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NCERT Solutions for Class 9 Maths Chapter 14 Statistics are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 14 Statistics.
Board  CBSE 
Textbook  NCERT 
Class  Class 9 
Subject  Maths 
Chapter  Chapter 14 
Chapter Name  Statistics 
Exercise  Ex 14.1, Ex 14.2, Ex 14.3, Ex 14.4. 
Number of Questions Solved  26 
Category  NCERT Solutions 
NCERT Solutions for Class 9 Maths Chapter 14 Statistics
Chapter 14 Statistics Ex 14.1
Question 1.
Give five examples of data that you collect from your daytoday life.
Solution.
Five examples of data that we can gather from our daytoday life are
 number of students in our class.
 number of fans in our class.
 electricity bills of our house for last two years.
 election results obtained from television or news paper.
 literacy rate figures obtained from educational survey.
Of course, remember that there can be many more different answers.
Question 2.
Classify the data in Q. 1 above as primary or secondary data.
Solution.
We know that, when the information was collected by the investigator herself or himself with a definite objective in her or his mind, the data obtained is called primary data.
∴ In the given data (in Q.l) examples (i), (ii) and (iii) are called primary data and when the information was gathered from a source which already had the information stored, the data obtained is called secondary data.
∴ In the given data (in Q.l) examples (iv) and (v) are called secondary data.
Chapter 14 Statistics Ex 14.2
Question 1.
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?
Solution.
The number of students who have a certain type of blood group is called the frequency of those blood groups. To make data more easily understandable, we write it in a table, as given below :
Blood Group  Number of students 
A  9 
B  6 
0  12 
AB  3 
Total  30 
From table, we observe that the higher frequency blood group i.e., most common blood group is O and the lowest frequency blood group i.e., rarest blood group is AB.
Question 2.
The distance (in km) of 40 engineers from their residence to their place of work were 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 05 (5 not included). What main features do you observe from this tabular representation?
Solution.
To present such a large amount of data, so that a reader can make sense of jt easily, we condense it into groups like 05, 510,…, 3035 (since, our data is from 5 to 32). These grouping are called ‘classes’ or ‘classintervals’ and their size is called the class size or class width which is 5 in this case. In each of these classes the least number is called the lower class limit and the greatest number is called the upper class limit e.g., in 05, 0 is the ‘lower class limit’ and 5 is the ‘upper class limit’. Now, using tally marks, the data (given) can be condensed in tabular form as follows :
Presenting data in this form simplifies and condenses data and enables us to observe certain important feature at a glance. This is called a grouped frequency distribution table. We observe that the classes in the table above are nonoverlapping.
Question 3.
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  89.0 
(i) Construct a grouped frequency distribution table with classes 8486, 8688 etc.
(ii) Which month or season do you think this data is about?
(iii) What is the range of this data?
Solution.
(i) We condense the given data into groups, like 8486, 8688, …98100 (since, our data is from 84.9 to 99.2). So, the class width in this case is 2. Now, the given data can be condensed in tabular form as follows :
(ii) From the table, we observe that the data appears to be taken in the rainy season as the relative humidity is high.
(iii) We know that,
Range = Upper limit of data – Lower limit of data = 99.2 – 84.9 = 14.3
Question 4.
The heights of 50 students, measured to the nearest centimeters 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 class intervals as 160165, 165170 etc.
(ii) What can you conclude about their heights from the table?
Solution.
(i) We condense the given data into groups like 150155, 155160…170175 (since, our data is from 150 to 172). The class width in this case is 5.
Now, the given data can be condensed in tabular form as follows :
(ii) From the table, our conclusion is that more than 50% of student (i.e., 12 + 9+14 = 3 5) are shorter than 165 cm height.
Question 5.
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 as 0.000.04, 0.040.08 and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million (ppm)?
Solution.
(i) We condense the given data into groups like 0.000.04, 0.040.08,…,0.200.24. (since, our data is from 0.01 to 0.22). The class width in this case is 0.04.
Now, the given data can be condensed in tabular form as follows :
Concentration of sulphur dioxide (in ppm)  Frequency 
0.00 – 0.04  4 
0.04 – 0.08  9 
0.08 – 0.12  9 
0.12 0.16  2 
0.16 – 0.20  4 
0.20 – 0.24  2 
Total  30 
(ii) The concentration of sulphur dioxide was more than 0.11 ppm for 2 + 4 + 2 = 8 days (by table.)
Question 6.
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.
Solution.
Firstly, we write the data in a table :
Number of heads  Frequency 
0  6 
1  10 
2  9 
3  5 
Total  30 
In above table, we observe that the repetition of ‘0’ in given data is 6 times, 1 as to 10 times, 2 as 9 times and 3 as 5 times. Also, the above table is called an ungrouped frequency distribution table or simply a frequency distribution table.
Question 7.
The value of π upto 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?
Solution.
Firstly, we write the data i.e., digits from 0 to 9 after the decimal point in a table below :
Digits  Frequency 
0  2 
1  5 
2  5 
3  8 
4  4 
5  5 
6  4 
7  4 
8  5 
9  8 
Total  50 
(i) From the table, we observe that the digit’s after the decimal points i.e., 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 repeated 2, 5, 5, 8, 4, 5, 4, 4, 5, 8 times, respectively.
(ii) From the table, we observe that the digits after the decimal point 3 and 9 are most frequently occurring i.e., 8 times. The digit ‘0’ is the least occurring i.e., only 2 times.
Question 8.
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?
Solution.
(i) We condense the given data into groups like 0 – 5, 5 – 10,…, 15 – 20 (since, our data is from 1 to 17). The class width in this case is 5.
Now, required grouped frequency distribution table is as follows :
Number of hours  Frequency 
0 – 5  10 
5 – 10  13 
10 – 15  5 
15 – 20  2 
Total  30 
(ii) From the table, we observe that the number of children is 2, who watched television for 15 or more hours a week.
Question 9.
A company manufactures car batteries of a particular type. The lives (in years) of 40 such 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.
Solution.
We condense the given data into groups like 2.0 – 2.5, 2.5 – 3.0,…4.5 – 5.0 (Since, our data is from
2.2 to 4.6). The class width in this case is 0.5.
Now, the given data can be condensed in tabular form as follows :
Chapter 14 Statistics Ex 14.3
Question 1.
A survey conducted by an organisation for the cause of illness and death among the women between the ages 1544 (in years) worldwide, found the following figures (in %):
S. No.  Causes  Female Fatality rate (in %) 
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 Igor 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) We draw the bar graph of this data in the following steps. Note that, the unit in the second column is percentage.
 We represent the causes (variable) on the horizontal axis choosing any scale, since width of the bar is not important but for clarity, we take equal widths for all bars and maintain equal gaps in between. Let one cause be represented by one unit
 We represent the female fatality rate (value) on the vertical axis. Here, we can choose the scale as 1 unit = 4%.
 To represent our first cause i.e., reproductive health conditions, we draw a rectangle bar with width 1 unit and height 31.8 units.
 Similarly, other heads are represented leaving a gap of 1 unit in between two consecutive bars.
Now, the graph is drawn in figure as given below :
(ii) From graph, we observe that ‘reproductive health conditions’ is the major cause of women’s ill health and death world wide because it has maximum percentage among the causes e., 31.8%.
(iii) Two other factors which play a major role in the cause in (ii) above are injuries and other causes.
Question 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:
Section  Number of girls per thousand boys 
Scheduled Caste (SC)  940 
Scheduled Tribe (ST)  970 
Non SC/ST  920 
Backward districts  950 
Nonbackward 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) We draw the bar graph of this data, note that the unit in the second column is number of girls per thousand boys.
 We represent the section on the horizontal axis choosing any scale, since width of the bar is not important but for clarity, we take equal width for all bars and maintain equal gaps in between. Let one section be represented by one unit.
 We represent the number of girls per thousand boys on the vertical axis.
Here, we can choose the scale as 1 unit = 100.
Now, the graph is drawn in figure as given below :
(ii) From graph, we observe that scheduled tribe (ST) number of girls is major section in different sections of Indian society, because it has maximum number of girls per thousand boys i.e., 970.
Question 3.
Given below are the seats won by different political parties in the polling outcome of a state assembly elections :
Political Party  A  B  C  D  E  F 
Seats Won  75  55  37  29  10  37 
(i) Draw a bar graph to represent the polling results.
(ii) Which political party won the maximum number of seats?
Solution.
(i) We draw the bar graph of this data, note that the unit in the second column is seats won by political party.
We represent the political party on the horizontal axis choosing any scale, since width of the bar is not important but for clarity, we take equal widths for all bars and maintain equal gaps in between. Let one political party be represented by one unit.
We represent the seats won on the vertical axis. Here, we can choose the scale as 1 unit = 10.
Now, the graph is drawn as follows :
(ii) Party A’ won the maximum number of seats i.e., 75.
Question 4.
The length of 40 leaves of a plant measured correct to one millimeter and the obtained data is represented in the following table
Length (in mm)  Number of leaves 
118126  3 
127135  5 
136 144  9 
145153  12 
154162  5 
163171  4 
172180  2 
(i) Draw a histogram to represent the given data. [Hint : First make the class intervals continuous.]
(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) We know that the areas of the rectangles are proportional to the frequencies in a histogram. Here, the given frequency distribution is not continuous. For continuous distribution, we get first interval as
(118 – 0.5) – (126 + 0.5)= 117.5126.5. The class width in this case is 9
[ ∵ ]
So, we get the following modified table of given data :
Length (in mm)  Frequency  Width of the class 
117.5126.5  3  9 
126.5135.5  5  9 
135.5144.5  9  9 
144.5153.5  12  9 
153.5162.5  5  9 
162.5171.5  4  9 
171.5180.5  2  9 
Now, we can draw the histogram for given data as follows :
(ii) Yes, other suitable graphical representation for the same data is frequency polygon.
(iii) No, because the maximum number of leaves have their lengths lying in the interval 145153.
Question 5.
The following table gives the lifetimes of 400 neon lamps :
Lifetime (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) The histogram for given table is as follows :
(ii) Number of lamps have a lifetime of more than 700 hours = 74 + 62 + 48 = 184
Question 6.
The following table gives the distribution of students of two sections according to the marks obtained by them :
Section A  Section B  
Marks  Frequency  Marks  Frequency 
010  3  010  5 
10 20  9  1020  19 
2030  17  2030  15 
3040  12  3040  10 
4050  9  4050  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 make modified table by given data as shown below :
Classes  Class Marks  Frequency (Section A)  Frequency (Section B) 
010  5  3  5 
1020  15  9  19 
2030  25  17  15 
3040  35  12  10 
4050  45  9  1 
Now, the required frequency polygons are as follows :
It is clear that from the polygon that the performance of section A is better in comparison of section B.
Question 7.
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 
16  2  5 
712  1  6 
1318  8  2 
1924  9  10 
2530  4  5 
3136  5  6 
3742  6  3 
4348  10  4 
4954  6  8 
5560  2  10 
Represent the data of both the teams on the same graph by frequency polygons.Represent the data of
[Hint : First make the class intervals continuous]
Solution.
First, we make the class intervals continuous then modified table of given data is as shown below :
Number of balls  Class marks  Team A  Team B 
0.56.5  3.5  2  5 
6.512.5  9.5  r  6 
12.518.5  15.5  8  2 
18.524.5  21.5  9  10 
24.530.5  27.5  4  5 
30.536.5  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 
54.560.5  57.5  2  10 
Now, frequency polygon for both teams are given below :
Question 8.
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 
12  5 
23  3 
35  6 
57  12 
710  9 
1015  10 
1517  4 
Draw a histogram to represent the data above.
Solution.
Here, the widths of the rectangles are varying. So, we need to make certain modifications in the lengths of the rectangles, so that the areas are became proportional to the frequencies. The minimum class size is 1. Length of rectangle (Adjusted frequency)
=
Then modified table of given data is shown below :
Age (in years) 
Number of children (Frequency)  Width of the class  Length of the rectangle 
12  5  1  1/1 x 5= 5 
23  3  1  1/1 x 3 = 3 
35  6  2  1/2 x 6 = 3 
57  12  2  1/2 x 12 = 6 
710  9  3  1/3 x 9 = 3 
1015  10  5  1/ 5 x 10 = 2 
1517  4  2  1 / 2 x 4 = 2 
So, the histogram with varying width is given below :
Question 9.
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 :
Solution.
Number of Letters  Number of Surnames 
14  6 
46  30 
68  44 
812  16 
1220  4 
(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.
Solution.
(i) We know that, the areas of the rectangles are proportional to the frequencies in a histogram. Here, the widths of the rectangles are varying. So, we need to make certain modifications in the lengths of the rectangles. So that the areas are again proportional to the frequencies.
 Select a class interval with the minimum class size. The minimum class size is 2.
 The lengths of the rectangles are then modified to be proportionate to the class size 2.
Since, we have calculated these lengths for interval of 2 letters in each case, we may call these lengths as ‘proportion of surnames per 2 mark interval’.
Here, we make a modified table by given data with minimum class size 2.
Number of letters  Number of surnames  Width of the class 
Length of the rectangle 
14  6  3  
46  30  2  
68  44  2  
812  16  4  
1220  4  8 
So, the correct histogram with varying width is given below :
(ii) The class interval in which the maximum number of surnames lie is 68.
Chapter 14 Statistics Ex 14.4
Question 1.
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.
Solution.
Question 2.
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.
Solution.
Question 3.
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
Solution.
Question 4.
Find the mode of 14, 25,14, 28, 18,17,18,14, 23, 22,14 and 18.
Solution.
The given data is, 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18 Arranging the data in ascending order, we have 14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28
Here 14 occurs most frequently (4 times).
∴ Mode = 14
Question 5.
Find the mean salary of 60 workers of a factory from the following table:
Salary (in Rs.)  Number of workers 
3000  16 
4000  12 
5000  10 
6000  8 
7000  6 
8000  4 
9000  3 
1000  1 
Total  60 
Solution.
Salary (In Rs.)  Number of workers  f_{i}x_{i}_{ } 
3000  16  48000 
4000  12  48000 
5000  10  50000 
6000  8  48000 
7000  6  42000 
8000  4  32000 
9000  3  27000 
1000  1  10000 
Total 

Question 6.
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.
Solution.
(i) Mean marks in a test in mathematics.
(ii) Average beauty
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