Mean of UnGrouped Data

The mean (or average) of observations, as we know, is the sumproduct of the values of all the observations divided by the totalhalf of the 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

= 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,

= ∑_i=1ⁿf_ix_if_i

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

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

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

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 .

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 class-limit would be considered in the next class, e.g., 4 students who have obtained marks would be considered in the class-interval 40-55 and not in 25-40.

With this convention in our mind, let us form a grouped frequency distribution table.

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

=∑f_ix_i∑f_i = 186030 =

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

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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 differentsame.

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 meanmode, while 62 an approximate meanmode.

Sometimes when the numerical values of xi and fi are large, finding the product of xi and fi 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 fi's but we can change each xi 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 xi's.

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The first step is to choose one among the xi's as the assumed mean, and denote it by ‘a’. Also, to further reduce our calculation work, we may take ‘a’ to be that xi which lies in the centre of x1,x2,.... xn. 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 di between a and each of the xi's, that is, the deviation of ‘a’ from each of the xi's.

i.e., di = xi - a = xi –

The third step is to find the product of di with the corresponding fi , and take the sum of all the fi di's. The calculations are shown in the below table.

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

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Activity 1 : From the Direct Method table, find the mean by taking each of xi(i.e., 17.5, 32.5,and so on) as ‘a’.

What do you observe? You will find that the mean determined in each case is the same, i.e., 62. (Why?)

So, we can say that the value of the mean obtained does notdoes depend on the choice of ‘a’.

Observe that in the above given table, the values in Column 4 are all multiples of 1530.

So, if we divide the values in the entire Column 4 by 15, we would get smaller numbers to multiply with fi. (Here, 15 is the class size of each class interval.)

So, let ui = xi−ah, where a is the assumed mean and h is the class size.

Now, we calculate ui in this way and continue as before (i.e., find fiui and then Σfiui).

Taking h = 15, let us form the following table:

Let = ΣfiuiΣfiHere, again let us find the relation between and .

We have, ui = xi−ah

Therefore, = Σfixi−ahΣfi = 1h [Σfixi−aΣfiΣfi]

= 1h [ΣfixiΣfi - aΣfiΣfi]

= 1h[-a]

So, h = - a

i.e., = a + h

So, = a + h(ΣfiuiΣfi)

Now, substituting the values of a, h, Σfiui and Σfi from Table 14.5, we get

= + × (2930)

= 47.5 + =

So, the mean marks obtained by a student is 62.

The method discussed above is called the Step-deviationdirect method.

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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.

Solution:

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

Here we take a = 50, h = 10, then di = xi -50 and ui = xi−5010xi+5010

We now find di and ui and put them in Table

Table

From the table above, we obtain ∑fi = , ∑fi xi = ,

∑fi di = , ∑fi ui = .

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Using the direct method, =∑f_ix_i∑f_i = 139035 = . (Upto two decimal places)

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 .

So the choice of method to be used depends on the numerical values of xi and fi.

If xi fi are sufficiently small, then the direct method is an appropriate choice. If xi fi are numerically large numbers, then we can go for the assumed meandirect method or step-deviationdirect method.

If the class sizes are unequal, and xi are large numerically, we can still apply the step-deviation method by taking h to be a suitable divisor of all the di's.

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?

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

So, = −10645⋅

Therefore, = = 200 + 20 10645 = 200 – 47.1146 = 152.89-152.

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!

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