recall.
 We usually find a specific property in the numerical data collected in a survey 
that the scores have a tendency to cluster around a particular score. This score is 
a representative number of the group. The number is called the measure of central 
tendency.
 In the previous standards we have studied the measures of central tendency, 
namely the mean, median and mode, for ungrouped data.
Francis Galton
   · Measures of a central tendency- 
   mean, median and mode from grouped frequency table. 
   · Graphical representation of statistical data -
    histogram, frequency polygon, pie diagram
Let’s study.
130
 Activity 1 : Measure the height in cm of all students in your class. We find that 
the heights of many students cluster near a specific number. 
Activity 2 : Collect a number of fallen leaves of a peepal tree. Distribute the 
leaves among the students and ask them to measure the lengths of them. Record 
the lengths. We notice that their lengths tend to cluster around a number. 
 
 Now we are going to do some more study of the mean, median and mode. Let 
us know the symbols and the terminology required for it. 
 The mean of statistical data =
The sum of all scores 
Total no. of scores
=  
         (Here x
i
 is the i 
th 
score)
 Mean is denoted by X and it represents the average of the given data. 
  X  = 
Let’s learn.
  Mean from classified frequency distribution
 When the number of scores in a data is large, it becomes tedious to write all 
numbers in the above formula and take their sum. So we use some different methods 
to find the sum. 
 Sometimes, the large data collected from an experiment is presented in a table 
in the grouped form. In such a case, we cannot find the exact mean of statistical data. 
Hence, let us study a method which gives the approximate mean, or a number nearby.
   Direct method
 Let us study the method by an example.
Ex. : The following table shows the frequency distribution of the time required for
 each worker to complete a work . From the table find the mean time required to
 complete the job for a worker.
Time (Hrs.) for each to complete the work
15-19 20-24 25-29 30-34 35-39
No. of workers
10 15 12 8 5
x
i
i
N
1
N
x
i
i
N
1
N
Page 3


129
6
Statistics
Statistics is useful in many fields of life: for example, agriculture, economics, 
commerce, medicine, botany, biotechnology, physics, chemistry, education, sociology, 
administration etc. An experiment can have many outcomes. To assess the possibility 
of possible outcomes, one has to carry out the experiment on a large scale and keep 
the record meticulously. Possibilities of different outcomes can be assessed using 
the record. For this purpose, rules are formulated in statistics.
Francis Galton (1822-1911) has done much of fundamental work in 
statistics.He used to prepare questionnaires, distribute 
them among people and request them to fill them up. 
He collected information from a number of people 
and recorded their backgrounds, financial situations, 
likes and dislikes, health etc. on a large scale. By that 
time, it was known that the fingerprints of different 
people are different. He collected finger-prints of a 
large number of people and invented a method of their 
classification. Using statistical methods, he showed that the possibility of 
finger prints of two different people being identical is nearly zero. This result 
made it possible to identify a person from his finger-prints. This method of 
identifying criminals was accepted in the judiciary. He had done much work 
in the field of anthropology of humans and other animals also. 
Let’s recall.
 We usually find a specific property in the numerical data collected in a survey 
that the scores have a tendency to cluster around a particular score. This score is 
a representative number of the group. The number is called the measure of central 
tendency.
 In the previous standards we have studied the measures of central tendency, 
namely the mean, median and mode, for ungrouped data.
Francis Galton
   · Measures of a central tendency- 
   mean, median and mode from grouped frequency table. 
   · Graphical representation of statistical data -
    histogram, frequency polygon, pie diagram
Let’s study.
130
 Activity 1 : Measure the height in cm of all students in your class. We find that 
the heights of many students cluster near a specific number. 
Activity 2 : Collect a number of fallen leaves of a peepal tree. Distribute the 
leaves among the students and ask them to measure the lengths of them. Record 
the lengths. We notice that their lengths tend to cluster around a number. 
 
 Now we are going to do some more study of the mean, median and mode. Let 
us know the symbols and the terminology required for it. 
 The mean of statistical data =
The sum of all scores 
Total no. of scores
=  
         (Here x
i
 is the i 
th 
score)
 Mean is denoted by X and it represents the average of the given data. 
  X  = 
Let’s learn.
  Mean from classified frequency distribution
 When the number of scores in a data is large, it becomes tedious to write all 
numbers in the above formula and take their sum. So we use some different methods 
to find the sum. 
 Sometimes, the large data collected from an experiment is presented in a table 
in the grouped form. In such a case, we cannot find the exact mean of statistical data. 
Hence, let us study a method which gives the approximate mean, or a number nearby.
   Direct method
 Let us study the method by an example.
Ex. : The following table shows the frequency distribution of the time required for
 each worker to complete a work . From the table find the mean time required to
 complete the job for a worker.
Time (Hrs.) for each to complete the work
15-19 20-24 25-29 30-34 35-39
No. of workers
10 15 12 8 5
x
i
i
N
1
N
x
i
i
N
1
N
131
Class
(Time-
hours)
Class 
mark
x
i
Frequency 
(No. of 
Workers)
f
i
Class mark ´ 
Frequency
x
i  
f
i
15-19
20-24
25-29
30-34
35-39
17
22
27
32
37
10
15
12
8
5
170
330
324
256
185
Total å f
i
 = 50 å x
i  
f
i 
= 
1265
Solution :
 (1) Vertical columns are drawn as 
shown in the table.
(2) Classes are written in the first 
column. 
(3) The class mark x
i
 is in the second  
column. 
(4) In the third column, the number 
of workers, that is frequency (f
i
) 
is written.
(5) In the fourth column, the product 
(x
i 
´
 
f
i
) for each class is written. 
(6) Then 
i
N


1
x
i  
f
i
 is written.
(7) The mean is found using the 
formula 
Mean = X  = 
å x
i  
f
i
 
N
  = 
1265
50
 = 25.3 
\
 å f
i  
=  N
 The mean time required to complete the work for a worker = 25.3 hrs. (Approx)
ÒÒÒ?? Solved Examples ÒÒÒ
Ex. (1) The percentage of marks of 50 students in a test is given in the following table. 
Find the mean of the percentage. 
Percentage of marks       0-20     20-40     40-60     60-80      80-100
No. of students              3 7 15 20 5
Solution : The following table is prepared as per steps. 
Class 
(Percentage of 
marks)
Class 
mark 
x
i
Frequency 
(No. of 
students)
f
i
Class mark ´ 
frequency
x
i 
f
i
0-20
20-40
40-60
60-80
80-100
10
30
50
70
90
3
7
15
20
5
    30
  210
  750
1400
  450
Total 
N = å f
i
 = 50 å x
i  
f
i 
=  2840
X = 
å x
i  
f
i
å f
i 
   
    = 
2840
50
 
    = 56.8
  \ The mean of 
the percentage 
= 56.8
Page 4


129
6
Statistics
Statistics is useful in many fields of life: for example, agriculture, economics, 
commerce, medicine, botany, biotechnology, physics, chemistry, education, sociology, 
administration etc. An experiment can have many outcomes. To assess the possibility 
of possible outcomes, one has to carry out the experiment on a large scale and keep 
the record meticulously. Possibilities of different outcomes can be assessed using 
the record. For this purpose, rules are formulated in statistics.
Francis Galton (1822-1911) has done much of fundamental work in 
statistics.He used to prepare questionnaires, distribute 
them among people and request them to fill them up. 
He collected information from a number of people 
and recorded their backgrounds, financial situations, 
likes and dislikes, health etc. on a large scale. By that 
time, it was known that the fingerprints of different 
people are different. He collected finger-prints of a 
large number of people and invented a method of their 
classification. Using statistical methods, he showed that the possibility of 
finger prints of two different people being identical is nearly zero. This result 
made it possible to identify a person from his finger-prints. This method of 
identifying criminals was accepted in the judiciary. He had done much work 
in the field of anthropology of humans and other animals also. 
Let’s recall.
 We usually find a specific property in the numerical data collected in a survey 
that the scores have a tendency to cluster around a particular score. This score is 
a representative number of the group. The number is called the measure of central 
tendency.
 In the previous standards we have studied the measures of central tendency, 
namely the mean, median and mode, for ungrouped data.
Francis Galton
   · Measures of a central tendency- 
   mean, median and mode from grouped frequency table. 
   · Graphical representation of statistical data -
    histogram, frequency polygon, pie diagram
Let’s study.
130
 Activity 1 : Measure the height in cm of all students in your class. We find that 
the heights of many students cluster near a specific number. 
Activity 2 : Collect a number of fallen leaves of a peepal tree. Distribute the 
leaves among the students and ask them to measure the lengths of them. Record 
the lengths. We notice that their lengths tend to cluster around a number. 
 
 Now we are going to do some more study of the mean, median and mode. Let 
us know the symbols and the terminology required for it. 
 The mean of statistical data =
The sum of all scores 
Total no. of scores
=  
         (Here x
i
 is the i 
th 
score)
 Mean is denoted by X and it represents the average of the given data. 
  X  = 
Let’s learn.
  Mean from classified frequency distribution
 When the number of scores in a data is large, it becomes tedious to write all 
numbers in the above formula and take their sum. So we use some different methods 
to find the sum. 
 Sometimes, the large data collected from an experiment is presented in a table 
in the grouped form. In such a case, we cannot find the exact mean of statistical data. 
Hence, let us study a method which gives the approximate mean, or a number nearby.
   Direct method
 Let us study the method by an example.
Ex. : The following table shows the frequency distribution of the time required for
 each worker to complete a work . From the table find the mean time required to
 complete the job for a worker.
Time (Hrs.) for each to complete the work
15-19 20-24 25-29 30-34 35-39
No. of workers
10 15 12 8 5
x
i
i
N
1
N
x
i
i
N
1
N
131
Class
(Time-
hours)
Class 
mark
x
i
Frequency 
(No. of 
Workers)
f
i
Class mark ´ 
Frequency
x
i  
f
i
15-19
20-24
25-29
30-34
35-39
17
22
27
32
37
10
15
12
8
5
170
330
324
256
185
Total å f
i
 = 50 å x
i  
f
i 
= 
1265
Solution :
 (1) Vertical columns are drawn as 
shown in the table.
(2) Classes are written in the first 
column. 
(3) The class mark x
i
 is in the second  
column. 
(4) In the third column, the number 
of workers, that is frequency (f
i
) 
is written.
(5) In the fourth column, the product 
(x
i 
´
 
f
i
) for each class is written. 
(6) Then 
i
N


1
x
i  
f
i
 is written.
(7) The mean is found using the 
formula 
Mean = X  = 
å x
i  
f
i
 
N
  = 
1265
50
 = 25.3 
\
 å f
i  
=  N
 The mean time required to complete the work for a worker = 25.3 hrs. (Approx)
ÒÒÒ?? Solved Examples ÒÒÒ
Ex. (1) The percentage of marks of 50 students in a test is given in the following table. 
Find the mean of the percentage. 
Percentage of marks       0-20     20-40     40-60     60-80      80-100
No. of students              3 7 15 20 5
Solution : The following table is prepared as per steps. 
Class 
(Percentage of 
marks)
Class 
mark 
x
i
Frequency 
(No. of 
students)
f
i
Class mark ´ 
frequency
x
i 
f
i
0-20
20-40
40-60
60-80
80-100
10
30
50
70
90
3
7
15
20
5
    30
  210
  750
1400
  450
Total 
N = å f
i
 = 50 å x
i  
f
i 
=  2840
X = 
å x
i  
f
i
å f
i 
   
    = 
2840
50
 
    = 56.8
  \ The mean of 
the percentage 
= 56.8
132
Ex. (2) The maximum temperatures in °C of 30 towns, in the last summer, is shown in 
the following table. Find the mean of the maximum temperatures. 
Max. temp. 24-28 28-32 32-36 36-40  40-44            
No. of towns
4 5 7 8 6
Solution :
Class 
(Temp.  °C)
Class mark 
x
i
Frequency 
(No. of towns) 
f
i
Class mark ´ 
frequency
x
i  
f
i
24-28
28-32
32-36
36-40
40-44
26
30
34
38
42
4
5
7
8
6
104
150
238
304
252
Total N = å f
i
 = 30 å x
i 
f
i 
= 1048
 Mean =  X  = 
å x
i  
f
i
å f
i 
   = 
1048
30
 = 34.9 °C 
    Assumed mean method
 In the examples solved above, we see that some times the product x
i 
f
i 
is 
large. Hence it becomes difficult to calculate the mean by direct method. So let us 
study another method, called the 'assumed mean method'. Finding the mean becomes 
simpler if we use addition and division in this method. 
For example, we have to find the mean of the scores 40, 42, 43, 45, 47 and 48. 
 The obeservation of the scores reveals that the mean of the data is more than 
40. So let us assume that the mean is 40. 40-40 = 0, 42 - 40 = 2, 43-40 = 3,  
45-40 = 5, 47 - 40 = 7, 48 - 40 = 8 These are called 'deviations'. Let us find 
their mean. Adding this mean to the assumed mean, we get the mean of the data. 
 That is, mean = assumed mean + mean of the deviations 
 X  = 40 + 
02 3 578
6
   




     = 40 + 
25
6
  = 40 + 4
1
6
     = 44
1
6
 
Page 5


129
6
Statistics
Statistics is useful in many fields of life: for example, agriculture, economics, 
commerce, medicine, botany, biotechnology, physics, chemistry, education, sociology, 
administration etc. An experiment can have many outcomes. To assess the possibility 
of possible outcomes, one has to carry out the experiment on a large scale and keep 
the record meticulously. Possibilities of different outcomes can be assessed using 
the record. For this purpose, rules are formulated in statistics.
Francis Galton (1822-1911) has done much of fundamental work in 
statistics.He used to prepare questionnaires, distribute 
them among people and request them to fill them up. 
He collected information from a number of people 
and recorded their backgrounds, financial situations, 
likes and dislikes, health etc. on a large scale. By that 
time, it was known that the fingerprints of different 
people are different. He collected finger-prints of a 
large number of people and invented a method of their 
classification. Using statistical methods, he showed that the possibility of 
finger prints of two different people being identical is nearly zero. This result 
made it possible to identify a person from his finger-prints. This method of 
identifying criminals was accepted in the judiciary. He had done much work 
in the field of anthropology of humans and other animals also. 
Let’s recall.
 We usually find a specific property in the numerical data collected in a survey 
that the scores have a tendency to cluster around a particular score. This score is 
a representative number of the group. The number is called the measure of central 
tendency.
 In the previous standards we have studied the measures of central tendency, 
namely the mean, median and mode, for ungrouped data.
Francis Galton
   · Measures of a central tendency- 
   mean, median and mode from grouped frequency table. 
   · Graphical representation of statistical data -
    histogram, frequency polygon, pie diagram
Let’s study.
130
 Activity 1 : Measure the height in cm of all students in your class. We find that 
the heights of many students cluster near a specific number. 
Activity 2 : Collect a number of fallen leaves of a peepal tree. Distribute the 
leaves among the students and ask them to measure the lengths of them. Record 
the lengths. We notice that their lengths tend to cluster around a number. 
 
 Now we are going to do some more study of the mean, median and mode. Let 
us know the symbols and the terminology required for it. 
 The mean of statistical data =
The sum of all scores 
Total no. of scores
=  
         (Here x
i
 is the i 
th 
score)
 Mean is denoted by X and it represents the average of the given data. 
  X  = 
Let’s learn.
  Mean from classified frequency distribution
 When the number of scores in a data is large, it becomes tedious to write all 
numbers in the above formula and take their sum. So we use some different methods 
to find the sum. 
 Sometimes, the large data collected from an experiment is presented in a table 
in the grouped form. In such a case, we cannot find the exact mean of statistical data. 
Hence, let us study a method which gives the approximate mean, or a number nearby.
   Direct method
 Let us study the method by an example.
Ex. : The following table shows the frequency distribution of the time required for
 each worker to complete a work . From the table find the mean time required to
 complete the job for a worker.
Time (Hrs.) for each to complete the work
15-19 20-24 25-29 30-34 35-39
No. of workers
10 15 12 8 5
x
i
i
N
1
N
x
i
i
N
1
N
131
Class
(Time-
hours)
Class 
mark
x
i
Frequency 
(No. of 
Workers)
f
i
Class mark ´ 
Frequency
x
i  
f
i
15-19
20-24
25-29
30-34
35-39
17
22
27
32
37
10
15
12
8
5
170
330
324
256
185
Total å f
i
 = 50 å x
i  
f
i 
= 
1265
Solution :
 (1) Vertical columns are drawn as 
shown in the table.
(2) Classes are written in the first 
column. 
(3) The class mark x
i
 is in the second  
column. 
(4) In the third column, the number 
of workers, that is frequency (f
i
) 
is written.
(5) In the fourth column, the product 
(x
i 
´
 
f
i
) for each class is written. 
(6) Then 
i
N


1
x
i  
f
i
 is written.
(7) The mean is found using the 
formula 
Mean = X  = 
å x
i  
f
i
 
N
  = 
1265
50
 = 25.3 
\
 å f
i  
=  N
 The mean time required to complete the work for a worker = 25.3 hrs. (Approx)
ÒÒÒ?? Solved Examples ÒÒÒ
Ex. (1) The percentage of marks of 50 students in a test is given in the following table. 
Find the mean of the percentage. 
Percentage of marks       0-20     20-40     40-60     60-80      80-100
No. of students              3 7 15 20 5
Solution : The following table is prepared as per steps. 
Class 
(Percentage of 
marks)
Class 
mark 
x
i
Frequency 
(No. of 
students)
f
i
Class mark ´ 
frequency
x
i 
f
i
0-20
20-40
40-60
60-80
80-100
10
30
50
70
90
3
7
15
20
5
    30
  210
  750
1400
  450
Total 
N = å f
i
 = 50 å x
i  
f
i 
=  2840
X = 
å x
i  
f
i
å f
i 
   
    = 
2840
50
 
    = 56.8
  \ The mean of 
the percentage 
= 56.8
132
Ex. (2) The maximum temperatures in °C of 30 towns, in the last summer, is shown in 
the following table. Find the mean of the maximum temperatures. 
Max. temp. 24-28 28-32 32-36 36-40  40-44            
No. of towns
4 5 7 8 6
Solution :
Class 
(Temp.  °C)
Class mark 
x
i
Frequency 
(No. of towns) 
f
i
Class mark ´ 
frequency
x
i  
f
i
24-28
28-32
32-36
36-40
40-44
26
30
34
38
42
4
5
7
8
6
104
150
238
304
252
Total N = å f
i
 = 30 å x
i 
f
i 
= 1048
 Mean =  X  = 
å x
i  
f
i
å f
i 
   = 
1048
30
 = 34.9 °C 
    Assumed mean method
 In the examples solved above, we see that some times the product x
i 
f
i 
is 
large. Hence it becomes difficult to calculate the mean by direct method. So let us 
study another method, called the 'assumed mean method'. Finding the mean becomes 
simpler if we use addition and division in this method. 
For example, we have to find the mean of the scores 40, 42, 43, 45, 47 and 48. 
 The obeservation of the scores reveals that the mean of the data is more than 
40. So let us assume that the mean is 40. 40-40 = 0, 42 - 40 = 2, 43-40 = 3,  
45-40 = 5, 47 - 40 = 7, 48 - 40 = 8 These are called 'deviations'. Let us find 
their mean. Adding this mean to the assumed mean, we get the mean of the data. 
 That is, mean = assumed mean + mean of the deviations 
 X  = 40 + 
02 3 578
6
   




     = 40 + 
25
6
  = 40 + 4
1
6
     = 44
1
6
 
133
 Using the symbols-
 A- for assumed mean; d- for deviation and d - for the mean of the deviations, 
the formula for mean of the given data can be briefly written as    X = A + d .  
 Let us solve the same example taking 43 as assumed mean. For this, let us find 
the deviations by subtracting 43 from each score. 
 40 - 43 = -3, 42 - 43 = -1, 43 - 43 = 0, 45 - 43 = 2, 47 - 43 = 4, 48 - 43 = 5 
 The sum of the deviations = -3 -1 + 0 + 2 + 4 + 5 = 7
 Now,  X = A + d  
       = 43 + 
7
6





  (as the number of deviations is 6)
       = 43 + 1
1
6
 = 44
1
6
 Note that; use of assumed mean method reduces the work of calculations. 
 Also note that; taking any score, or any other convenient number as asssumed 
mean does not change the mean of the data. 
Ex. :  The daily sale of 100 vegetable vendors is given in the following table. Find the 
mean of the sale by assumed mean method. 
Daily sale (Rupees)             1000-1500    1500-2000        2000-2500 2500-3000
No. of vendors                        15 20                                      35 30
Solution : Assumed mean = A = 2250, d
i
 = x
i 
-  A is the deviation. 
Class 
Daily sale (Rupees)
Class 
mark 
x
i
d
i
 = x
i
- A
= x
i
- 2250
Frequency (No. 
of vendors)
f
i
Frequency ´ 
deviation
f
i 
d
i
1000-1500
1500-2000
2000-2500
2500-3000
1250
1750
2250 A
2750
-1000
-500
0
500
15
20
35
30
-15000
-10000
0
15000
Total N = å f
i
 = 100 å f
i
 d
i
 
 
=  -10000