Chapter 25.7 - Measures of Central Tendency

In the previous section we completed the discussion on Frequency Polygons. In this section we will see 'Measures of Central Tendency'. So first we have to see what 'Central Tendency' is. We will start with 'Averages'. Consider the following example:

■ A farmer has 5 plants of a particular kind. They give a special fruit which is considered very costly. The yield from each plant in the previous year was as follows:
• First plant yielded 10 fruits
• Second plant yielded 7 fruits
• Third plant yielded 13 fruits
• Fourth plant yielded 20 fruits
• Fifth plant yielded 15 fruits
What is the average yield?
Solution:
We have done such problems in our earlier classes. We have seen that: 
Average of a certain number of observations = ^{Sum of all observations}⁄_{Total number of observations}.
In this problem, the sum of all the observations = 10 +7 + 13 +20 +15 = 65
So the average in this problem is ⁶⁵⁄₅ = 13

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Let us see one application of the above calculated average:

A friend of the farmer would like to grow the same fruits. He wants to know whether it would be profitable or not. He has space to grow 8 such plants. But just by having more plants, one cannot guarantee profit. Because, more number of plants will require more manure, more labour hours (especially if the plants need special care) etc., So the friend wants to know how many fruits he would get from one plant. 
But all the plants will not give the same number of fruits. Some plants give more, and some plants give less. In such a situation, we take the average. For a preliminary estimate, the average can be taken as equal to the yield from a single plant. 
It may be noted that, many detailed calculations involving complex theories are involved before finalising any business activity. We will study more about them in higher classes on ‘Business administration’.

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In science and engineering problems, another term is used for ‘average’. It is the ‘mean’. It is denoted as x . It is read as ‘x bar’.   So we can write:
■ Mean of a certain number of observations = x = ^{Sum of all observations}⁄_{Total number of observations}.

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Now we will write this in the form of a simple formula. 

1. In our present problem, we have 5 observations. We will first denote them as x₁, x₂, . . . , x₅
2. So the sum of 5 observations is x₁ + x₂ + x₃ + x₄ + x₅
3. We can write: x = ^{(x₁ +x₂ +x₃ +x₄ +x₅ )}⁄₅  = ⁶⁵⁄₅ = 13 
4. Now we will see a method to write the formula in an even simpler way:
• The subscripts 1, 2, 3, . . . can be denoted by the letter ‘i’. So:
    ♦ when i = 1, x_i denotes the first observation x₁
    ♦ when i = 2, x_i denotes the second observation x₂
    ♦ - - - 
    ♦ - - - 
    ♦ when i = 5, x_i denotes the fifth observation x₅
5. The Greek symbol Σ (for the letter Sigma) is used for summation.
In computer spread sheet programs also, this symbol indicates summation. An example is shown below:

So, the sum x₁ + x₂ + x₃ + x₄ + x₅ is written as:

It is read as: 'The sum of x_i as i varies from 1 to 5'
6. Similarly consider a case of 30 observations. We want x₁ + x₂ + x₃ + . . . + x₃₀. It can be written as:

It is read as: 'The sum of x_i as i varies from 1 to 30'
7. Now consider a case of n observations. We want  x₁ + x₂ + x₃ + . . . + x_n. It can be written as:

It is read as: 'The sum of x_i as i varies from 1 to n'
8. Now coming back to mean in our present problem, we can write:

9. In general, for n observations, we can write:
Eq.25.1:

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Consider the following problem:

The daily wages of five friends who work in a factory are:
Rs. 350, 400, 350, 450 and 450.
What is the mean daily wage of a worker in this group?
Solution:

1. We have:

2. First let us calculate the numerator:

= 350 +400 +350 +450 +450 = 2000


3. The denominator = number of observations = n = 5
4.  So x  = ²⁰⁰⁰⁄₅ = Rs.400

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Let us see if there is an easier method:

In step 2, we did the summation. Some values are occurring more than once. We can group such values as follows:

5. (2 × 350) + 400 + (2 × 450) = 700 + 400 + 900 = 2000 (The same sum as in step 2)

6. Now let us see the raw data in a tabular form:

7. We can see that, the frequency values can help us to group the values as in step (5) so that, summation can be done quickly

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Another example:

The daily wages of 20 workers in a factory are:

What is the mean daily wage of a worker in this group?
Solution:
Note that the raw data is given to us in the form of a frequency distribution table. This is because there are 20 observations, and such a large number of observations should always be presented in the form of a frequency distribution table. Let us start our calculations to find the mean x:
1. We have:

2. First let us calculate the numerator:

This is the sum of all observations. As in the previous example, the summation can be quickly done, if we group repeating values. So we can write:

= (2 × 300) + (4 × 300) + . . . + (4 × 500)


3. The above grouping and summation can be conveniently written in a tabular form:

So we get the summation result as Rs.8200
4. The denominator = number of observations = n = 20
5.  So x  = ⁸²⁰⁰⁄₂₀ = Rs.410

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From the above two examples, it is clear that the frequency helps us to group observations and thus to speed up the summation. We can write this:

1. Sum of all observations = f₁x₁ + f₂x₂ + f₃x₃ + . . . + f_nx_n
This can be written in short form as:
Sum of all observations = 

2. Now, the total number of observations is obviously equal to the sum of all frequencies. This is in fact seen as the 'total' at the bottom of the frequency column. So we can write:
Total number of observations = f₁ + f₂ + f₃ + . . . + f_n
This can be written in short form as:

Total number of observations =

3. We have:  = x = ^{Sum of all observations}⁄_{Total number of observations}.
So we can write:
Eq.25.2:

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In the next section we will see some solved examples.

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Chapter 25.5 - Solved examples on Histograms

In the previous section we discussed about Histograms. In this section we will see some solved examples.
Solved example 25.13
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table (a):

(i) Draw a histogram to represent the given data. [Hint: First make the class intervals continuous]
(ii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
Solution:
The class intervals in the given table are not continuous. There is a gap of '1 mm' between all the class intervals. So our first task is to make the classes continuous. Table (b) gives the required continuous classes.
A sample calculation: 
Consider the gap between 145-153 and 154-162
We will use the general method that we saw earlier:
Step 1: Gap =  lower limit of 2^nd – upper limit of 1^st = 154 - 153 = 1 
Step 2: New upper limit of 1^st  = Original upper limit  + ^Gap⁄₂ = 153 + 1/2 = 153 + 0.5 = 153.5
Step 3: New lower limit of 2^nd = Original lower limit  - ^Gap⁄₂  = 154 - 1/2 = 154 - 0.5 = 153.5
Based on table (b), the histogram is prepared as shown in fig.25.11 below:

Fig.25.11

A sample rectangle:
• Consider the third rectangle: It corresponds to the third class interval 135.5-144.5 in table (b). It's frequency in the table (b) is 9. 
• In the histogram, we see that the rectangle of this class interval has a height above the horizontal line through 8. In fact, it is half way between 8 and 10. That means, it is at 9. 
• The reader is advised to draw the whole histogram himself/herself on a fresh graph paper.  
• When a suitable scale is fixed, the thin subdivision lines in the graph paper will enable us to mark the correct heights of the bars
Part (ii)
No. It is not correct to conclude that the maximum number of leaves are 153 mm long.
To write the reason, let us write from the beginning:
1. In this experiment, the length of each of the 40 leaves are measured.
2. It is difficult to show those 40 lengths in a table. Even if it is a 'frequency distribution table', it will be a lengthy table
3. So they adopted the 'grouped frequency distribution table'. In this, there are groups which are known as class intervals.
4. Consider the example of the class 126.5-135.5
• The leaves falling in this class may have lengths any where from 126.5 to 135.5, excluding 135.5.
• That means, several lengths are possible in this class
5. Similarly, consider the class 144.5-153.5. Several lengths are possible in this class.
■ We cannot say: 'all the 12 leaves in this class are of 153 cm length'

Solved example 25.14
The following table gives the life times of 400 neon lamps:

(i) Represent the given information with the help of a histogram. 
(ii) How many lamps have a life time of more than 700 hours?
Solution:
The required histogram is given below in fig.25.12:

Fig.25.12

A sample rectangle:
• Consider the fifth rectangle: It corresponds to the fifth class interval 700-800 in table. It's frequency in the table is 74. 
• In the histogram, we see that the rectangle of this class interval has a height just above the horizontal line through 70. The excess above the horizontal line is '4'.
• The reader is advised to draw the whole histogram himself/herself on a fresh graph paper.  
• When a suitable scale is fixed, the thin subdivision lines in the graph paper will enable us to mark the correct heights of the bars
Part (ii)
• The number of bulbs which have a life time of 700 hours or more will be represented by the rectangles which lie on the right side of the '700 mark' on the x-axis
• So the required number is: 74 +62 +48 = 184

Solved example 25.15
A random survey of the number of children of various age groups playing in a park was found as follows:

Draw a histogram to represent the data above
Solution:
In the given table, the class intervals are not equal. So we need to find adjusted frequencies. They are tabulated in the table below:

A sample calculation:
We will use the general method that we saw earlier:
1. Find the smallest width in the problem
2. Leave all the intervals with the smallest width as such
3. New frequency of each of the other intervals is given by:
^{original frequency of that interval}⁄_n 
Where n = ^{width of that interval}⁄_{smallest width in the problem}
• The smallest width is 1. So we will leave the first and second class intervals as such. The frequencies of those classes do not need any adjustments
• We will take the interval 10-15 as a sample
• 'n' for that interval =  ^{width of that interval}⁄_{smallest width in the problem} = 5/1 = 5
• New frequency of that interval = ^{original frequency of that interval}⁄_n  = 10/5 = 2 
The final histogram is shown below:

Fig.25.13

A sample rectangle:
• Consider the fifth rectangle: It corresponds to the fifth class interval 7-10 in the modified table. It's width is 3 and frequency is 3. 
• In the histogram, we see that the rectangle of this class interval has a height same as the horizontal line through 3.
• The width of the base on the x-axis is also 3. That is., from 7 to 10
• The reader is advised to draw the whole histogram himself/herself on a fresh graph paper.

Solved example 25.16
100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie
Solution:
In the given table, the class intervals are not equal. So we need to find adjusted frequencies. They are tabulated in the table below:

A sample calculation:
We will use the general method that we saw earlier:
1. Find the smallest width in the problem
2. Leave all the intervals with the smallest width as such
3. New frequency of each of the other intervals is given by:
^{original frequency of that interval}⁄_n 
Where n = ^{width of that interval}⁄_{smallest width in the problem}
• The smallest width is 2. So we will leave the second and third class intervals as such. The frequencies of those classes do not need any adjustments
• We will take the interval 8-12 as a sample
• 'n' for that interval =  ^{width of that interval}⁄_{smallest width in the problem} = 4/2 = 2
• New frequency of that interval = ^{original frequency of that interval}⁄_n  = 16/2 = 8 

The final histogram is shown below:

Fig.25.14

A sample rectangle:
• Consider the fourth rectangle: It corresponds to the fourth class interval 8-12 in the modified table. It's width is 4 and frequency is 8. 
• In the histogram, we see that the rectangle of this class interval has a height above 5 and below 10. It is near to 10 than to 5.
• The width of the base on the x-axis is 4. That is., from 8 to 12
• The reader is advised to draw the whole histogram himself/herself on a fresh graph paper.
• When a suitable scale is fixed, the thin subdivision lines in the graph paper will enable us to mark the correct heights of the bars

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So we have completed the discussion on histograms. In the next section we will see 'frequency polygons'.

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Chapter 25.3 - Pictorial representation of Data - Bar graphs

In the previous section we completed the discussion on 'how data is represented by tables'. In this section, we will see how data can be represented by graphs. It is said that one picture is better than a thousand words. This is very true when data is represented by pictorial representations like graphs and pie charts. The main features of the data will easily become clear when using graphs. We will now see various types of graphs.

Bar Graphs

We have already learned about bar graphs in our earlier classes. Details here. Let us look at their main features:
• All bars have the same width
• The spacing between all bars are the same
• The base of each bar rests on the horizontal axis. That is., the ‘x-axis’
• Variables (eg: expenses, sales, profits etc.,) are marked on the x-axis
• The values of the variables are marked on the vertical axis. That is., the ‘y-axis’
• The heights of the bars depend upon the values taken by the variables

Let us see an example:
In a particular section of Class IX, 40 students were asked about the months of their birth and the following graph was prepared from the data so obtained:

Observe the bar graph given above and answer the following questions:
(i) How many students were born in the month of November?
(ii) In which month were the maximum number of students born? 

Solution:
The variable in this problem is ‘Month of Birth’. For getting the value of the variable at a particular point (here, point is the name of the month), we look at the corresponding point on the y-axis. The white horizontal dotted lines will help us to find this value from the y-axis.
Part (i) 
1. First we take the point ‘November’ on the x-axis. 
2. The height of the bar at November is 4. So the value of the variable is 4
3. Thus we get the answer: 4 students were born in the month of november
Part (ii)
1. Here we have to find the maximum y value first. Looking at the y-axis, we find that, the maximum y value is 6. 
2. The x value corresponding to it is August
3. Thus we get the answer: The Maximum number of students were born in the month of August.

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Now we will see an example which demonstrates the construction of a bar graph:
A family with a monthly income of Rs 20,000 had planned the following expenditures per month under various heads:

Draw a bar graph for the data above.
Solution:
We can construct the bar graph by the following steps:
• There must be one bar for each item
• Width of all bars must be the same
• Spacing between all bars must be the same
• Height of each bar will be equal to the corresponding value in the ‘Expenditure’ column of the given data table
• Each unit on the y-axis represents Rs. 1000
    ♦ Example: The height of education column is 5 units. That means, the actual expenditure is Rs. 5000
• The completed graph is shown below:

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So we have completed the discussion on bar graphs. We can write this:
• First we obtain the raw data from the field
• Next we form the ‘frequency distribution table’
• Finally, based on the frequency distribution table, we construct the bar graph
■ So we can say: Bar graph correspond to ‘frequency distribution table’
• For example, in the first section of this chapter, we saw a frequency distribution table related to the marks scored by students in a class. See it here. And just below that, the corresponding bar graph was given.

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Now we will see some solved examples
Solved example 25.10
A survey conducted by an organisation for the cause of illness and death among the women between the ages 15 - 44 (in years) worldwide, found the following figures (in %):

(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
Solution:
Part (i)
The given data can be represented graphically using a 'Bar graph' as shown in fig.25.2 below.

Fig.25.2

Each item should be given a bar.
A sample bar:
• Consider item no.2: Neuropsychiatric conditions. It's frequency in the given data table is 25.4. 
• In the bar graph, we see that the bar of this item has a height just above the horizontal line through 25. The excess above the horizontal line is '0.4'. 
• The reader is advised to draw the whole graph himself/herself on a fresh graph paper.  
• When a suitable scale is fixed, the thin subdivision lines in the graph paper will enable us to mark the correct heights of the bars
Part (ii)
The tallest bar is that of item 1. So we can write this:

Reproductive health conditions is the major cause of women’s ill health and death worldwide

Solved example 25.11
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below:

(i) Represent the information above by a bar graph.

(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Solution:
The required bar graph is shown in fig.25.3 below:

Fig.25.3

A sample bar:
• Consider item no.5: Non-backward districts. It's frequency in the given data table is 920. 
• In the bar graph, we see that the bar of this item has a height exactly same as the horizontal line through 920.
• The reader is advised to draw the whole graph himself/herself on a fresh graph paper.  
• When a suitable scale is fixed, the thin subdivision lines in the graph paper will enable us to mark the correct heights of the bars
• But in this case, we will not need to consider the subdivision lines. If the scale chosen is “10 no. of girls = 1 cm”, the top edge of all the bars will fall on main lines of the graph paper
Part (ii)
We can arrive at many conclusions from the above bar graph:
• Surveys were conducted separately on different sections of the society
• Number of girls and number of boys were obtained from each section
• Based on those numbers, the following result was obtained for each section:
    ♦ The number of girls per 1000 boys
• According to World health organisation (WHO) guide lines, If there are 1000 boys, the number of girls should be greater than 1000
    ♦ But the above survey results show that this is not achieved
• In some sections, the number of girls is far less than 1000. Example: Non SC/ST and Non-backward districts. There are only 920 girls per 1000 boys.
• The condition is even worse in urban section. The number is only 910. It has the shortest bar.

• In Scheduled tribe section, there are 970 girls per 1000 boys. So it is a more desirable result obtained so far. It has the tallest bar.

Solved example 25.12
Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
Solution:
The required bar graph is shown in fig.25.4 below:

Fig.25.4

A sample bar:
• Consider 'C':  It's frequency in the given data table is 37. 
• In the bar graph, we see that the bar of this item has a height between 30 and 40. It is nearly 40. That is., it is above the 'midpoint 35' between 30 and 40.
• The reader is advised to draw the whole graph himself/herself on a fresh graph paper.  
• When a suitable scale is fixed, the thin subdivision lines in the graph paper will enable us to mark the correct heights of the bars
• If the scale chosen is “10 seats = 1 cm”, the top edge of  bar of C will be 7 subdivisions above 30.

Part (ii)
The tallest bar is that of Party A. So it has won the election. The number of seats is 75

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In the next section we will see histograms.

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Chapter 25.2 - Solved examples on Grouped Frequency Distribution Tables

In the previous section we saw two solved examples on grouped frequency distribution tables. In this section, we will see more solved examples.

Solved example 25.3
The relative humidity (in %) of a certain city for a month of 30 days was as follows:

(i) Construct a grouped frequency distribution table with classes 84 - 86, 86 - 88, etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Solution:
Part (i)
1. The smallest value is 84.9. So the first interval can be 84-86.
2. It is given that the classes can be 84-86, 86-88 etc., So the class width is the same, which is equal to 2. Thus the classes that we use in this problem are: 84-86, 86-88, 88-90, 90-92, ... , 98-100
3. The largest value is 99.2. So 98-100 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:

Sample calculation: 
• Consider the class-interval 94-96. The values from 94 to 96 in the raw data are:
95.3, 94.2, 95.1, 95.1, 95.2, 95.7
• All together, there are 6 values
• Thus, the frequency of the class-interval 94-96 is 6
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
Part (ii)
The relative humidity is high on all days of the given month. So the recordings are made in a rainy season
Part (iii)
The smallest value is 84.9. The largest value is 99.2. So the range is 99.2 -84.9 = 14.3

Solved example 25.4
The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:

(i) Represent the data given above by a grouped frequency distribution table, taking
the class intervals as 160 - 165, 165 - 170, etc.
(ii) What can you conclude about their heights from the table?
Solution:
Part (i)
1. It is given that the classes can be 160 - 165, 165 - 170 etc., So the class width is the same, which is equal to 5. 
2. The smallest value is 150. So the first interval will be 150-155.
2. Thus the classes that we use in this problem are: 150-155, 155-160, ...
3. The largest value is 173. So 170-175 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:

Sample calculation: 
• Consider the class-interval 155-160. The heights from 155 to 160 in the raw data are:
158, 156, 160, 159, 156, 158, 160, 159, 159, 158, 159
• All together, there are 11 heights. But 160 cm should be included in the next upper class 160-165. There are two 160 values.
• So only 9 distances are eligible to be included in the class-interval 155-160
• Thus, the frequency of the class-interval 155-160 is 9
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
Part (ii)
• The heights range from 150 to 172. So an approximate average will be (150+172)/2 = 322/2 = 161
• Let this be rounded to 165 cm
• Then the number of students whose height is equal to or greater than the average 165 cm = 10 +5 = 15
• 15 is less than half of the total 50. So we can say this: 
Less than 50% of the students have a height equal to or greater than 165 cm

Solved example 25.5
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:

(i) Make a grouped frequency distribution table for this data with class intervals as 0.00 - 0.04, 0.04 - 0.08, and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?

Solution:
Part (i)
1. It is given that the classes can be 0.00 - 0.04, 0.04 - 0.08 etc., So the class width is the same, which is equal to 0.04 
2. The smallest value is 0.01. So the first interval will be 0.00 - 0.04.
2. Thus the classes that we use in this problem are: 0.00 - 0.04, 0.04 - 0.08, ...
3. The largest value is 0.22. So 0.20-0.24 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:

Sample calculation: 
• Consider the class-interval 0.12-0.16. The concentrations from 0.12 to 0.16 in the raw data are:
0.16, 0.12 and 0.13
• All together, there are 3 values. But 0.16 should be included in the next upper class 0.16-0.20
• So only 2 values are eligible to be included in the class-interval 0.12-0.16
• Thus, the frequency of the class-interval 0.12-0.16 is 2
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
Part (ii)
1. The concentration greater than 0.11 is 0.12
2. A value of 0.12 will be included in the class 0.12-0.16 
3. It will not be included in the class 0.08-0.12
4. So we need to consider only the last three classes in the table
5. The answer is 2 +4 +2 = 8 days

Solved example 25.6
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:

Prepare a frequency distribution table for the data given above.

Solution:
In this problem, we do not have to fix up the classes. Because we are not asked to make a 'grouped frequency distribution table'. What we are asked is a 'frequency distribution table'. It is shown below:

Sample calculation:
The number '2' occurs 9 times in the raw data. So the frequency corresponding to '2' is 9

Solved example 25.7
The value of π upto 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digit
Solution:
Let us first separate the 50 digits that occur after the decimal place. The separated digits are written in a tabular form below:

In this problem, we do not have to fix up the classes. Because we are not asked to make a 'grouped frequency distribution table'. What we are asked is a 'frequency distribution table'. It is shown below:

• From the table we can see that 3 and 9 are the most frequently occurring digits
• Also 0 is the least frequently occurring digit.

Solved example 25.8
Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:

(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 - 10.
(ii) How many children watched television for 15 or more hours a week?
Solution:
Part (i)
1. It is given that one of the classes can be 5-10, and the class width is to be 5
2. The smallest value is 1 hour. So the first interval will be 0-5
2. Thus the classes that we use in this problem are: 0-5, 5-10, ...
3. The largest value is 17. So 15-20 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:

Sample calculation: 
• Consider the class-interval 10-15. The hours from 10 to 15 in the raw data are:
12, 10, 12, 15, 14 and 12
• All together, there are 6 values. But 15 should be included in the next upper class 15-20
• So only 5 values are eligible to be included in the class-interval 10-15
• Thus, the frequency of the class-interval 10-15 is 5
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.
Part (ii)
All the values equal to or greater than 15 will be in the last class 15-20. So the required answer is 2

Solved example 25.9
A company manufactures car batteries of a particular type. The lives (in years) of 40 such batteries were recorded as follows:

Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 - 2.5
Solution:
Part (i)
1. It is given that the first class is to be 2-2.5, and the class width is to be 0.5
2. The smallest value is 2.2. So the first interval of 2-2.5 is appropriate
2. Thus the classes that we use in this problem are: 2-2.5, 2.5-3.0, ...
3. The largest value is 4.6. So 4.5-5.0 will be the last class.
4. Based on the above facts, the grouped frequency distribution table is constructed as shown below:

Sample calculation: 
• Consider the class-interval 4-4.5. The values from 4 to 4.5 in the raw data are:
4.1, 4.5, 4.4, 4.3 and 4.2
• All together, there are 5 values. But 4.5 should be included in the next upper class 4.5-5.0
• So only 4 values are eligible to be included in the class-interval 4.5-5.0
• Thus, the frequency of the class-interval 4.5-5.0 is 4
■ The above sample calculation is given only to show that the numbers given in the frequency column are correct. To obtain those numbers, we must follow the 'method of tally marking'. We saw the details about it earlier here.

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In the next section we will see graphs.

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