Class 9 – Statistics
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Introduction: We already learn how to find mean, median and mode of ungrouped data. In this synopsis we are going to learn about mean of grouped data. In this synopsis we are going to learn about mean of grouped data. Before going to discuss about mean of ungrouped data, we just recall mean of a regrouped data. Mean of grouped data: We find the mean of grouped data in three ways. 1. Direct method 2. Shortcut (or) assumed mean method 3. Step deviation method Direct method: As we know the mean (or average) of observations is the sum of the values of all the observations divided by the total number of observations. Let be observations with respective frequencies This means that observation Occurs times, occurs times, and so on. Now, the sum of the values of all the observations and the number of observations . So, the mean of the data is given by 
Recall that we can write this in short, using the Greek letter Σ which means summation i.e., Example 1. The following table shows the weights of 15 members of an athletic team in a school.
Find the mean weight. Solution: Form the above data, we may prepare the table given below:
Mean weight = = = 45.2 kg. 

Example 2. The marks obtained by a set of students in an examination are given below:
Marks 
5 
10 
15 
20 
25 
30 
Number of students 
6 
4 
6 
12 
x 
4 
If the mean of the above data is 18, calculate the numerical value of x.
Solution: From the above data, we may prepare the table given below:
Marks 
Number of athletes (frequency) 

5 10 15 20 25 30 
6 4 6 12 X 4 
30 40 90 240 25x 120 

= (32 + x) 
Mean = =
But Mean = 18.
= 18 Þ 520 + 25x = 18 (32 + x) Þ 7x = 56 Þ x = 8
Hence, x = 8.
Shortcut Method: If the values of x or f are large, then direct method to calculate mean is not used because it will be timeconsuming and difficult. In such a case, we take deviations, from an arbitrary point a called assumed mean, then
, where di = xi a.
Example: Find the mean of given data in assumed mean method.
Class 
1025 
2540 
4055 
5570 
7085 
85100 
F 
2 
3 
7 
6 
6 
6 
Solution:
Class interval 
Number student (fi) 
Class Marks (xi) 
di = xi – 47.5 
fidi 
1025 2540 4055 5570 7085 85100 
2 3 7 6 6 6

17.5 32.5 47.5(a) 62.5 77.5 92.5 
30 15 0 15 30 45 
60 45 0 90 180 270

Total 
So, form the above table, the mean of the deviations, =
Now, let us find the relation between d and
Since, in obtaining di we subtracted a from each xi so, in order to get the mean We need to add a to . This can be explained mathematically as:
Mean of deviations, =
So,

d =  a
Therefore
Substituting the values of a, and from the table, we get
x = 47.5 +
Therefore, the mean of the marks obtained by the students is 62.
Any number can be taken as assumed mean a but for the sale of fi.
Any number can be taken as assumed mean a but fo the sake of simplicity, it is generally taken as that value of the variable which has the greatest frequency in the frequency distribution or which is near about the middle of the frequency distribution.
Step deviation Method: When we use shortcut method to find mean, then the deviations, (di) are divisible by a common number (h). In such a case, the mean is reduced by taking
Where
Example: Using step deviation method, calculate the mean for the following data:
Height (in cm) 
135140 
140145 
145150 
150155 
155160 
160165 
165170 
170175 
Number of boys 
4 
9 
18 
28 
24 
10 
5 
2 
Solution: Here, class size (c) = 5. Take assumed mean (A) = 152.5 Thus, from the given data, we may prepare the table given below:
ClassInterval 
Classmark xi 
Frequency fi 
fiui 

135140 140145 145150 150155 155165 165170 170175 
137.5 142.5 147.5 152.7 = A 157.5 1652.5 167.5 172.5 
4 9 18 28 24 10 5 2 
3 2 1 0 1 2 3 4 
12 18 18 0 24 20 15 8 

Mean = = (152.5 + 0.95) = 153.45 cm.
Example: The following table gives the frequency distribution of the percentage of marks obtained by 2300 students in a competitive examination.
Marks obtained 
1120 
2130 
3140 
4150 
5160 
6170 
7180 
Number of students 
141 
221 
439 
529 
495 
322 
153 
Find the mean percentage of marks of the group.
Solution: First we convert the given frequency distribution in continuous form (Exclusive class interval).
Adjustment factor =
So, we may prepare the table given below:
ClassInterval 
ClassMark xi 
Frequency fi 
fiui 

10.520.5 20.530.5 30.540.5 40.550.5 50.560.5 60.570.5 70.580.5 
15.5 25.5 35.5 45.5 = A 55.5 65.5 75.5 
141 221 439 529 495 322 153 
3 2 1 0 1 2 3 
423 442 439 0 495 644 459 

Mean = = = 45.5 + 1.278 =46.778
Step deviation method is a very short method and should always be used for gropued data where class interval sizes are equal.
The mean of 10 observations is 16.3. By an error, one observation is registered as 32 instead of 23. Find the correct mean.
Solution: Here n = 10,
But this value of is incorrect as 32 is registered instead of 23.