Mean of Grouped Data

The mean (or average) of observations, as we know, is the sum of the values of all the observations divided by the total number of observations. From Class IX, recall that if x1, x2,. . ., xn are observations with respective frequencies f1, f2, . . ., fn, then this means observation x1 occurs f1 times, x2 occurs f2 times, and so on.

Now, the sum of the values of all the observations = f1x1 + f2x2 + ....fnxn, and the number of observations = f1 + f2 + .....fn.

So, the mean x̄ of the data is given by

x̄ = f₁x₁ + f₂x₂ + .... f_nx_nf₁ + f₂ +... +f_n

Recall that we can write this in short form by using the Greek letter ∑(capital sigma) which means summation. That is,

x̄ = ∑_i=1ⁿf_ix_if_i

which, more briefly, is written as x̄ = ∑f_ix_if_i , if it is understood that i varies from 1 to n.

Let us apply this formula to find the mean in the following example.

Example 1 :

The marks obtained by 30 students of Class X of a certain school in a Mathematics paper consisting of 100 marks are presented in table below. Find the mean of the marks obtained by the students.

Marks obtained(x_i)	10	20	36	40	50	56	60	70	72	80	88	92	95
Number of students(f_i)	1	1	3	4	3	2	4	4	1	1	2	3	1

Solution: Recall that to find the mean marks, we require the product of each x_i with the corresponding frequency f_i. So, let us put them in a column as shown in Table.

Marks obtained(x_i)	Number of students(f_i)	f_ix_i
10	1
20	1
36	3
40	4
50	3
56	2
60	4
70	4
72	1
80	1
88	2
92	3
95	1
Total	∑f_i =	f_ix_i

Now = x̄ =∑f_ix_i∑f_i = 177930 =

Therefore, the mean marks obtained is

In most of our real life situations, data is usually so large that to make a meaningful study it needs to be condensed as grouped data. So, we need to convert given ungrouped data into grouped data and devise some method to find its mean.

Let us convert the ungrouped data of Example 1 into grouped data by forming class-intervals of width, say 15. Remember that, while allocating frequencies to each class-interval, students falling in any upper class-limit would be considered in the next class, e.g., 4 students who have obtained 40 marks would be considered in the classinterval 40-55 and not in 25-40. With this convention in our mind, let us form a grouped frequency distribution table.

Class interval	10-25	25-40	40-55	55-70	70-85	85-100
Number of students(f_i)	2	3	7	6	6	6

Now, for each class-interval, we require a point which would serve as the representative of the whole class. It is assumed that the frequency of each classinterval is centred around its mid-point. So the mid-point (or class mark) of each class can be chosen to represent the observations falling in the class. Recall that we find the mid-point of a class (or its class mark) by finding the average of its upper and lower limits. That is

Class mark = Upper class limit + Lower class limit2

With reference to Table, for the class 10-25, the class mark is 10 + 252 =

Similarly, we can find the class marks of the remaining class intervals. We put them in Table. These class marks serve as our x_is. Now, in general, for the ith class interval, we have the frequency f_i corresponding to the class mark x_i We can now proceed to compute the mean in the same manner as in Example 1.

Class interval	Number of students(f_i)	Class mark(x_i)	f_ix_i
10-25	2	17.5
25-40	3
40-55	7
55-70	6
70-85	6
85-100	6
Total	f_i=		f_ix_i

The sum of the values in the last column gives us Σ f_ix_i. So, the mean x of the given data is given by

x̄ =∑f_ix_i∑f_i = 186030 =

This new method of finding the mean is known as the Direct Method.

We observe that Tables are using the same data and employing the same formula for the calculation of the mean but the results obtained are different. Can you think why this is so, and which one is more accurate? The difference in the two values is because of the mid-point assumption in Table 59.3 being the exact mean, while 62 an approximate mean.

Sometimes when the numerical values of x_i and f_i are large, finding the product of x_i and f_i becomes tedious and time consuming. So, for such situations, let us think of a method of reducing these calculations.

We can do nothing with the f_i's but we can change each x_i to a smaller number so that our calculations become easy. How do we do this? What about subtracting a fixed number from each of these x_i's

The first step is to choose one among the x_i's as the assumed mean, and denote it by ‘a’. Also, to further reduce our calculation work, we may take ‘a’ to be that x_iwhich lies in the centre of x₁,x₂,.... x_n. So, we can choose a = 47.5 or a = 62.5. Let us choose a = 47.5.

The next step is to find the difference d_i between a and each of the x_i's, that is, the deviation of ‘a’ from each of the x_i's.

i.e., d_i = x_i - a = x_i – 47.5

The third step is to find the product of d_i with the corresponding f_i , and take the sum of all the f_i d_i's. The calculations are shown in Table.

Class interval	Number of students(f_i)	Class mark(x_i)	d_i = x_i– 47.5	f_id_i
10-25	2	17.5	-30
25-40	3	32.5
40-55	7	47.5
55-70	6	62.5
70-85	6	77.5
85-100	6	92.5
Total	f_i=30			f_id_i=

So, from Table 13.4, the mean of the deviations, x̄ =∑f_id_i∑f_i.

Example 2

The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India. Find the mean percentage of female teachers by all the three methods discussed in this section.

Percentage of female teachers	15-25	25-35	35-45	45-55	55-65	65-75	75-85
Number of States/U.T.	6	11	7	4	4	2	1

Solution:

Let us find the class marks, x_i of each class, and put them in a column:

Percentage of female teachers	Number of States/U.T.	x_i
15 - 25	6	20
25 - 35	11
35 - 45	7
45 - 55	4
55 - 65	4
65 - 75	2
75 - 85	1

Here we take a = 50, h = 10, then d_i = x_i-50 and u_i = x_i - 50 10

We now find d_i and u_i and put them in Table

Percentage of female teachers	Number of States/U.T.	x_i	d_i = x_i-50	u_i = x_i-50 / 10	f_ix_i	f_id_i	f_iu_i
15 - 25	6	20	-30	-3	120	-180	-18
25 - 35	11	30	-20
35 - 45	7	40	-10
45 - 55	4	50	0
55 - 65	4	60	10
65 - 75	2	70	20
75 - 85	1	80	30
Total	35

From the table above, we obtain ∑f_i =35, ∑f_ix_i = 1390,

∑f_id_i = – 360, ∑f_iu_i = -36.

Using the direct method, x̄ =∑f_ix_i∑f_i = 139035 = .

Therefore, the mean percentage of female teachers in the primary schools of rural areas is 39.71.

Remark : The result obtained by all the three methods is the same. So the choice of method to be used depends on the numerical values of x_i and f_i. If x_i f_i are sufficiently small, then the direct method is an appropriate choice. If x_i f_i are numerically large numbers, then we can go for the assumed mean method or step-deviation method. If the class sizes are unequal, and x_i are large numerically, we can still apply the step-deviation method by taking h to be a suitable divisor of all the d_i's.

Example 3

The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify?

Number of wickets	20-60	60-100	100-150	150-250	250-350	350-450
Number of bowlers.	7	5	16	12	2	3

Here, the class size varies, and the x_i's are large. Let us still apply the stepdeviation method with a = 200 and h = 20. Then, we obtain the data as in Table.

Number of wickets taken	Number of bowlers(f_i)	x_i	u_i=d_i/20
20 - 60	7	40
60 - 100	5	80
100 - 150	16	125	]
150 - 250	12	200
250 - 350	2	300
350 - 450	3	400

So,(\bar{u}) = −10645⋅ Therefore, x̄ = = 200 + 20 10645 = 200– 47.11 = .

This tells us that, on an average, the number of wickets taken by these 45 bowlers in one-day cricket is 152.89.

Now, let us see how well you can apply the concepts discussed in this section!

Activity 2 :

Divide the students of your class into three groups and ask each group to do one of the following activities.

Collect the marks obtained by all the students of your class in Mathematics in the latest examination conducted by your school. Form a grouped frequency distribution of the data obtained.
Collect the daily maximum temperatures recorded for a period of 30 days in your city. Present this data as a grouped frequency table.
Measure the heights of all the students of your class (in cm) and form a grouped frequency distribution table of this data.

After all the groups have collected the data and formed grouped frequency distribution tables, the groups should find the mean in each case by the method which they find appropriate.

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Mean of Grouped Data

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Glossary

Mean of Grouped Data

Example 3