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 2nd – upper limit of 1st = 154 - 153 = 1
Step 2: New upper limit of 1st = Original upper limit + Gap⁄2 = 153 + 1/2 = 153 + 0.5 = 153.5
Step 3: New lower limit of 2nd = 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:
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:
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:
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:
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
So we have completed the discussion on histograms. In the next section we will see 'frequency polygons'.
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 2nd – upper limit of 1st = 154 - 153 = 1
Step 2: New upper limit of 1st = Original upper limit + Gap⁄2 = 153 + 1/2 = 153 + 0.5 = 153.5
Step 3: New lower limit of 2nd = 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 |
• 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 |
• 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:
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:
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 |
• 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:
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 |
• 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
No comments:
Post a Comment