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 14.1



Now, ` bar x = (sum f_i x_i)/( sum f_i) = 1779/30 = 59.3`

Therefore, the mean marks obtained is 59.3.

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 fin1.1d 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 (see Table 14.2).



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,

`text (Class mark ) = ( text (Upper class limit) + text (Lower class limit) )/2`

With reference to Table 14.2, for the class 10-25, the class mark is ` (10+25)/2` , i.e,


17.5. Similarly, we can find the class marks of the remaining class intervals. We put
them in Table 14.3. These class marks serve as our `x_i`’s. Now, in general, for the `i^(th)`
class interval, we have the frequency `f_i `corresponding to the class mark xi. We can
now proceed to compute the mean in the same manner as in Example 1.



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

`bar x = (sum f_i x_i )/( sum f_i) = (1860.0)/30 = 62`

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


We observe that Tables 14.1 and 14.3 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 14.3, 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? Let us try this method.

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 `x_i`
which lies in the centre of `x_1, x_2, . . ., 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 di with the corresponding `f_i`, and take the sum
of all the `f_i d_i`’s. The calculations are shown in Table 14.4.




So, from Table 14.4, the mean of the deviations, `bar d = (sum f_i d_i )/( sum f_i )`

Now, let us find the relation between `bar d` and `bar x` .

Since in obtaining `d_i`, we subtracted ‘a’ from each `x_i`, so, in order to get the mean
`bar x` , we need to add ‘a’ to `bar d` . This can be explained mathematically as:

Mean of deviations, `bar d = (sum f_i d_i )/(sum f_i)`


So, `bar d = (sum f_i (x_i -a) )/(sum f_i )`

`= (sum f_i x_i)/(sum f_i) - ( sum f_i a)/( sum f_i )`

`= bar x - a (sum f_i )/(sum f_i)`

`= bar x - a`

So, `bar x = a+ bar d`

i.e., `bar x = a + (sum f_i d_i )/( sum f_i )`

Substituting the values of `a , sum f_i d_i ` and `sum f_i` from Table 14.4, we get

`bar x = 47.5 + 435/30 = 47.5 +14.5 =62`,

Therefore, the mean of the marks obtained by the students is 62.

The method discussed above is called the Assumed Mean Method.

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Solution:

Let us find the class marks, `x_i`, of each class, and put them in a column
(see Table 14.6):



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 14.7.



From the table above, we obtain `Σ f_i = 35, Σ f_i x_i = 1390`,

`Σ f_i d_i = – 360, Σ f_i u_i = –36`.

Using the direct method, `bar x = (sum f_i x_i )/( sum f_i) = 1390/35 = 39.71`

Using the assumed mean method,

`bar x = a + (sum f_i d_i )/(sum f_i ) =50 + ( -360)/(35) = 39.71`

Using the step-deviation method,

`bar x = a + ( (sum f_i u_i )/( sum f_i ) ) xx h = 50 + ( (-36)/(35) ) xx 10 = 39.71`

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

`text ( 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` and `f_i` are
sufficiently small, then the direct method is an appropriate choice. If `x_i` and `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.

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Solution:

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 14.8.



So, `bar u = (-106)/45`. Therefore, `bar x = 200 +20 ( (-106)/45) = 200 -47.11 =152.89`

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!

`text ( Activity 2 : )`

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

1. 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.

2. Collect the daily maximum temperatures recorded for a period of 30 days in your
city. Present this data as a grouped frequency table.

3. 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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