Statistics
Class 9th Mathematics CBSE Solution
- A survey conducted by an organisation for the cause of illness and death among…
- The following number of goals were scored by a team in a series of 10 matches:…
- The blood groups of 30 students of Class VIII are recorded as follows: A, B, O,…
- Give five examples of data that you can collect from your day-to-day life.…
- Classify the data in Q.1 above as primary or secondary data.
- The following data on the number of girls (to the nearest ten) per thousand…
- In a mathematics test given to 15 students, the following marks (out of 100)…
- The distance (in km) of 40 engineers from their residence to their place of…
- Given below are the seats won by different political parties in the polling…
- The relative humidity (in %) of a certain city for a month of 30 days was as…
- The following observations have been arranged in ascending order. If the median…
- The heights of 50 students, measured to the nearest centimetres, have been…
- The length of 40 leaves of a plant are measured correct to one millimetre, and…
- Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18
- A study was conducted to find out the concentration of sulphur dioxide in the…
- The following table gives the life times of 400 neon lamps: Life time (in…
- Find the mean salary of 60 workers of a factory from the following table:…
- The following table gives the distribution of students of two sections…
- Give one example of a situation in which (i) The mean is an appropriate measure…
- Three coins were tossed 30 times simultaneously. Each time the number of heads…
- The runs scored by two teams A and B on the first 60 balls in a cricket match…
- The value of up to 50 decimal places is given below:
- A random survey of the number of children of various age groups playing in a…
- Thirty children were asked about the number of hours they watched TV programmes…
- A company manufactures car batteries of a particular type. The lives (in years)…
- 100 surnames were randomly picked up from a local telephone directory and a…
Exercise 14.1
Question 1.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 %):
(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.
Answer:
The parts of the questions are solved below:
(i) The data is represented below graphically:
(ii) From the above graphical data, we observe that reproductive health condition is the major cause of women’s ill health and death worldwide.
Explanation: From the bar graph, bar of reproductive health condition is the highest.
(iii) The two factors that are responsible for the ill health and death of women worldwide due to reproductive health conditions are mentioned below:
The lack of proper care and understanding.
Lack of medical facilities in the various regions of the world.
Question 2.
The following number of goals were scored by a team in a series of 10 matches:
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.
Answer:
Given data is:
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
(i) Mean:
= 
= 
=
= 2.8
Hence, the mean of the given data is 2.8
(ii) Median:
In order to find the mean of the given data, we will arrange the data in ascending order:
0, 1, 2, 3, 3, 3, 3, 4, 4, 5
Total number of data = 10
Now,
Median= 
= 
=
= 
= 3
Hence, the median of the following data is 3.
(iii) Mode:
Since 3 occur most of the times that is 4 times.
Hence, the mode of the provided data is 3.
Question 3.
The blood groups of 30 students of Class VIII are recorded as follows:
A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O,
A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O
Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?
Answer:
The frequency refers to the number of students having common blood group. We will represent the data in table:
Most common blood group (Highest frequency) = O
Rarest blood group (Lowest frequency) = AB
Question 4.
Give five examples of data that you can collect from your day-to-day life.
Answer:
Five examples from day to day life are mentioned below:
i. Daily expenditures of household material.
ii. Amount of rainfall in a year.
iii. Bill of electricity.
iv. Marks obtained by students in distinguished subjects.
v. Various survey and poll results.
Question 5.
Classify the data in Q.1 above as primary or secondary data.
Answer:
The above data after classifying into secondary and primary data is:
Primary data:
i. Daily expenditures of household materials.
ii. Bill of electricity.
iii. Marks obtained by students in distinguished subjects.
Secondary data:
i. Amount of rainfall in a year.
ii. Various survey and poll results.
Question 6.
The following data on the number of girls (to the nearest ten) per thousand boys indifferent 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.
Answer:
The parts of the questions are solved below:
(i) The bar graph of the above information is represented below:
(ii) It can be observed from the above bar graph that the maximum number of girls per thousand boys is in Scheduled Tribe that is 970.
Also,
The Backward Districts and Rural Areas have more number of girls per thousand boys than Non-Backward Districts and Urban Areas.
Question 7.
In a mathematics test given to 15 students, the following marks (out of 100) are recorded:
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Answer:
(i) Mean:
Using 
(ii) Median:
To find the median, arrange the given data in ascending order, as follows:
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Since, the number of terms is 15, an odd number, we find out the median by finding the marks obtained by 
student, which is 8th student.
Therefore, the median marks will be 52.
(iii) Again, the data 52 occurs most frequently that is, three times.
Hence, the mode of the given data would be 52.
Question 8.
The distance (in km) of 40 engineers from their residence to their place of work was found as follows:
5 3 10 20 25 11 13 7 12 31
19 10 12 17 18 11 32 17 16 2
7 9 7 8 3 5 12 15 18 3
12 14 2 9 6 15 15 7 6 12
Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as 0-5 (5 not included). What main features do you observe from this tabular representation?
Answer:
The given data is very large. Hence, we will construct a group frequency of class size 5.
Therefore,
The class interval will be 0-5, 5-10, 10-15, 15-20 and so on
The data in the table is represented as:
The classes in the table are not overlapping.
Now we can clearly see that 36 out of 40 engineers live under 20 km distance and rest of the 4 engineers live at a distance of more than 20 km.
Question 9.
Given below are the seats won by different political parties in the polling outcome ofa state assembly elections:
(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
Answer:
The parts of the questions are solved below:
(i) The given polling results are represented by the bar graph below:
(ii) The party named A has won the maximum number of seats.
Question 10.
The relative humidity (in %) of a certain city for a month of 30 days was as follows:
98.1 98.6 99.2 90.3 86.5 95.3 92.9 96.3 94.2 95.1
89.2 92.3 97.1 93.5 92.7 95.1 97.2 93.3 95.2 97.3
96.2 92.1 84.9 90.2 95.7 98.3 97.3 96.1 92.1 89
(i) Construct a grouped frequency distribution table with classes 84 - 86, 86 - 88, etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Answer:
(i) The given data is very large. Therefore, we will be construct a group frequency of class size 2.
Therefore,
Class interval will be:
84 – 86, 86 – 88, 88 – 90, 90 – 92 and so on.
The data is represented in the table as:
(ii) The humidity is very high in the data which is observed during rainy season. Hence, it must be rainy season.
(iii) Range of data = Maximum value of data – Minimum value of data
= 99.2 – 84.9
= 14.3
Question 11.
The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Answer:
The given data arranged in ascending order is as
29, 32, 48, 50, x, x+2, 72, 78, 84, 95
There are 10 terms.
So, there are two middle terms that are:
The 
That is the 5th and 6th terms.
Hence, the median is the mean of the values of the 5th and 6th terms.
Therefore,
But median = 63 (Given)
Therefore,
x + 1 = 63
x = 63 – 1
= 62
Question 12.
The heights of 50 students, measured to the nearest centimetres, have been found to be as follows:
161 150 154 165 168 161 154 162 150 151
162 164 171 165 158 154 156 172 160 170
153 159 161 170 162 165 166 168 165 164
154 152 153 156 158 162 160 161 173 166
161 159 162 167 168 159 158 153 154 159
(i) Represent the data given above by a grouped frequency distribution table, taking the class intervals as 160 - 165, 165 - 170, etc.
(ii) What can you conclude about their heights from the table?
Answer:
(i) The data with class interval 160-165, 165-170 and so on is represented in the table as:
(ii) From the given data, it can be concluded that 35 students that is more than 50% are shorter than 165 cm.
Question 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:
(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 leave are 153 mm long? Why?
Answer:
The parts of the questions are solved below:
(i) The data is represented in a discontinuous class interval. So, at first we will make it continuous.
The difference is 1.
So, we subtract � = 0.5 from the lower limit and add 0.5 to the upper limit.
Now, the above information is represented by the histogram below:
(ii) Yes, the above given data can also be represented by the frequency polygon.
(iii) No, it is incorrect to conclude that the maximum number of leaves are lying between length of 144.5 – 153.5
Explanation: Maximum length of leave lies between 144.5 - 153.5, it is not necessarily 153 mm. It can be any value between the range.
Question 14.
Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18
Answer:
Arranging the given data in ascending order as follows, we get:
14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28
Here,
14 occur most frequently that is 4 times
Hence, the mode of the given data is 14
Question 15.
A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for 30 days is as follows:
0.03 0.08 0.08 0.09 0.04 0.17
0.16 0.05 0.02 0.06 0.18 0.20
0.11 0.08 0.12 0.13 0.22 0.07
0.08 0.01 0.10 0.06 0.09 0.18
0.11 0.07 0.05 0.07 0.01 0.04
(i) Make a grouped frequency distribution table for this data with class intervals as0.00 - 0.04, 0.04 - 0.08, and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11parts per million?
Answer:
(i) The data with class interval 0.00-0.04, 0.04-0.08 and so on is represented in the table as:
(ii) 2 + 4 + 2 = 8
Hence, 8 days have the concentration of sulphur dioxide more than 0.11 parts per million
Question 16.
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?
Answer:
The parts of the questions are solved below:
(i) The above given information is represented with the help of histogram below:
(ii) 74 + 62 + 48 = 184
Hence, 184 lamps have a life time of more than 700 hours.
Question 17.
Find the mean salary of 60 workers of a factory from the following table:
Answer:
Calculation of mean:
Therefore,
where fi is the frequency of ith row
and xi is the data value at the ith row
Mean= 
Thus, the mean salary of 60 workers is Rs 5083.33(approx).
Question 18.
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.
Answer:
The class marks can be found by
= 
Now,
For section A,
For section B,
Now, the frequency polygon for the given data:
Section A has more number of high scoring students and less number of low scorers.
Section B has less number of high scoring students and more number of low scorers.
Question 19.
Give one example of a situation in which
(i) The mean is an appropriate measure of central tendency.
(ii) The mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency.
Answer:
(i) The mean is an appropriate measure because of its unique value and can be used to compare different groups of data
for example,
Take the marks obtained by the student in 3 subjects - English - 40/50, Maths - 25/50, Science 50/50
Now if we calculate the mean of the marks obtained by the student, it will be 38.3/50
And this will give the appropriate measure of the central tendency of the data.
(ii) For the measurement of qualitative characteristics. For instance say beauty, honesty, intelligence, etc., mean cannot be used. But you can use the median for distinguishing that.
Question 20.
Three coins were tossed 30 times simultaneously. Each time the number of heads occurring was noted down as follows:
0 1 2 2 1 2 3 1 3 0
1 3 1 1 2 2 0 1 2 1
3 0 0 1 1 2 3 2 2 0
Prepare a frequency distribution table for the data given above.
Answer:
The frequency distribution table for the data given above can be represented as follows:
Question 21.
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.
[Hint: First make the class intervals continuous.]
Answer:
The data is represented in a discontinuous class interval. So, at first, we will make it continuous.
The difference is 1.
Hence, we subtract μ = 0.5 from the lower limit and add 0.5 to the upper limit.
Now,
We will draw frequency polygon for the given data:
Question 22.
The value of π up to 50 decimal places is given below:
3.14159265358979323846264338327950288419716939937510
(i) Make a frequency distribution of the digits from 0 to 9 after the decimal point.
(ii) What are the most and the least frequently occurring digits?
Answer:
The parts of the questions are solved below:
(i) The frequency is given as follows:
(ii) The digit having the least frequency occurs the least and the digit with highest frequency occurs the most.
Hence,
We can say that:
0 has its frequency as 2 and thus occurs least frequently while 3 and 9 have their frequency 8 and hence occur most frequently.
Question 23.
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.
Answer:
Here the class intervals are of different lengths.
So we have to calculate the adjusted frequency.
The required size of interval = minimum size of the interval
= 2 - 1
= 1
So the adjusted frequency is:
The class interval with minimum class size 1 is selected and the length of the rectangle is proportional to it.Taking the age of the children on x-axis and proportion of children per 1 year interval on the y-axis, the following histogram can be drawn:
Question 24.
Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows:
1 6 2 3 5 12 5 8 4 8
10 3 4 12 2 8 15 1 17 6
3 2 8 5 9 6 8 7 14 12
(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as 5 - 10.
(ii) How many children watched television for 15 or more hours a week?
Answer:
The parts of the questions are solved below:
(i) The distribution table for the given data, taking class width 5 and one of the class intervals as 5-10 is as follows:
(ii) We conclude from the given table that two children watch television for 15 or more hours a week.
Question 25.
A company manufactures car batteries of a particular type. The lives (in years) of 40such batteries were recorded as follows:
2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5
3.5 2.3 3.2 3.4 3.8 3.2 4.6 3.7
2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8
3.5 3.2 3.9 3.2 3.2 3.1 3.7 3.4
4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6
Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the interval 2 - 2.5
Answer:
A grouped frequency table using class intervals of size 0.5 starting from the interval 2-2.5 is constructed below:
Question 26.
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 surname lie
Answer:
The class interval in the data is having varying width.
(i) We know that the area of the rectangle is proportional to the frequencies in the histogram.
The class interval with minimum class size 2 is selected and the length of the rectangle is proportional to it. The proportion of the surnames per 2 letters interval can be calculated as:
(ii) The class interval in which the maximum number of surname lie is 6-8.