Class 8 – Statistics
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Introduction: Statistics is the study of collecting, organizing and analyzing sets of data in order to reveal information. Data: A collection of numerical facts about objects or events is called data. For example. The percentage of marks scored by 10 students of a class in a test are: 72, 84, 82, 96, 94, 98, 99, 67, 92 and 93. The set of these numerical values is the data related to the marks obtained by 10 students in a class test. Ungrouped data: The data obtained in original form are called raw data or ungrouped data. Clearly, the data shown above is a raw data. Grouped data: If the data extends over a wide range, such tables become lengthy and unwieldy. In such cases, statisticians usually condense the data into more usable form, i.e., in groups or classes. For example, consider the following ungrouped marks out of 50 awarded to 30 students. 3, 5, 8, 15, 25, 30, 16, 7, 35, 40, 40, 30, 15, 14, 21, 23, 22, 25, 27, 29, 32, 15, 1, 9, 11, 14, 42, 43, 49, 8. 
Let us arrange them in the following array, i. e., in ascending order 1, 3, 5, 7, 8, 8, 9, 11, 14, 14, 15, 15, 16, 14, 21, 22, 23, 25, 25, 27, 29, 30, 30, 32, 35, 40, 40, 42, 43, 49. Arrangement in ascending or descending order is called an array. We can classify these marks into groups or classes as shown in the table given below. The groups 0 – 5, 5 – 10 etc. are called the class intervals. Each class interval is bounded by two figures which are called the class limits. The class limits of the first class. For example, the numbers 0 and 5 and those of the second class are 5 and 10. 

The larger of these numbers namely 5 is called the upper limit, and the smaller number 0 is called the lower limit. Similarly, the upper and the lower limits of the second class are 10 and 5 respectively, and so on.
It may be noted that:
i) The upper limit of one class coincides with the lower limit of the next class.
ii) The class 0 and 5 means 0 and less than 5; 510 means 5 and less than 10, and soon. Here the upper limit 5 belongs to the next class 510 and not to the class05. Similar is the case with other classes.
Types of frequency Distributions: The frequency distribution of the form 010, 1020, 2030,… in which the upper limit of one class interval is the same as the lower limit of the next class interval is called an exclusive distribution. There is another type of frequency distribution of the form 09, 1019, 2029,…. in which the upper limit of one class does not coincide with the lower limit of the next class interval. Such a frequency distribution is called an inclusive distribution.In this type of distribution both the lower and upper limits of a class belong to that class only, i.e., both 1 and 5 belong to the class 15.
i) In exclusive classes 05, 510, 1015, …… 0 is lower limit (boundary) and 5 is upper limit (boundary) of the class 05.
ii) In exclusive classes 110, 1120, 2130, …….‘1’ is lower limit and ‘10’ is upper limit of the class 110. But lower boundary of class 1120 = upper boundary of 1120 =
Class size, Class – Mark and Class Frequency:
i) The difference between the upper and the lower boundaries of the same class (or) the difference between two successive upper limits (lower limits) class is called the size or width of the class interval. For example, in the table, the size of the class interval
05 is 50 = 5, of 510 is 105 = 5, of 1015 is 1510 = 5, and so on.
ii) The midvalue of a classinterval is called its class mark. Thus, the class mark of the class interval 10 – 15 is (Average of lower and upper limits of the class is the mid value of that class interval).
The frequency of a class interval is called its class frequency. Thus, for the above given table the class frequency of the class interval 05 is 2, of 510is 5, of 1015 is 3 and so on.
Example: In an examination 40 boys secured the following marks:
8,11, 20, 37, 40, 15, 29, 31, 27, 8,
7,13, 29, 25, 42, 37, 30, 10, 9, 27,
18,25, 9, 2, 17, 47, 32, 11, 29, 6,
15,41, 37, 10, 40, 21, 39, 13, 15, 3.
Represent the data by a frequency table.
Solution: Arranging the given marks in ascending order, we have
2, 3,6, 7, 8, 8, 9, 9, 10, 10, 11, 11, 11, 13, 13, 15, 15, 15, 18, 20, 21, 25, 25,27, 27, 29, 29, 29 , 30, 31, 32, 37, 37, 37, 39, 40, 40, 41, 42, 47.
Classes 
Frequency (f) 
010 
8 
1020 
11 
2030 
9 
3040 
7 
4050 
5 
Total 
40 
Here,the maximum marks secured = 47,
the minimum marks secured = 2
∴ Range = 47 – 2 = 45
Suppose we want to form 5 classes, then the class interval should be ,i.e., 9 or roughly 10 as we prefer 5 or 10 or 20, etc. Therefore the class intervals are 0 – 10, 10 – 20, 20 – 30, 30 – 40, 40 – 50 (Exclusive classes).
The frequency distribution is as given in the table above.
Size of a class interval
Mean of grouped data:
Mean: The average of all the numbers given is called the arithmetic mean denoted by .
We can calculate mean by many methods.
i) Direct method
ii) Assumed mean or deviation method.
a) Direct Method: The mean is given by ,where
xi = mid value of the ith class interval
fi = frequency of ith class interval
b) Assumed Mean Method (or) Deviation Method:
Steps:
i) Calculate class marks (xi) of given class intervals
ii) Choose a suitable number ‘a’, lying in the middle of xi, known as assumed mean.
iii) For each class mark (xi)subtract ‘a’ to find a new variable ‘y’ such that .
iv) Multiply the new variables (the value of yi obtained in step (iii) by the corresponding frequency.
v) Take sum of all the products obtained in step (iv).
vi) Divide the sum obtained in step (v) by the sum of frequencies.
vii) The result of step (vi) is the mean of new variables yi.
viii) The mean of given data is ,where
Example: Find the mean of following grouped frequency distribution by
a)Direct method b) Assumed Mean Method.
Class interval 
20 – 30 
30 – 40 
40 – 50 
50 – 60 
60 – 70 
Frequency (fi) 
6 
12 
19 
8 
5 
Solution:
a) Direct method
Class Interval (CI) 
Freq.(fi) 
Class mark (xi) 
fixi 
20 – 30 
6 
150 

30 – 40 
12 
420 

40 – 50 
19 
855 

50 – 60 
8 
440 

60 – 70 
5 
325 

50 
2190 
Mean
Here
.
∴ Mean of given data =43.8
b) Assumed Mean Method
Class Interval (CI) 
Freq. (fi) 
Class mark (xi) 
Deviations 

20 – 30 
6 
25 
20 
120 
30 – 40 
12 
35 
10 
120 
40 – 50 
19 
45 (a) 
0 
0 
50 – 60 
8 
55 
10 
80 
60 – 70 
5 
65 
20 
100 
Total 
50 
60 
Let assumed mean ‘a’ = 45
Mean
Here n= 5,
Step deviation method: When the values of xi and fi are large and the values of xi are equally spaced, we use the step deviation method for finding the mean.
Steps:
1. Choose a suitable value of xi in the middle as the assumed mean and denote it by ‘a’.
2. Calculate h = upper limit – lower limit.
3. Calculate for each i.
4. Calculate fiui for each i and hence find ∑fiui
5. Calculate the mean using the formula