Exercise 12.1 Practice
Graphical Representation of Data
Q1: Bar Graph
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A survey conducted by an organization regarding the causes of illness and death among 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?
| S.No. | Causes | Female Fatality Rate (%) |
|---|---|---|
| 1 | Reproductive health conditions | 31.8 |
| 2 | Neuro-psychiatric conditions | 25.4 |
| 3 | Injuries | 12.4 |
| 4 | Cardiovascular conditions | 4.3 |
| 5 | Respiratory conditions | 4.1 |
| 6 | Other causes | 22.0 |
(ii) Which condition is the major cause of women's ill health and death worldwide?
(i) Bar Graph:
Represent Causes on the x-axis and Fatality Rate on the y-axis.
Represent Causes on the x-axis and Fatality Rate on the y-axis.
(ii) Major Cause:
By observing the graph, the tallest bar corresponds to Reproductive health conditions (31.8%).
By observing the graph, the tallest bar corresponds to Reproductive health conditions (31.8%).
Major Cause: Reproductive health conditions
Q2: Bar Graph
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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.
| Section | Number of girls per thousand boys |
|---|---|
| Scheduled Caste (SC) | 940 |
| Scheduled Tribe (ST) | 970 |
| Non SC/ST | 920 |
| Backward districts | 950 |
| Non-backward districts | 920 |
| Rural | 930 |
| Urban | 910 |
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
(i) Bar Graph:
Plot Sections on x-axis and Number of girls on y-axis.
Scale: 1 unit = 10 girls (starting break from 0 to 900).
Plot Sections on x-axis and Number of girls on y-axis.
Scale: 1 unit = 10 girls (starting break from 0 to 900).
(ii) Conclusions:
1. ST section has the maximum number of girls (970) per thousand boys.
2. Urban section has the minimum number of girls (910) per thousand boys.
1. ST section has the maximum number of girls (970) per thousand boys.
2. Urban section has the minimum number of girls (910) per thousand boys.
Graph drawn. ST has max, Urban has min.
Q3: Bar Graph
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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?
| Political Party | A | B | C | D | E | F |
|---|---|---|---|---|---|---|
| Seats Won | 75 | 55 | 37 | 29 | 10 | 37 |
(ii) Which political party won the maximum number of seats?
(i) Bar Graph:
x-axis: Political Party, y-axis: Seats Won.
x-axis: Political Party, y-axis: Seats Won.
(ii) Maximum Seats:
Party A won 75 seats, which is the maximum.
Party A won 75 seats, which is the maximum.
Party A
Q4: Histogram
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The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:
(i) Draw a histogram to represent the given data.
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
| Length (in mm) | Number of Leaves |
|---|---|
| 118 - 126 | 3 |
| 127 - 135 | 5 |
| 136 - 144 | 9 |
| 145 - 153 | 12 |
| 154 - 162 | 5 |
| 163 - 171 | 4 |
| 172 - 180 | 2 |
(ii) Is there any other suitable graphical representation for the same data?
(iii) Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?
(i) Histogram:
First, convert the data into continuous class intervals.
Gap = $127 - 126 = 1$. Half gap = $0.5$.
New intervals: $117.5 - 126.5$, $126.5 - 135.5$, etc.
Draw rectangles with these bases and heights equal to frequencies.
First, convert the data into continuous class intervals.
Gap = $127 - 126 = 1$. Half gap = $0.5$.
New intervals: $117.5 - 126.5$, $126.5 - 135.5$, etc.
Draw rectangles with these bases and heights equal to frequencies.
(ii) Other Representation:
Yes, a Frequency Polygon can also be used.
Yes, a Frequency Polygon can also be used.
(iii) Conclusion:
No. The interval 145-153 has the maximum frequency (12), but it does not mean all or most leaves are exactly 153 mm. They are distributed within the range.
No. The interval 145-153 has the maximum frequency (12), but it does not mean all or most leaves are exactly 153 mm. They are distributed within the range.
(ii) Frequency Polygon, (iii) No
Q3: Frequency Polygon
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The following table gives the distribution of students of two sections according to the marks obtained by them:
Represent the marks of the students of both the sections on the same graph by two frequency polygons.
| Section A | Section B | ||
|---|---|---|---|
| Marks | Frequency | Marks | Frequency |
| 0 - 10 | 3 | 0 - 10 | 5 |
| 10 - 20 | 9 | 10 - 20 | 19 |
| 20 - 30 | 17 | 20 - 30 | 15 |
| 30 - 40 | 12 | 30 - 40 | 10 |
| 40 - 50 | 9 | 40 - 50 | 1 |
Steps:
1. Find the Class Marks ($x_i$) for each interval. Class Mark = $\frac{\text{Upper} + \text{Lower}}{2}$.
Class Marks: 5, 15, 25, 35, 45.
1. Find the Class Marks ($x_i$) for each interval. Class Mark = $\frac{\text{Upper} + \text{Lower}}{2}$.
Class Marks: 5, 15, 25, 35, 45.
Points to Plot:
Section A: $(5, 3), (15, 9), (25, 17), (35, 12), (45, 9)$.
Section B: $(5, 5), (15, 19), (25, 15), (35, 10), (45, 1)$.
Section A: $(5, 3), (15, 9), (25, 17), (35, 12), (45, 9)$.
Section B: $(5, 5), (15, 19), (25, 15), (35, 10), (45, 1)$.
Plot these points on a graph and join them with straight lines for each section separately.
Graph plotted with two polygons.
Q7: Frequency Polygon
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The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:
Represent the data of both the teams on the same graph by frequency polygons.
| Number of balls | Team A | Team B |
|---|---|---|
| 1 - 6 | 2 | 5 |
| 7 - 12 | 1 | 6 |
| 13 - 18 | 8 | 2 |
| 19 - 24 | 9 | 10 |
| 25 - 30 | 4 | 5 |
| 31 - 36 | 5 | 6 |
| 37 - 42 | 6 | 3 |
| 43 - 48 | 10 | 4 |
| 49 - 54 | 6 | 8 |
| 55 - 60 | 2 | 10 |
Step 1: Convert class intervals to continuous.
Gap = 1. Adjustment = 0.5.
Intervals: 0.5-6.5, 6.5-12.5, etc.
Gap = 1. Adjustment = 0.5.
Intervals: 0.5-6.5, 6.5-12.5, etc.
Step 2: Find Class Marks.
Class Mark = $\frac{\text{Upper} + \text{Lower}}{2}$.
3.5, 9.5, 15.5, ..., 57.5.
Class Mark = $\frac{\text{Upper} + \text{Lower}}{2}$.
3.5, 9.5, 15.5, ..., 57.5.
Step 3: Plot points for Team A and Team B and join them.
Frequency Polygons drawn.
Q8: Histogram (Varying Width)
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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.
| Age (in years) | Number of Children |
|---|---|
| 1 - 2 | 5 |
| 2 - 3 | 3 |
| 3 - 5 | 6 |
| 5 - 7 | 12 |
| 7 - 10 | 9 |
| 10 - 15 | 10 |
| 15 - 17 | 4 |
Note: The class widths are not uniform (1, 1, 2, 2, 3, 5, 2).
We must calculate the Adjusted Frequency.
$\text{Adjusted Freq} = \frac{\text{Frequency}}{\text{Class Width}} \times \text{Min Class Width}$.
Minimum Class Width = 1.
We must calculate the Adjusted Frequency.
$\text{Adjusted Freq} = \frac{\text{Frequency}}{\text{Class Width}} \times \text{Min Class Width}$.
Minimum Class Width = 1.
Calculations:
1-2: $\frac{5}{1} \times 1 = 5$
2-3: $\frac{3}{1} \times 1 = 3$
3-5: $\frac{6}{2} \times 1 = 3$
5-7: $\frac{12}{2} \times 1 = 6$
7-10: $\frac{9}{3} \times 1 = 3$
10-15: $\frac{10}{5} \times 1 = 2$
15-17: $\frac{4}{2} \times 1 = 2$
1-2: $\frac{5}{1} \times 1 = 5$
2-3: $\frac{3}{1} \times 1 = 3$
3-5: $\frac{6}{2} \times 1 = 3$
5-7: $\frac{12}{2} \times 1 = 6$
7-10: $\frac{9}{3} \times 1 = 3$
10-15: $\frac{10}{5} \times 1 = 2$
15-17: $\frac{4}{2} \times 1 = 2$
Draw the histogram with heights equal to these adjusted frequencies.
Histogram drawn with adjusted frequencies.
Q9: Histogram (Varying Width)
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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.
| Number of letters | Number of surnames |
|---|---|
| 1 - 4 | 6 |
| 4 - 6 | 30 |
| 6 - 8 | 44 |
| 8 - 12 | 16 |
| 12 - 20 | 4 |
(ii) Write the class interval in which the maximum number of surnames lie.
(i) Histogram:
Class widths vary: 3, 2, 2, 4, 8.
Min width = 2.
Adjusted Frequencies:
1-4: $\frac{6}{3} \times 2 = 4$
4-6: $\frac{30}{2} \times 2 = 30$
6-8: $\frac{44}{2} \times 2 = 44$
8-12: $\frac{16}{4} \times 2 = 8$
12-20: $\frac{4}{8} \times 2 = 1$
Class widths vary: 3, 2, 2, 4, 8.
Min width = 2.
Adjusted Frequencies:
1-4: $\frac{6}{3} \times 2 = 4$
4-6: $\frac{30}{2} \times 2 = 30$
6-8: $\frac{44}{2} \times 2 = 44$
8-12: $\frac{16}{4} \times 2 = 8$
12-20: $\frac{4}{8} \times 2 = 1$
(ii) Max Surnames:
The interval 6-8 has the maximum frequency (44).
The interval 6-8 has the maximum frequency (44).
(ii) 6 - 8