Chapter 37.6 - The Ogive curve

In the previous section we completed a discussion about median. In this section we will see Ogive curves.
• We know that one picture can replace a thousand words. We have seen examples of this in our previous discussions in statistics (part II). There we saw bar graphs, histograms, frequency polygons etc., 
• We saw that those graphs help to get an understanding about the data at a glance. 
• In our present discussion, we have not yet drawn any graphs. So let us try to 'represent pictorially', what we have learned so far in this chapter. We will write it in steps:

1. Consider the table 37.39 that we saw earlier. It is shown again below:

Table.37.39

• Let us plot the cumulative frequencies along the y axis
• Then the class intervals will be plotted along the x axis
• But we see that each class interval has two values: a lower limit and an upper limit. Which one will we use for plotting?
2. Let us choose the upper limits
• Then the coordinates of the points on the graphs will be: 
(10,5), (20,8), (30,12), (40,15), . . . , (100,53)
• The graph thus obtained is shown in fig.37.1 below:

Fig.37.1

• Note that the scale need not be the same for both the axes
• The curve (shown in magenta colour) joining the points is called a cumulative frequency curve or an ogive
3. So we know how to draw an ogive. Let us now see it's features:
• Suppose we want to know the cumulative frequency corresponding to a class 70-80
• Then we take the point on the graph at which the x coordinate is 80
• At that point, the y coordinate is 38. 
• That means the cumulative frequency at that point is 38
• So the cumulative frequency corresponding to class 70-80 is 38
• That means, 38 observations in the given data have a value lesser than 80
4. In our present example, the observations are: 'marks obtained by 53 students'
• So we can write:
38 students have marks less than 80
• If the observations are 'weights', we can write:
38 students have weights less than 80 kg
• If the observations are 'lengths of leaves', we can write:
38 leaves have length less than 80 mm'
5. So it is a convenient way to get a quick understanding about the data
• Note that we use less than 80. This is because, if there is a mark equal to 80, it will not be included in the class 70-80. It will be included only in 80-90
• So all marks equal to and greater than 80 will be on the right side of (80,38)
6. So which ever point we consider in the graph in fig.37.1 above, we get valuable information. We will break down the information as follows:
(i) The coordinates of the point will be in the form: (x coordinate, y coordinate)
(ii) 'y coordinate' number of students have scored less than 'x coordinate' marks
An example:
• Take class mark 50 on the x axis
• The y coordinate corresponding to it is 18. So we have: (50,18)
■ We can write:
18 students have scored less than 50 marks
7. So the graph in fig.37.1 is marked as 'Less than ogive'
8. (80, 38) and (50,18) are points already written on the ogive. We can get information on other points also:
(i) Consider the score 75. We want to know how many students got less than 75
(ii) For that, mark a point R at (75,0) on the x axis. 
(iii) Draw a vertical dashed line (shown in white colour in fig.37.2 below) through R

Fig.37.2

(iv) Let it intersect the ogive at Q
(v) Draw a horizontal white dashed line through Q
(vi) Let it intersect the y axis at P
(vii) Note down the 'y coordinate' of P. It is 33.5
• We can write:
About 33 students have scored less than 75 marks
• Note that, if we are using a graph paper, we will not need to draw the dashed lines. They will be already present in the form of thin lines. 

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So we have seen the Less than ogive. Is there a 'More than ogive' ? Let us see:
1. Consider the table 37.40 that we saw earlier. It is from the same problem. It is shown again below:

Table.37.40

• Let us plot the cumulative frequencies along the y axis
• Then the class intervals will be plotted along the x axis
• But we see that each class interval has two values: a lower limit and an upper limit. Which one will we use for plotting?
2. Let us choose the lower limits
• Then the coordinates of the points on the graphs will be: 
(0,53), (10,48), (20,45), (30,41), . . . , (90,8)
• The graph thus obtained is shown in fig.37.3 below:

Fig.37.3

• Note that the scale need not be the same for both the axes
• The curve (shown in yellow colour) joining the points is a cumulative frequency curve or an ogive
3. So we know how to draw this new type of ogive also. Let us now see it's features:
• Suppose we want to know the cumulative frequency corresponding to a class 70-80
• Then we take the point on the graph at which the x coordinate is 70
• At that point, the y coordinate is 24. 
• That means the cumulative frequency at that point is 24
• So the cumulative frequency corresponding to class 70-80 is 24
• That means, 24 observations in the given data have a value more than 70
4. In our present example, the observations are: marks obtained by 53 students
• So we can write:
24 students have marks more than 70
• If the observations are 'weights', we can write:
24 students have weights more than 70 kg
• If the observations are 'lengths of leaves', we can write:
24 leaves have length more than 70 mm
5. So it is a convenient way to get a quick understanding about the data
• Note that we use more than 70. 
6. So which ever point we consider in the graph in fig.37.2 above, we get valuable information. We will break down the information as follows:
(i) The coordinates of the point will be in the form: (x coordinate, y coordinate)
(ii) 'y coordinate' number of students have scored more than 'x coordinate' marks
An example:
• Take class mark 50 on the x axis
• The y coordinate corresponding to it is 35. So we have: (50,35)
■ We can write:
35 students have scored more than 50 marks
7. So the graph in fig.37.3 is marked as 'More than ogive'
8. (70, 24) and (50,35) are points already written on the ogive. We can get information on other points also. 
• For that, draw the required vertical and horizontal dashed lines just as we saw in the case of less than ogive.

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Now we will see an interesting application of the 'less than ogive'. We will write it in steps:
1. In fig.37.2 above, we started to work from point R which is on the x axis. 
• Then we reached Q and then finally P, which is on the y axis. 
• The point P gave us the 'number of students'
2. How about working in a reverse order?
• If we start from a point P on the y axis, we will be starting with a 'particular number of students'
• When we reach the x axis, we will get a 'particular score' Q
• The no. of students at P will have scored less than the marks at Q
3. Do we have any 'particular number of students' which might be of interest?
• We certainly do. 
• 'Half the number' is often an important point in any data. It is related to 'the median'
• In our present case, the median is (⁵³⁄₂) = 26.5
4. So we mark P (0,26.5) on the y axis and start from there. The path to R on the x axis is shown in fig.37.4 below:

Fig.37.4

• We get R (66.4,0). So we can write:
About 26 students scored less than 66.4 The other 26 scored more than 66.4
• In other words, 66.4 is the median score
■ So it is clear: To find the median, mark (0,ⁿ⁄₂) on the y axis and work towards the x axis
5. We proved it using a less than ogive. Will it work on a more than ogive?
The path from P to R is marked on the more than ogive in fig.37.5 below:

Fig.37.5

• We get the same result: The median is 66.4
7. So it means that point Q has the same coordinates in both less than and more than ogives
That is., Q is common in both the curves
In other words, Q is the point of intersection of the two curves. This is shown in the fig.37.6 below:

Fig.37.6

8. So, instead finding (ⁿ⁄₂) and then drawing horizontal and vertical dashed lines, we can do the following steps:
(i) Draw both the ogives 
(ii) Mark the point of intersection Q
(iii) Draw a vertical dashed line through Q. Let it meet the x axis at R
(iv) then the x coordinate of R is the median

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An example:
The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution :

Table.37.53

Draw both ogives for the data above. Hence obtain the median profit.
Solution:
1. We are given a cumulative frequency table. 
• The cumulative frequencies in the second column can be used to draw the more than ogive. But those cumulative frequencies will give only the y coordinates of the points. 
• To know the x coordinates, we need class intervals. 
2. So we will convert the given 'cumulative frequency distribution table' into a 'grouped frequency distribution table':
• Consider the first row of the given table:
All the shops are making a profit of more than or equal to 5 lakhs 
• Consider the second row of the given table:
28 shops are making a profit of more than or equal to 10 lakhs 
• 'More than or equal to 10' will include 'more than or equal to 5' also
• So number of shops making a profit of more than 5 lakhs but less than 10 lakhs is (30-28) = 2
• So the frequency corresponding to class 5-10 will be 2
• This is shown in the first row of our new table 37.54 below
3. Consider the third row of the given table:
• 16 shops are making a profit of more than or equal to 15 lakhs 
• 'More than or equal to 15' will include 'more than or equal to 10' also
• So number of shops making a profit of more than 10 lakhs but less than 15 lakhs is (28-16) = 12
• So the frequency corresponding to class 10-15 will be 12
• This is shown in the second row of our new table 37.54 below
4. Consider the fourth row of the given table:
• 14 shops are making a profit of more than or equal to 20 lakhs 
• 'More than or equal to 20' will include 'more than or equal to 15' also
• So number of shops making a profit of more than 15 lakhs but less than 20 lakhs is (16-14) = 2
• So the frequency corresponding to class 15-20 will be 2
• This is shown in the third row of our new table 37.54 below

Table.37.54

In this way we can continue up to the last row. But once we understand the pattern, the columns can be easily filled up.
5. Now we can draw the ogives. 
(i) First we will draw the less than ogive.
• We know that, for the less than ogive, the x coordinates are the upper limits of the class intervals
• y coordinates are the cumulative frequencies of the less than type. This is given in the third column of the above table 37.53
    ♦ As usual, this column is easily filled up using the pattern
• So the coordinates will be: (10,2), (15,14), (20,16), . . . , (40,30)
• The curve joining these points is the less than ogive of this problem. It is shown in magenta colour in fig.37.7 below:

Fig.37.7

• A sample:
    ♦ (30,23) is a point on the less than ogive
    ♦ So there are 23 shops which make a profit of less than 23 lakhs
(ii) Now we will draw the more than ogive.
• We know that, for the more than ogive, the x coordinates are the lower limits of the class intervals
• y coordinates are the cumulative frequencies of the more than type. They were given to us in the question. They are shown again in the fourth column of the above table 37.53
• So the coordinates will be: (5,30), (10,28), (15,16), . . . , (35,3)
• The curve joining these points is the more than ogive of this problem. It is shown in yellow colour in fig.37.7 above.
• A sample:
    ♦ (25,10) is a point on the more than ogive
    ♦ So there are 10 shops which make a profit of more than 25 lakhs
6. To find the median:
• Let the two ogives intersect at Q. From Q, draw a vertical dashed line. It will intersect the x axis at R
• The coordinates of R are (17.5,0)
• So the median profit is 17.5 lakhs
• Note that, the y coordinate of Q is 15.
    ♦ This 15 is equal to (ⁿ⁄₂) = (³⁰⁄₂)
7. Conclusion:
• If we arrange all the 30 shops in the increasing order of profit, the shop which makes a profit of 17.5 lakhs will come in the middle.
• There will be about 15 shops on the left of this shop. Each of their profits will be less than 17.5 lakhs
• There will be about 15 shops on the right of this shop. Each of their profits will be more than 17.5 lakhs

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The term ‘ogive’ is pronounced as ‘ojeev’ and is derived from the word 'ogee'. An ogee is a curved shape consisting of  concave and convex arcs. It was used to form arches of buildings in the 14th and 15th century Gothic styles. Our ogive curves in statistics resemble those curves, and so the name was given.

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In the next section, we will see some solved examples.

The link below gives the notes on statistics of class 11:

Mean deviation and Standard deviation.

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Chapter 37.3 - Mode of the Data

In the previous section we completed a discussion on mean. We also saw some solved examples demonstrating all three methods to find mean. In this section we will discuss about mode.

We have seen some basics about mode in part II (Details here). Let us see a new example:
Example 5:
The wickets taken by a bowler in 10 cricket matches are as follows:
5, 3, 4, 2, 0, 2, 1, 6, 2, 3
Find the mode of this data
Solution:
1. Let us analyse the data:
• In the first match he took 2 wickets
• In the second match he took 3 wickets
• In the third match he took 4 wickets
_ _ _
_ _ _
• In the tenth match he took 3 wickets
2. We have to find the mode
• That is., the item which occur the most number of times
• Is '5' the mode?
    ♦ '5' occurs once. If there are any other items which occur more than once, '5' cannot be the mode
    ♦ The next item '3' is occurs 2 times. So '5' cannot be the mode
• Then is '3' the mode?
    ♦ '3' occurs twice. If there are any other items which occur more than twice, '3' cannot be the mode
_ _ _
_ _ _
3. Instead of going on like this, we can speed up the work by writing the frequency of each item (or making an 'ungrouped frequency distribution table'):
We get:
Frequency of 5 = 1
Frequency of 3 = 2
Frequency of 4 = 1
Frequency of 2 = 3
Frequency of 0 = 1
Frequency of 6 = 1
• The ungrouped frequency distribution table of the above data will be as shown below:

Table.37.25

• So the maximum frequency is 3
    ♦ Item having this maximum frequency is '2'
• Thus the mode of the given data is '2'

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• What we saw above is an 'ungrouped frequency distribution table'. 
    ♦ We know that, such a table is prepared when the data is small. 
• If the data is large, we will be given a 'grouped frequency distribution table'.
• In example 1 above, if we are given the data as a grouped frequency distribution table (with width of class intervals 2), it will look like as in table 37.26 below:

Table.37.26

• In this table, the maximum frequency is 5
    ♦ This maximum frequency is possessed by the class interval: '2 - 4'  
■ So which item has the maximum frequency?
• It is not possible to answer this question
• Within the class interval '2 - 4', the items possible are '2' and '3'
• But there is no way to find the frequency of each of them. 
    ♦ This is because, we are given a 'grouped frequency distribution table'.
• So we have to develop a new method to find the mode when 'grouped frequency distribution tables' are given to us

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• When a 'grouped frequency distribution table' is given to us, we can immediately write the class interval which has the 'largest frequency'. 
    ♦ This class interval is called the modal class
• The 'actual item' which has the largest frequency is hidden inside the modal class. 
    ♦ It can be calculated using the formula:

Where:
l = lower limit of the modal class
h = width of the class interval (assuming all classes are of the same width)
f₁ = frequency of the modal class
f₀ = frequency of the class preceding the modal class
f₂ = frequency of the class succeeding the modal class

Example 6:
A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:

Table.37.27

Find the mode of this data
Solution:
1. The modal class is the class interval having the highest frequency
• So in this problem, the class interval '3 - 5' is the modal class. It has the highest frequency of '8'
2. Now we can calculate the mode using the formula:

l = lower limit of the modal class = 3
h = width of the class interval (assuming all classes are of the same width) = 2
f₁ = frequency of the modal class = 8
f₀ = frequency of the class preceding the modal class = 7
f₂ = frequency of the class succeeding the modal class = 2
Substituting all the values, we get:
mode = 3 + (^8-7⁄_2×8-7-2)×2 = 3 + (¹⁄_16-9)×2 = 3 + (²⁄₇) = 3.286

Example 7:
Consider the first example on mean that we did at the beginning of this chapter. The data was given in table 37.3. It is the marks distribution of 30 students in a mathematics examination. The mean was calculated as 62. Now find the mode of this data. Also compare and interpret the mode and the mean.
Solution:
• For convenience, the table 37.3 is shown again below:

Table.37.3

1. The modal class is the class interval having the highest frequency
• So in this problem, the class interval '40 - 55' is the modal class. It has the highest frequency of '7' 
2. Now we can calculate the mode using the formula:

l = lower limit of the modal class = 40
h = width of the class interval (assuming all classes are of the same width) = 15
f₁ = frequency of the modal class = 7
f₀ = frequency of the class preceding the modal class = 3
f₂ = frequency of the class succeeding the modal class = 6
Substituting all the values, we get:
mode = 40 + (^7-3⁄_2×7-3-6)×15 = 40 + (⁴⁄_14-9)×15 = 40 + 12 = 52
3. We have already obtained the mean as 62. Now we get the mode as 60
• So we can write:
    ♦ The average mark of the class is 62
    ♦ The mark obtained by the largest number of students is 52
• That is., in a table showing the marks of all the 30 students, values near 52 will appear more than others

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Now we will see some solved examples
Solved example 37.7
The following table shows the ages of the patients admitted in a hospital during a year:

Table.37.27

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
1. First we will find the mean:

Table.37.28

We will use the step deviation method:
• u is given by the formula:

• The numerator is the value at the bottom end of the eighth column. It is 43
• The denominator is the value at the bottom end of the second column. It is 80
• So we get u = ⁴³⁄₈₀ 
• Thus x = a + hu = [30 + (10 × ⁴³⁄₈₀)] = 35.375
2. Now we will find the mode
(i) The modal class is the class interval having the highest frequency
• So in this problem, the class interval '35 - 45' is the modal class. It has the highest frequency of '23' 
(ii) Now we can calculate the mode using the formula:

l = lower limit of the modal class = 35
h = width of the class interval (assuming all classes are of the same width) = 10
f₁ = frequency of the modal class = 23
f₀ = frequency of the class preceding the modal class = 21
f₂ = frequency of the class succeeding the modal class = 14
Substituting all the values, we get:
mode = 35 + (^23-21⁄_2×23-21-14)×10 = 35 + (²⁄_46-35)×10 = 36.82
3. So we can write:
• The average age of all the patients admitted at the hospital in a year is 35.3
• The people around an age of 36.8 are the most who are admitted in that year
    ♦ In other words, the number of patients around an age of 36.8 is greater than others

Solved example 37.8
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Table.37.29

Determine the modal lifetimes of the components
Solution:
1. The modal class is the class interval having the highest frequency
• So in this problem, the class interval '60 - 80' is the modal class. It has the highest frequency of '61' 
2. Now we can calculate the mode using the formula:

l = lower limit of the modal class = 60
h = width of the class interval (assuming all classes are of the same width) = 20
f₁ = frequency of the modal class = 61
f₀ = frequency of the class preceding the modal class = 52
f₂ = frequency of the class succeeding the modal class = 38
Substituting all the values, we get:
mode = 60 + (^61-52⁄_2×61-52-38)×20 = 60 + (⁹⁄_122-90)×20 = 60 + 5.625 = 65.625

Solved example 37.9
The following data gives the distribution of total monthly household expenditure of 200 families of a village. 

Table.37.30

Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:

1. First we will find the mean:

Table.37.31

We will use the step deviation method:
• u is given by the formula:

• The numerator is the value at the bottom end of the eighth column. It is -35
• The denominator is the value at the bottom end of the second column. It is 200
• So we get u = ^-35⁄₂₀₀ 
• Thus x = a + hu = [2750 + (500 × ^-35⁄₂₀₀)] = 2662.5
2. Now we will find the mode
(i) The modal class is the class interval having the highest frequency
• So in this problem, the class interval '1500 - 2000' is the modal class. It has the highest frequency of '40' 
(ii) Now we can calculate the mode using the formula:

l = lower limit of the modal class = 1500
h = width of the class interval (assuming all classes are of the same width) = 500
f₁ = frequency of the modal class = 40
f₀ = frequency of the class preceding the modal class = 24
f₂ = frequency of the class succeeding the modal class = 33
Substituting all the values, we get:
mode = 1500 + (^40-24⁄_2×40-24-33)×500 = 1500 + (¹⁶⁄_80-57)×500 = 1847.83
3. So we can write:
• The average expenditure of all the 200 families of the village is 2662.5
• The 'number of families having an expense of around 1847.83' is greater than others


Solved example 37.10
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. 

Table.37.32

Find the mode and mean of this data. Interpret the two measures.
Solution:
1. First we will find the mean:

Table.37.33

We will use the step deviation method:
• u is given by the formula:

• The numerator is the value at the bottom end of the eighth column. It is -23
• The denominator is the value at the bottom end of the second column. It is 35
• So we get u = ^-23⁄₃₅ 
• Thus x = a + hu = [32.5 + (5 × ^-23⁄₃₅)] = 29.2
2. Now we will find the mode
(i) The modal class is the class interval having the highest frequency
• So in this problem, the class interval '30 - 35' is the modal class. It has the highest frequency of '10' 
(ii) Now we can calculate the mode using the formula:

l = lower limit of the modal class = 30
h = width of the class interval (assuming all classes are of the same width) = 5
f₁ = frequency of the modal class = 10
f₀ = frequency of the class preceding the modal class = 9
f₂ = frequency of the class succeeding the modal class = 3
Substituting all the values, we get:
mode = 30 + (^10-9⁄_2×10-9-3)×5 = 30 + (¹⁄_20-12)×5 = 30.6
3. So we can write the conclusion:
• We are given the 'number of students per teacher'
• If a value in the data is low, it indicates a better condition because, then the teacher is in charge of a lesser number of students, and so, each of those students will get better attention
• However, in this problem, we are not dealing with such aspects. We want the mean and the mode
• The mean value is 29.2.
    ♦ So on an average, each teacher is in charge of 29.2 students
• The mode is 30.6
    ♦ So the number of teachers who are in charge of 30.6 students are the highest

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In the next section, we will discuss about median.

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Chapter 37.2 - Calculation of mean - Solved examples

In the previous section we saw Step deviation method. We also saw some solved examples demonstrating all three methods. In this section we will see a few more solved examples.

Solved example 37.2
Consider the following distribution of daily wages of 50 workers of a factory.

Table.37.14

Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
• We will try all the three methods and then decide which method is appropriate. This will help us to select a method when we encounter such problems in the future. The three methods are:
(i) Direct method, (ii) Assumed mean method and (iii) Step deviation method
    ♦ Method (i) has Σf_ix_i in the numerator
    ♦ Method (ii) has Σf_id_i in the numerator
    ♦ Method (iii) has Σf_iu_i in the numerator
• All the three numerators can be conveniently calculated in a single table. It is shown below as table 37.15
• a is taken as 150, since it is the middle x_i
• h is taken as 20, since it is the class width

Table.37.15

1. Direct method:
• x is given by the formula:

• The numerator is calculated at the bottom end of the sixth column. It's value is 7260
• The denominator is calculated at the bottom end of second column. It's value is 50
• So we get x = ⁷²⁶⁰⁄₅₀ = 145.2
2. Assumed mean method:
• d is given by the formula:

• The numerator is calculated at the bottom end of the seventh column. It's value is -240
• The denominator is calculated at the bottom end of second column. It's value is 50
• So we get d = ^-240⁄₅₀ = - 4.8
• Thus x = a + d = 150 - 4.8 = 145.2
■ This is the same value obtained by method 1
3. Step deviation method:
• u is given by the formula:

• The numerator is calculated at the bottom end of the eighth column. It's value is -12
• The denominator is calculated at the bottom end of second column. It's value is 50
• So we get u = ^-12⁄₅₀ = - 0.24
• Thus x = a + hu = 150 + (20 × -0.24) = 150 - 4.8 = 145.2
■ This is the same value obtained by methods 1 & 2
■ In this problem, the values in the x_i column are not small. They are hundreds
• Because of these high values, we get thousands in the sixth column. It is not convenient to find all those thousands and then do calculations with them. 
• So we must reduce x_i by subtracting 'a' from it. Thus we get d_i. But then, we get hundreds in the seventh column. It is not convenient to deal with hundreds either
• So we must reduce d_i by dividing it with 'h'. We get small numbers in the eighth column
■ Upon seeing hundreds in the x_i column, we can immediately decide to use the step deviation method. We need not fill up the sixth and seventh columns 
■ The significance of the result is clear in this problem:
The average (or mean) wages of workers in that factory is 145.2 per day
• The factory owner has to deal with two items: Daily Income and Daily Expense 
• Under Daily expense, he has two items:
    ♦ Total cost of material per day
    ♦ Total cost of labour per day
• There are 50 workers. So the 'total cost of labour per day' will be approximately equal to (50×145.2) = 7260
• To obtain accurate values, we will have to apply more advanced principles of statistics. We will see them in higher classes

Solved example 37.3
The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

Table.37.16

Solution:
• We will try all the three methods and then decide which method is appropriate. This will help us to select a method when we encounter such problems in the future. The three methods are:
(i) Direct method, (ii) Assumed mean method and (iii) Step deviation method
    ♦ Method (i) has Σf_ix_i in the numerator
    ♦ Method (ii) has Σf_id_i in the numerator
    ♦ Method (iii) has Σf_iu_i in the numerator
• All the three numerators can be conveniently calculated in a single table. It is shown below as table 37.17
• a is taken as 18, since it is the middle x_i
• h is taken as 2, since it is the class width

Table.37.17

1. Direct method:
• x is given by the formula:

• The numerator is calculated at the bottom end of the sixth column. It's value is (752+20f)
• The denominator is calculated at the bottom end of second column. It's value is (44+f)
• So we get x = ^(752+20f)⁄_(44+f) 
• But this mean is given to us as 18. So we can write:
^(752+20f)⁄_(44+f) = 18 ⟹ (752+20f) = 18(44+f) ⟹ (752+20f) = (792+18f) ⟹ 2f = 40 ⟹ f = 20

2. Assumed mean method:
• d is given by the formula:

• The numerator is calculated at the bottom end of the seventh column. It's value is (-40+2f)
• The denominator is calculated at the bottom end of second column. It's value is (44+f)
• So we get d = ^(-40+2f)⁄_(44+f)
• Thus x = a + d = [18 + ^(-40+2f)⁄_(44+f)]
• But this mean is given to us as 18. So we can write:
[18 + ^(-40+2f)⁄_(44+f)] = 18 ⟹ [^(-40+2f)⁄_(44+f)] = 0 ⟹ (-40+2f) = 0 ⟹ 2f = 40 ⟹ f = 20 
■ This is the same value obtained by method 1
3. Step deviation method:
• u is given by the formula:

• The numerator is calculated at the bottom end of the eighth column. It's value is (-20+f)
• The denominator is calculated at the bottom end of second column. It's value is (44+f)
• So we get u = ^(-20+f)⁄_(44+f)
• Thus x = a + hu = [18 + (2 × ^(-20+f)⁄_(44+f))]
• But this mean is given to us as 18. So we can write:

[18 + (2 × ^(-20+f)⁄_(44+f))] = 18 ⟹ [(2 × ^(-20+f)⁄_(44+f))] = 0 ⟹ [^(-20+f)⁄_(44+f)] = 0 
⟹ (-20+f) = 0 ⟹ f = 20 
■ This is the same value obtained by methods 1 & 2
■ In this problem, the step deviation method is appropriate because, as we see in the eighth column, it is easier to do calculations with small numbers. So there is no need to fill up the sixth and seventh columns.

Solved example 37.4
Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

Table.37.18

Solution:
• We will try all the three methods and then decide which method is appropriate. This will help us to select a method when we encounter such problems in the future. The three methods are:
(i) Direct method, (ii) Assumed mean method and (iii) Step deviation method
    ♦ Method (i) has Σf_ix_i in the numerator
    ♦ Method (ii) has Σf_id_i in the numerator
    ♦ Method (iii) has Σf_iu_i in the numerator
• All the three numerators can be conveniently calculated in a single table. It is shown below as table 37.19
• a is taken as 75.5, since it is the middle x_i
• h is taken as 3, since it is the class width

Table.37.19

1. Direct method:
• x is given by the formula:

• The numerator is calculated at the bottom end of the sixth column. It's value is 2277
• The denominator is calculated at the bottom end of second column. It's value is 30
• So we get x = ²²⁷⁷⁄₃₀ = 75.9

2. Assumed mean method:
• d is given by the formula:

• The numerator is calculated at the bottom end of the seventh column. It's value is 12
• The denominator is calculated at the bottom end of second column. It's value is 30
• So we get d = ¹²⁄₃₀ = 0.4
• Thus x = a + d = 75.5 + 0.4 = 75.9
■ This is the same value obtained by method 1
3. Step deviation method:
• u is given by the formula:

• The numerator is calculated at the bottom end of the eighth column. It's value is 4
• The denominator is calculated at the bottom end of second column. It's value is 30
• So we get u = ⁴⁄₃₀
• Thus x = a + hu = [75.5 + (3 × ⁴⁄₃₀)] = 75.9
■ This is the same value obtained by methods 1 & 2
■ In this problem, the step deviation method is appropriate because, as we see in the eighth column, it is easier to do calculations with small numbers. So there is no need to fill up the sixth and seventh columns.

Solved example 37.5
In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Table.37.20

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution:
• The class intervals given to us are not continuous. There is a gap of '1' between all the class intervals
    ♦ For example, the gap between the first and second intervals is (53-52) = 1
• We know the method to make the intervals continuous. (Details here)
• The new table 37.21 is given below:

Table.37.21

• Now we can begin to solve the problem. We will try all the three methods and then decide which method is appropriate. This will help us to select a method when we encounter such problems in the future. The three methods are:
(i) Direct method, (ii) Assumed mean method and (iii) Step deviation method
    ♦ Method (i) has Σf_ix_i in the numerator
    ♦ Method (ii) has Σf_id_i in the numerator
    ♦ Method (iii) has Σf_iu_i in the numerator
• All the three numerators can be conveniently calculated in a single table. It is shown below as table 37.22
• a is taken as 57, since it is the middle x_i
• h is taken as 3, since it is the class width

Table.37.22

1. Direct method:
• x is given by the formula:

• The numerator is calculated at the bottom end of the sixth column. It's value is 22875
• The denominator is calculated at the bottom end of second column. It's value is 400
• So we get x = ²²⁸⁷⁵⁄₄₀₀ = 57.1875

2. Assumed mean method:
• d is given by the formula:

• The numerator is calculated at the bottom end of the seventh column. It's value is 12
• The denominator is calculated at the bottom end of second column. It's value is 30
• So we get d = ⁷⁵⁄₄₀₀ = 0.1875
• Thus x = a + d = 57 + 0.1875 = 57.1875
■ This is the same value obtained by method 1
3. Step deviation method:
• u is given by the formula:

• The numerator is calculated at the bottom end of the eighth column. It's value is 4
• The denominator is calculated at the bottom end of second column. It's value is 30
• So we get u = ²⁵⁄₄₀₀
• Thus x = a + hu = [57 + (3 × ²⁵⁄₄₀₀)] = 57.1875
■ This is the same value obtained by methods 1 & 2
■ In this problem, the step deviation method is appropriate because, as we see in the eighth column, it is easier to do calculations with small numbers. So there is no need to fill up the sixth and seventh columns.

Solved example 37.6
The mean of the following distribution is 50. Also Σf = 120. Find the missing frequencies f1 and f2.

Table.37.23

Solution:
• We will try all the three methods and then decide which method is appropriate. This will help us to select a method when we encounter such problems in the future. The three methods are:
(i) Direct method, (ii) Assumed mean method and (iii) Step deviation method
    ♦ Method (i) has Σf_ix_i in the numerator
    ♦ Method (ii) has Σf_id_i in the numerator
    ♦ Method (iii) has Σf_iu_i in the numerator
• All the three numerators can be conveniently calculated in a single table. It is shown below as table 37.24
• a is taken as 50, since it is the middle x_i
• h is taken as 20, since it is the class width

Table.37.24

• From the table, it is clear that, the step deviation method is appropriate because, it involves smaller numbers in the eighth column. So we will use it:
Step deviation method:
• u is given by the formula:

• The numerator will be the value at the bottom end of the sixth column. It is (4-f1+f2)
• The denominator will be the value at the bottom end of the second column. It is (68+f1+f2)
• So we get u = ^(4-f1+f2)⁄_(68+f1+f2) 
• Thus x = a + hu = [50 + (20 × ^(4-f1+f2)⁄_(68+f1+f2))]
• But this mean is given to us as 50. So we can write:
[50 + (20 × ^(4-f1+f2)⁄_(68+f1+f2))] = 50 ⟹ [(20 × ^(4-f1+f2)⁄_(68+f1+f2))] = 0
⟹ ^(4-f1+f2)⁄_(68+f1+f2) = 0 ⟹ (4-f1+f2) = 0 ⟹ f1- f2 = 4
• Now consider the denominator. In the denominator, there will be Σf. But Σf is given as 120
So we can write: Σf = (68+f1+f2) = 120 ⟹ (f1+f2) = 52
• Thus we have two equations:
(i) f1 + f2 = 52
(ii) f1 - f2 = 4
• From (ii) we get: f2 = (f1-4)
• substituting this in (i), we get: 
f1 + f1 - 4 = 52 ⟹ 2f1 = 56 ⟹ f1 = 28
• Substituting this value of f1 in (ii) we get:
f2 = 28 - 4 = 24

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In the next section, we will discuss about mode.

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