For previous exercise Statistics – Exercise 14.2 – Class IX
- 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 %):
|Sl. No.||Causes||Female Fertility Rate(%)|
|1||Reproductive health conditions||31.8|
(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) Here we are representing causes on x – axis and female fertility rate on y –axis, the graph of the information given above can be constructed as follows:
All the rectangular bars are of the same width and have equal spacing between them.
(ii) Reproductive health condition is the major cause of female infertility and death worldwide as 31.8% are affected by it.
(iii)The two main factors which play a major role in the women’s illness which is mentioned in (ii) above being the major cause are
- Lack of medical facilities
- Lack of proper knowledge about health issues
- 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|
(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 section 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)
Here, all the rectangular bars are of the same length an have equal spacing in between them.
(ii) It can be observed that maximum number of girls per thousand boys is ST and minimum number of girls per thousand boys is Urban.
Also number of girls per thousand boys is greater in rural areas than that in urban areas backward districts than that in non backward districts, SC and ST than that in non SC/ST.
- 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)Take polling results on x-axis and seats won on y-axis and choose scale such as 1unit = 10seats for y-axis, the required graph o the above information can be constructed as follows:
Here, all the rectangular bars are of the same length an have equal spacing in between them
(ii) Political party named A won with maximum seats 75.
- 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?
(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 claass 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 as above. Here, 1 unit on y-axis represents 2 leaves.
(ii) other suitable graphical representation of this data is frequency polygon.
(iii) No. As maximum number of leaves (i.e., 12) has their length in between 144.5 mm and 153.5mm. It is not necessary that all have their lengths as 153 mm.
- The following table gives the life times of 400 neon lamps:
|Life time (in hours)||Number of lumps|
|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 life time of more than 700 hours?
(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 as follows.
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 sum of the number of neon lamps having lifetime as 800 – 800, 800 – 900 and 900 – 1000
Therefore, the number of neon lamps having their lifetime more than 700 hours is 74 + 62 + 48 =184.
- The following table gives the distribution of students of two sections according to the marks obtained by them:
|Section A||Section B|
|0 – 10||3||0 – 10||5|
|10 – 20||9||10 – 20||19|
|20 – 30||17||20 – 30||15|
|30 – 40||1||30 – 40||10|
|40 – 50||9||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.
We can find the class marks of the given class intervals by using the following formula.
Class mark = Upper class limit + Lower class limit/2
|Section A||Section B|
|Marks||Class marks||Frequency||Marks||Class marks||Frequency|
|0 – 10||5||3||0 – 10||5||5|
|10 – 20||15||9||10 – 20||15||19|
|20 – 30||25||17||20 – 30||25||15|
|30 – 40||35||1||30 – 40||35||10|
|40 – 50||45||9||40 – 50||45||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 as follows.
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.
- 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.
[Hint : First make the class intervals continuous.]
It can be observed that the class intervals of the given data are not continuous.
There is a gap of 1 in between them. Therefore, 1/2 = 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 following formula.
Class mark = Upper class limit+Lower class limit/2
Continuous data with class mark of each class interval can be represented as follows.
|Number of balls||Class mark||Team A||Team B|
|1 – 6||3.5||2||5|
|7 – 12||9.5||1||6|
|13 – 18||15.5||8||2|
|19 – 24||21.5||9||10|
|25 – 30||27.5||4||5|
|31 – 36||33.5||5||6|
|37 – 42||39.5||6||3|
|43 – 48||45.5||10||4|
|49 – 54||51.5||6||8|
|55 – 60||57.5||2||10|
By taking class marks on x- axis and runs scored on y-axis , a frequency polygon can be constructed as follows:
- A random survey of the number of children of various age groups playing in a 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.
Here, it can be observed that the data has class intervals of varying width. The proportion of children per 1 year interval can be calculated as follows.
|Age (in years)||Frequency (Number of children)||Width of classes||Length of rectangle|
|1 – 2||5||1||5×1/1 = 5|
|2 – 3||3||1||3×1/1 = 3|
|3 – 5||6||2||6×1/2 = 3|
|5 – 7||12||2||12×1/2 = 6|
|7 – 10||9||3||9×1/2 = 3|
|10 – 15||10||5||10×1/5 = 2|
|15 – 17||4||2||4×1/2 = 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 as follows.
- 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 surnames lie.
(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.
|Number of letters||Frequency (Number of surnames)||Width of class||Length of rectangle|
|1 – 4||6||3||6×2/3= 4|
|4 – 6||30||2||30×2/2 = 30|
|6 – 8||44||2||44×2/2 = 44|
|8 – 12||16||4||16×2/4 = 8|
|12 – 20||4||8||4×2/8 = 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 such as 1 unit = 4 students for y axis, the histogram can be constructed as follows.
(ii) The class interval in which the maximum number of surnames lies is 6 − 8 as it has 44 surnames in it i.e., the maximum for this data.
For next exercise – Statistics – Exercise 14.4 – Class IX