Statistics
NCERT Solutions • Class 9 Maths • Chapter 12Exercise 12.1
1. A survey conducted by an organisation for the cause 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?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
(i) Represent the information given above graphically.
(ii) Which condition is the major cause of women’s ill health and death worldwide?
(iii) Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause.
(i) Graphical Representation:
A bar graph is drawn with ‘Causes’ on the x-axis and ‘Female Fatality Rate (%)’ on the y-axis.
The bars represent the given percentages: 31.8, 25.4, 12.4, 4.3, 4.1, 22.0.
A bar graph is drawn with ‘Causes’ on the x-axis and ‘Female Fatality Rate (%)’ on the y-axis.
The bars represent the given percentages: 31.8, 25.4, 12.4, 4.3, 4.1, 22.0.
(ii) Major Cause:
Observing the data, the maximum fatality rate is 31.8%.
Thus, Reproductive health conditions is the major cause of women’s ill health and death worldwide.
Observing the data, the maximum fatality rate is 31.8%.
Thus, Reproductive health conditions is the major cause of women’s ill health and death worldwide.
(iii) Contributing Factors:
1. Lack of proper medical facilities and hospitals.
2. Lack of education and awareness regarding reproductive health.
1. Lack of proper medical facilities and hospitals.
2. Lack of education and awareness regarding reproductive health.
2. 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.
(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
(i) Graphical Representation:
A bar graph is drawn with ‘Sections’ on the x-axis and ‘Girls per 1000 Boys’ on the y-axis.
Scale: 1 unit length = 10 girls.
A bar graph is drawn with ‘Sections’ on the x-axis and ‘Girls per 1000 Boys’ on the y-axis.
Scale: 1 unit length = 10 girls.
(ii) Conclusions:
1. The Scheduled Tribe (ST) section has the maximum number of girls (970) per 1000 boys.
2. The Urban section has the minimum number of girls (910) per 1000 boys.
3. Rural areas generally have a better sex ratio compared to urban areas.
1. The Scheduled Tribe (ST) section has the maximum number of girls (970) per 1000 boys.
2. The Urban section has the minimum number of girls (910) per 1000 boys.
3. Rural areas generally have a better sex ratio compared to urban areas.
3. 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?
(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
(i) Bar Graph:
X-axis: Political Parties (A, B, C, D, E, F)
Y-axis: Seats Won (Scale: 1 unit = 10 seats)
Heights: A(75), B(55), C(37), D(29), E(10), F(37).
X-axis: Political Parties (A, B, C, D, E, F)
Y-axis: Seats Won (Scale: 1 unit = 10 seats)
Heights: A(75), B(55), C(37), D(29), E(10), F(37).
(ii) Maximum Seats:
From the table/graph, Political Party A won the maximum number of seats (75).
From the table/graph, Political Party A won the maximum number of seats (75).
4. The length of 40 leaves of a plant are measured correct to one millimetre…
(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?
(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?
(i) Continuous Class Intervals:
Since the data is discontinuous (118-126, 127-135…), we convert it by subtracting 0.5 from lower limits and adding 0.5 to upper limits.
Draw a histogram using these continuous intervals.
Since the data is discontinuous (118-126, 127-135…), we convert it by subtracting 0.5 from lower limits and adding 0.5 to upper limits.
| Length (mm) | No. of Leaves |
|---|---|
| 117.5 – 126.5 | 3 |
| 126.5 – 135.5 | 5 |
| 135.5 – 144.5 | 9 |
| 144.5 – 153.5 | 12 |
| 153.5 – 162.5 | 5 |
| 162.5 – 171.5 | 4 |
| 171.5 – 180.5 | 2 |
(ii) Alternative Representation:
Yes, a Frequency Polygon is another suitable graphical representation for this data.
Yes, a Frequency Polygon is another suitable graphical representation for this data.
(iii) Conclusion Validity:
No, the conclusion is incorrect. The interval 144.5 – 153.5 has the maximum number of leaves (12). This implies their lengths fall within this range, not that they are all exactly 153 mm.
No, the conclusion is incorrect. The interval 144.5 – 153.5 has the maximum number of leaves (12). This implies their lengths fall within this range, not that they are all exactly 153 mm.
5. 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?
(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a life time of more than 700 hours?
(i) Histogram:
The intervals (300-400, 400-500…) are continuous.
X-axis: Lifetime (hours). Y-axis: Number of lamps.
Draw rectangular bars of equal width (100) and varying heights corresponding to frequencies.
The intervals (300-400, 400-500…) are continuous.
X-axis: Lifetime (hours). Y-axis: Number of lamps.
Draw rectangular bars of equal width (100) and varying heights corresponding to frequencies.
(ii) Lamps > 700 hours:
Sum of lamps in intervals 700-800, 800-900, and 900-1000.
$= 74 + 62 + 48$
$=$ $184$ lamps.
Sum of lamps in intervals 700-800, 800-900, and 900-1000.
$= 74 + 62 + 48$
$=$ $184$ lamps.
Statistics
NCERT Solutions • Class 9 Maths • Chapter 12Exercise 12.1 (Continued)
6. 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. From the two polygons compare the performance of the two sections.
Step 1: Calculate Class Marks
Class Mark ($x_i$) = $\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$
Step 2: Conclusion
By comparing the two polygons, we can see that the polygon for Section A is generally higher in the upper marks range (30-50) compared to Section B. Thus, Section A has performed better than Section B.
Class Mark ($x_i$) = $\frac{\text{Lower Limit} + \text{Upper Limit}}{2}$
| Class Interval | Class Mark | Sec A Freq | Sec B Freq |
|---|---|---|---|
| 0 – 10 | 5 | 3 | 5 |
| 10 – 20 | 15 | 9 | 19 |
| 20 – 30 | 25 | 17 | 15 |
| 30 – 40 | 35 | 12 | 10 |
| 40 – 50 | 45 | 9 | 1 |
Step 2: Conclusion
By comparing the two polygons, we can see that the polygon for Section A is generally higher in the upper marks range (30-50) compared to Section B. Thus, Section A has performed better than Section B.
7. 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.
Step 1: Make Class Intervals Continuous
The given data is discontinuous (1-6, 7-12). Gap = 1.
Adjustment: Subtract 0.5 from lower limit, add 0.5 to upper limit.
Plot these class marks on X-axis and Runs on Y-axis.
The given data is discontinuous (1-6, 7-12). Gap = 1.
Adjustment: Subtract 0.5 from lower limit, add 0.5 to upper limit.
| No. of Balls | Class Mark | Team A | Team B |
|---|---|---|---|
| 0.5 – 6.5 | 3.5 | 2 | 5 |
| 6.5 – 12.5 | 9.5 | 1 | 6 |
| 12.5 – 18.5 | 15.5 | 8 | 2 |
| 18.5 – 24.5 | 21.5 | 9 | 10 |
| 24.5 – 30.5 | 27.5 | 4 | 5 |
| 30.5 – 36.5 | 33.5 | 5 | 6 |
| 36.5 – 42.5 | 39.5 | 6 | 3 |
| 42.5 – 48.5 | 45.5 | 10 | 4 |
| 48.5 – 54.5 | 51.5 | 6 | 8 |
| 54.5 – 60.5 | 57.5 | 2 | 10 |
8. 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.
Note: Class widths vary.
Minimum Class Width = 1.
Adjusted Frequency = $\frac{\text{Frequency}}{\text{Class Width}} \times \text{Min Width (1)}$.
Minimum Class Width = 1.
Adjusted Frequency = $\frac{\text{Frequency}}{\text{Class Width}} \times \text{Min Width (1)}$.
| Age | Freq | Width | Adj Freq (Height) |
|---|---|---|---|
| 1 – 2 | 5 | 1 | $\frac{5}{1} \times 1 = 5$ |
| 2 – 3 | 3 | 1 | $\frac{3}{1} \times 1 = 3$ |
| 3 – 5 | 6 | 2 | $\frac{6}{2} \times 1 = 3$ |
| 5 – 7 | 12 | 2 | $\frac{12}{2} \times 1 = 6$ |
| 7 – 10 | 9 | 3 | $\frac{9}{3} \times 1 = 3$ |
| 10 – 15 | 10 | 5 | $\frac{10}{5} \times 1 = 2$ |
| 15 – 17 | 4 | 2 | $\frac{4}{2} \times 1 = 2$ |
9. 100 surnames were randomly picked… (i) Draw a histogram to depict the given information. (ii) Write the class interval in which the maximum number of surnames lie.
(i) Adjusted Frequency Calculation:
Minimum Class Width = 2 (e.g., 4-6, 6-8).
Formula: $\frac{\text{Frequency}}{\text{Class Width}} \times 2$.
Minimum Class Width = 2 (e.g., 4-6, 6-8).
Formula: $\frac{\text{Frequency}}{\text{Class Width}} \times 2$.
| Letters | Freq | Width | Adj Freq (Height) |
|---|---|---|---|
| 1 – 4 | 6 | 3 | $\frac{6}{3} \times 2 = 4$ |
| 4 – 6 | 30 | 2 | $\frac{30}{2} \times 2 = 30$ |
| 6 – 8 | 44 | 2 | $\frac{44}{2} \times 2 = 44$ |
| 8 – 12 | 16 | 4 | $\frac{16}{4} \times 2 = 8$ |
| 12 – 20 | 4 | 8 | $\frac{4}{8} \times 2 = 1$ |
(ii) Maximum Surnames:
By observing the original frequency table, the maximum frequency is 44.
This lies in the class interval 6 – 8.
By observing the original frequency table, the maximum frequency is 44.
This lies in the class interval 6 – 8.