Chapter 12: Statistics
Exercise 12.1
Q1
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 %):
| S.No. | Causes | Female fatality rate (%) |
|---|---|---|
| 1 | Reproductive health conditions | 31.8 |
| 2 | Neuropsychiatric conditions | 25.4 |
| 3 | Injuries | 12.4 |
| 4 | Cardiovascular conditions | 4.3 |
| 5 | Respiratory conditions | 4.1 |
| 6 | Other causes | 22.0 |
(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.
Solution:
(i) By representing causes on the x-axis and female fatality rate on the y-axis and choosing an appropriate scale (1 unit = 5% for y-axis), the graph of the information given above can be constructed.
(ii) Reproductive health condition is the major cause of women's ill health and death worldwide as 31.8% of women are affected by it.
(iii) The factors are as follows:
- Lack of medical facilities.
- Lack of correct knowledge of treatment.
Q2
The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below:
| 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 |
(i) Represent the information above by a bar graph.
(ii) In the classroom discuss what conclusions can be arrived at from the graph.
Solution:
(i) By representing section (variable) on x-axis and number of girls per thousand boys on y-axis, the graph of the information given above can be constructed by choosing an appropriate scale (1 unit = 100 girls for y-axis).
(ii) It can be observed that:
- The maximum number of girls per thousand boys (970) is for ST.
- The minimum number of girls per thousand boys (910) is for Urban areas.
- The number of girls per thousand boys is greater in rural areas than in urban areas, backward districts than in non-backward districts, and SC/ST than in non-SC/ST sections.
Q3
Given below are the seats won by different political parties in the polling outcome of a state assembly elections:
| Political Party | Seats Won |
|---|---|
| A | 75 |
| B | 55 |
| C | 37 |
| D | 29 |
| E | 10 |
| F | 37 |
(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
Solution:
(i) By taking polling results on x-axis and seats won as y-axis and choosing an appropriate scale (1 unit = 10 seats for y-axis), the required graph of the above information can be constructed.
(ii) Political party 'A' won the maximum number of seats.
Q4
The length of 40 leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:
| 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 |
(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?
Solution:
(i) It can be observed that the length of leaves is represented in a discontinuous class interval having a difference of 1 in between them. Therefore, \( 0.5 \) has to be added to each upper class limit and also have to subtract \( 0.5 \) from the lower class limits so as to make the class intervals continuous.
| Length (in mm) | Number 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 |
Taking the length of leaves on x-axis and the number of leaves on y-axis, the histogram of this information can be drawn. Here, 1 unit on y-axis represents 2 leaves.
(ii) Yes, another suitable graphical representation of this data is a frequency polygon.
(iii) No, it is not correct. The maximum number of leaves (i.e., 12) has their length in the interval 144.5 mm - 153.5 mm. It is not necessary that all of them have their lengths as exactly 153 mm.
Q5
The following table gives the life times of neon lamps:
| Life time (in hours) | Number of lamps |
|---|---|
| 300-400 | 14 |
| 400-500 | 56 |
| 500-600 | 60 |
| 600-700 | 86 |
| 700-800 | 74 |
| 800-900 | 62 |
| 900-1000 | 48 |
(i) Represent the given information with the help of a histogram.
(ii) How many lamps have a lifetime of more than 700 hours?
Solution:
(i) By taking life time (in hours) of neon lamps on x-axis and the number of lamps on y-axis, the histogram of the given information can be drawn. Here, 1 unit on y-axis represents 10 lamps.
(ii) It can be concluded that the number of neon lamps having their lifetime more than 700 is the sum of the number of neon lamps having their lifetime in the intervals 700-800, 800-900, and 900-1000.
Therefore, the number of lamps = \( 74 + 62 + 48 = 184 \).
Q6
The following table gives the distribution of students of two sections according to the marks obtained by them:
| Section A | |
|---|---|
| Marks | Frequency |
| 0-10 | 3 |
| 10-20 | 9 |
| 20-30 | 17 |
| 30-40 | 12 |
| 40-50 | 9 |
| Section B | |
|---|---|
| Marks | Frequency |
| 0-10 | 5 |
| 10-20 | 19 |
| 20-30 | 15 |
| 30-40 | 10 |
| 40-50 | 1 |
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.
Solution:
We can find the class marks of the given class intervals by using the formula:
\[ \text{Class mark} = \frac{\text{Upper class limit} + \text{Lower class limit}}{2} \]
| Marks | Class marks | Section A Frequency | Section B Frequency |
|---|---|---|---|
| 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 |
Taking class marks on x-axis and frequency on y-axis and choosing an appropriate scale (1 unit = 3 for y-axis), the frequency polygon can be drawn.
Conclusion: It can be observed that the performance of students of section 'A' is better than the students of section 'B' in terms of good marks (more students in higher mark ranges).
Q7
The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:
| 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 |
Represent the data of both the teams on the same graph by frequency polygons.
Solution:
It can be observed that the class intervals of the given data are not continuous (there is a gap of 1). Therefore, \( 0.5 \) has to be added to the upper class limits and \( 0.5 \) has to be subtracted from the lower class limits.
Also, class mark of each interval can be found by using the formula:
\[ \text{Class mark} = \frac{\text{Upper class limit} + \text{Lower class limit}}{2} \]
| Number 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 |
By taking class marks on x-axis and runs scored on y-axis, a frequency polygon can be constructed.
Q8
A random survey of the number of children of various age groups playing in park was found as follows:
| 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 |
Draw a histogram to represent the data above.
Solution:
Here, it can be observed that the data has class intervals of varying width. The proportion of children per 1 year interval (frequency density) needs to be calculated.
\[ \text{Length of rectangle} = \frac{\text{Frequency}}{\text{Width of class}} \times \text{Minimum class width} \]
(Minimum class width = 1)
| Age (in years) | Frequency | Width | Length of rectangle |
|---|---|---|---|
| 1-2 | 5 | 1 | \( \frac{5 \times 1}{1} = 5 \) |
| 2-3 | 3 | 1 | \( \frac{3 \times 1}{1} = 3 \) |
| 3-5 | 6 | 2 | \( \frac{6 \times 1}{2} = 3 \) |
| 5-7 | 12 | 2 | \( \frac{12 \times 1}{2} = 6 \) |
| 7-10 | 9 | 3 | \( \frac{9 \times 1}{3} = 3 \) |
| 10-15 | 10 | 5 | \( \frac{10 \times 1}{5} = 2 \) |
| 15-17 | 4 | 2 | \( \frac{4 \times 1}{2} = 2 \) |
Taking the age of children on x-axis and proportion of children per 1 year interval on y-axis, the histogram can be drawn.
Q9
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:
| Number of letters | Number of surnames |
|---|---|
| 1-4 | 6 |
| 4-6 | 30 |
| 6-8 | 44 |
| 8-12 | 16 |
| 12-20 | 4 |
(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surname lie.
Solution:
(i) Here, it can be observed that the data has class intervals of varying width. The proportion of the number of surnames per 2 letters interval can be calculated as follows.
Minimum class width = 2.
| Number of letters | Frequency | Width | Length of rectangle |
|---|---|---|---|
| 1-4 | 6 | 3 | \( \frac{6 \times 2}{3} = 4 \) |
| 4-6 | 30 | 2 | \( \frac{30 \times 2}{2} = 30 \) |
| 6-8 | 44 | 2 | \( \frac{44 \times 2}{2} = 44 \) |
| 8-12 | 16 | 4 | \( \frac{16 \times 2}{4} = 8 \) |
| 12-20 | 4 | 8 | \( \frac{4 \times 2}{8} = 1 \) |
By taking the number of letters on x-axis and the proportion of the number of surnames per 2 letters interval on y-axis and choosing an appropriate scale (1 unit = 4 surnames for y-axis), the histogram can be constructed.
(ii) The class interval in which the maximum number of surnames lies is 6-8 as it has 44 surnames in it (the maximum frequency).