Please Wait... While Loading Full Video#### Class 10 Chapter 14

### Introduction And Mean of Grouped Data

♦ Introduction

♦ Mean of Grouped Data

♦ Mean of Grouped Data

You have studied the classification of given data into ungrouped as well as grouped frequency distributions. You have also learnt to represent the data pictorially in the form of various graphs such as bar graphs, histograms (including those of varying widths) and frequency polygons.

In fact, you went a step further by studying certain numerical representatives of the ungrouped data, also called measures of central tendency, namely, mean, median and mode.

Here, we shall extend the study of these three measures, i.e., mean, median and mode from ungrouped data to that of grouped data.

We shall also discuss the concept of cumulative frequency, the cumulative frequency distribution and how to draw cumulative frequency curves, called ogives.

In fact, you went a step further by studying certain numerical representatives of the ungrouped data, also called measures of central tendency, namely, mean, median and mode.

Here, we shall extend the study of these three measures, i.e., mean, median and mode from ungrouped data to that of grouped data.

We shall also discuss the concept of cumulative frequency, the cumulative frequency distribution and how to draw cumulative frequency curves, called ogives.

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.

Recall that if `x_1, x_2,. . ., x_n` are observations with respective frequencies `f_1, f_2, . . ., f_n`, then this means observation `x_1` occurs `f_1` times, `x_2` occurs `f_2` times, and so on.

Now, the sum of the values of all the observations `= f_1 x_1 + f_2 x_2 + . . . + f_n x_n`, and

the number of observations `= f_1 + f_2 + . . . + f_n`.

So, the mean `bar x` of the data is given by

` color { red} { bar x = (f_1 x_1 +f_2 x_2 + ............+ f_n x_n )/( f_1 + f_2 +........+ f_n) } `

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

` color { blue } { bar x = ( sum_(i=1)^n (f_i x_i))/( sum_(i=1)^n f_i) }`

which, more briefly, is written as ` color { green } { bar x = (sum f_i x_i )/( sum f_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.

Recall that if `x_1, x_2,. . ., x_n` are observations with respective frequencies `f_1, f_2, . . ., f_n`, then this means observation `x_1` occurs `f_1` times, `x_2` occurs `f_2` times, and so on.

Now, the sum of the values of all the observations `= f_1 x_1 + f_2 x_2 + . . . + f_n x_n`, and

the number of observations `= f_1 + f_2 + . . . + f_n`.

So, the mean `bar x` of the data is given by

` color { red} { bar x = (f_1 x_1 +f_2 x_2 + ............+ f_n x_n )/( f_1 + f_2 +........+ f_n) } `

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

` color { blue } { bar x = ( sum_(i=1)^n (f_i x_i))/( sum_(i=1)^n f_i) }`

which, more briefly, is written as ` color { green } { bar x = (sum f_i x_i )/( sum f_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.

Q 3250134014

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.

Class 10 Chapter 14 Example 1

Mathematics paper consisting of 100 marks are presented in table below. Find the

mean of the marks obtained by the students.

Class 10 Chapter 14 Example 1

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.

From the Table 14.3 find the mean by taking each of `x_i` (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.

So, we can say that the value of the mean obtained does not depend on the choice of ‘a’.

Observe that in Table 14.4, the values in Column 4 are all multiples of 15. So, if we divide the values in the entire Column 4 by 15, we would get smaller numbers to multiply with `f_i`. (Here, 15 is the class size of each class interval.)

So, let `u_i = (x_i -a)/h` , 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 `f_i u_i` and

then ` Σ f_i u_i `). Taking `h = 15`, let us form Table 14.5.

Let `bar u = (sum f_i u_i )/(sum f_i )`

Here, again let us find the relation between `bar u` and `bar x` .

We have, ` u_i = (x_i -a )/h`

Therefore, ` color { orange } { bar u = (sum f_i (x_i -a)/h )/( sum f_i) =1/h [ (sum f_i x_i - a sum f_i )/(sum f_i ) ] } `

`color { violet } { = 1/h [ (sum f_i x_i )/(sum f_i) -a (sum f_i )/(sum f_i) ] } `

`= 1/h [bar x -a ]`

So, `h bar u = bar x - a`

i.e., `x = a + h bar u`

so, `bar x = a+h ( (sum f_i u_i )/(sum f_i ) )`

Now, substituting the values of `a, h, Σf_i u_i` and `Σ f_i` from Table 14.5, we get

`bar x = 47.5 +15 xx (29/30)`

`=47.5 +14.5 = 62`

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

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

We note that :

♦ the step-deviation method will be convenient to apply if all the `d_i`’s have a

common factor.

♦ The mean obtained by all the three methods is the same.

♦ The assumed mean method and step-deviation method are just simplified

forms of the direct method.

♦ The formula `bar x = a + h bar u` still holds if a and h are not as given above, but are

any non-zero numbers such that `u_i = (x_i -a )/h`

Let us apply these methods in another example.

So, we can say that the value of the mean obtained does not depend on the choice of ‘a’.

Observe that in Table 14.4, the values in Column 4 are all multiples of 15. So, if we divide the values in the entire Column 4 by 15, we would get smaller numbers to multiply with `f_i`. (Here, 15 is the class size of each class interval.)

So, let `u_i = (x_i -a)/h` , 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 `f_i u_i` and

then ` Σ f_i u_i `). Taking `h = 15`, let us form Table 14.5.

Let `bar u = (sum f_i u_i )/(sum f_i )`

Here, again let us find the relation between `bar u` and `bar x` .

We have, ` u_i = (x_i -a )/h`

Therefore, ` color { orange } { bar u = (sum f_i (x_i -a)/h )/( sum f_i) =1/h [ (sum f_i x_i - a sum f_i )/(sum f_i ) ] } `

`color { violet } { = 1/h [ (sum f_i x_i )/(sum f_i) -a (sum f_i )/(sum f_i) ] } `

`= 1/h [bar x -a ]`

So, `h bar u = bar x - a`

i.e., `x = a + h bar u`

so, `bar x = a+h ( (sum f_i u_i )/(sum f_i ) )`

Now, substituting the values of `a, h, Σf_i u_i` and `Σ f_i` from Table 14.5, we get

`bar x = 47.5 +15 xx (29/30)`

`=47.5 +14.5 = 62`

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

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

We note that :

♦ the step-deviation method will be convenient to apply if all the `d_i`’s have a

common factor.

♦ The mean obtained by all the three methods is the same.

♦ The assumed mean method and step-deviation method are just simplified

forms of the direct method.

♦ The formula `bar x = a + h bar u` still holds if a and h are not as given above, but are

any non-zero numbers such that `u_i = (x_i -a )/h`

Let us apply these methods in another example.

Q 3270634516

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.

Class 10 Chapter 14 Example 2

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.

Class 10 Chapter 14 Example 2

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.

Q 3210634519

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?

Class 10 Chapter 14 Example 3

one-day cricket matches. Find the mean number of wickets by choosing a suitable

method. What does the mean signify?

Class 10 Chapter 14 Example 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 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.

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.

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.