Chapter 14: Statistics

EXERCISE 14.1

1. Give five example of data that you can collect from your day-to-day life .

Solution:  There are five examples of data that we can collect from our day-to-day life :

(i) Heights of 20 students of the class IX .

(ii) Number of absentees in each day in our class for a month .

(iii) Number of members in the families of our classmates .

(iv) Heights of 15 plants in or around our school .

(v) Electricity bills of our house for last two years .

2. Classify the data in Q.1 above as primary or secondary data .

Solution:  Primary data :  (i) , (ii) , (iii) and (iv)

Secondary data : (v)

[ Primary data : 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 .

Secondary data : When the information was gathered from a source which already had the information stored, the data obtained is called secondary data .] 

EXERCISE 14.2

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: We construct the frequency distribution table :

Blood group	Frequency
A	9
B	6
O	12
AB	3
Total	30

 The most common blood group is O and the rarest blood group is AB .

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 0-5 (5 not included). What main features do you observe from this tabular representation?
Solution: We construct the frequency distribution table :

Class interval	Frequency
0 – 5 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35	5 11 11 9 1 1 2
Total	40

 

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
(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?
 

 

 

 

 

 

Solution: (i) We construct the frequency distribution table :

Relative humidity (in %)	Frequency
84 – 86 86 – 88 88 – 90 90 – 92 92 – 94 94 – 96  96 – 98 98 – 100	1 1 2 2 7 6 7 4
Total	30

 

(ii) The data appears to be taken in the rainy season( as the relative humidity is high) .

(iii) The range =99.2-84.9=14.3

4. 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?
Solution:  (i) We construct the frequency distribution table :

Heights (in cm)	Frequency
150 – 155 155 – 160 160 – 165 165 – 170 170 – 175	12 9 14 10 5
Total	50

(ii)  One conclusion that we can draw from the above table is that more than 50% of students are shorter than 165 cm .

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 0.00 - 0.04 , 0.04 - 0.08 , and so on.
(ii) For how many days, was the concentration of sulphur dioxide more than 0.11 parts per million?
 

 

 

Solution: (i) We construct the frequency distribution table :

Concentration of sulphur dioxide  (in ppm)	Frequency
0.00 – 0.04 0.04 – 0.08 0.08 – 0.12 0.12 – 0.16 0.16 – 0.20 0.20 – 0.24	4 9 9 2 4 2
Total	30

(ii) The concentration of sulphur dioxide was more than 0.11 ppm for 8 days .

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: We construct the frequency distribution table :

Number of heads	Frequency
0 1 2 3	6 10 9 5
Total	30

 

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: (i) We construct the frequency distribution table :

Digits	Frequency
0 1 2 3 4 5 6 7 8 9	2 5 5 8 4 5 4 4 5 8
Total	50

(ii) The most frequently occurring digits are 3 and 9 .

 The least frequently occurring digits is 0 .

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 construct the frequency distribution table :

Number of hours	Frequency
0 – 5 5 – 10 10 – 15 15 – 20	10 13 5 2
Total	30

(ii)  2 children

 

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 construct the frequency distribution table :

Life of batteries (in years)	Frequency
2.0 – 2.5 2.5 – 3.0 3.0 – 3.5 3.5 – 4.0 4.0 – 4.5 4.5 – 5.0	2 6 14 11 4 3
Total	40

 

EXERCISE 14.3

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 %):

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 :

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
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:

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:

4. 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 – 126 127 – 135 136 – 144 145 – 153 154 – 162 163 – 171 172 – 180	3 5 9 12 5 4 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:

(ii) Yes, there are other suitable graphical representations for the same data. One such representation is a frequency polygon.

A frequency polygon is a graph that uses line segments to connect the midpoints of each class interval. It helps visualize the distribution of data and can be used when we want to emphasize the continuous nature of the data.

(iii) No, it is not correct to conclude that the maximum number of leaves are 153 mm long. The histogram only represents the frequency of leaves falling within each class interval. The class interval "144.5 – 153.5" has the highest frequency (12 leaves), but it does not necessarily mean that the maximum length of leaves is 153 mm. There could be leaves longer than 153 mm that are not captured in this particular dataset.

 

 

 

5. The following table gives the life times of 400 neon lamps:

Life time (in hours)	Number of lamps
300 – 400 400 – 500 500 – 600 600 – 700 700 – 800 800 – 900 900 – 1000	14 56 60 86 74 62 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?
Solution:

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
0 – 10	3	0 – 10	5
10 – 20	9	10 – 20	19
20 – 30	17	20 – 30	15
30 – 40	12	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.
Solution:  We construct the data table :

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	12	30 – 40	35	10
40 – 50	45	9	40 – 50	45	1

 

We draw a frequency polygon by plotting the class-marks along the horizontal

axis, the frequencies along the vertical-axis, and then plotting and joining the points

B(145, 5), C(155, 10), D(165, 20), E(175, 9), F(185, 6) and G(195, 2) by line segments.

So, the resultant frequency polygon will be ABCDEFGH

 

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
1 – 6 7 – 12 13 – 18 19 – 24 25 – 30 31 – 36 37 – 42 43 – 48 49 – 54 55 – 60	2 1 8 9 4 5 6 10 6 2	5 6 2 10 5 6 3 4 8 10

Represent the data of both the teams on the same graph by frequency polygons.
[Hint : First make the class intervals continuous.]

Solution:  We have ,

The lower limit of the class interval 7 – 12 = 7

The upper limit of the class interval 1 – 6 = 6

The difference = 7 – 6 = 1

So , half difference

=12=0.5

Now, The lower limit = 1 – 0.5 = 0.5

And the upper limit = 6 + 0.5 = 6.5

Class marks

=6.5+0.52=72=3.5

We construct the data table :

Number of balls	Class marks	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
43.5 – 48.5	45.5	10	4
48.5 – 54.5	51.5	6	8
54.5 – 60.5	57.5	2	10

 We draw a frequency polygon by plotting the class-marks along the horizontal

axis, the frequencies along the vertical-axis, and then plotting and joining the points

B(145, 5), C(155, 10), D(165, 20), E(175, 9), F(185, 6) and G(195, 2) by line segments.

So, the resultant frequency polygon will be ABCDEFGH

8. 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 2 – 3 3 – 5 5 – 7 7 – 10 10 – 15 15 – 17	5 3 6 12 9 10 4

Draw a histogram to represent the data above.
Solution:

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:

Number of letters	Number of surnames
1 – 4 4 – 6 6 – 8 8 – 12 12 – 20	6 30 44 16 4

(i) Draw a histogram to depict the given information.
(ii) Write the class interval in which the maximum number of surnames lie.

Solution:

EXERCISE 14.4

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: We arrange the data in ascending order :

0 , 1 , 2 , 3 , 3 , 3 , 3 , 4 , 4 , 5

Mean

x=0+1+2+3+3+3+3+4+4+510

=2810=2.8

For median :  Here, n=10

Median

=n2th+n2+1th2 observation

=102th+102+1th2 observation

=5th+6th2 observation

=3+32

=62=3

For mode : Here 3 occurs most frequently (i.e., four times) . So, the mode is 3 .

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: We arrange the data in the following form :

39 , 40 , 40 , 41 , 42 , 46 , 48 , 52 , 52 , 52 , 54 , 60 , 62 , 96 , 98

For mean :

Meanx

=39+40×2+41+42+46+48+52×3+54+60+62+96+9815

=82215=54.8

For median : Here, n=15

 

Median

=n+12thobservation

=15+12thobservation

= 162thobservation

= 8th observation=52

 

For mode :

39 , 40 , 40 , 41 , 42 , 46 , 48 , 52 , 52 , 52 , 54 , 60 , 62 , 96 , 98

Here, 52 occurs most frequently ,i.e., three times. So, the mode is 52 .

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:   Given , the data :

29 , 32 , 48 , 50 , x , x + 2 , 72 , 78 , 84 , 95

Here, n=10

Median

=n2th+n2+1th2 observation

⇒63=102th+102+1th2 observation

⇒63=5th+6th2 observation

⇒63=x+x+22

⇒63=2x+22

⇒63=2x+12

⇒63=x+1

⇒x=63-1

⇒x=62

Therefore, the value of x is 62 .

 

4. Find the mode of 14 , 25 , 14 , 28 , 18 , 17 , 18 , 14 , 23 , 22 , 14 , 18.
Solution:  We construct the table :

Data	Frequency
14 17 18 22 23 25 28	4 1 3 1 1 1 1

Here, 14 occurs most frequently, i.e., four times.

Therefore, the mode of the data is 14 .

 

5. Find the mean salary of 60 workers of a factory from the following table:

Salary (in Rs)	Number of workers
3000 4000 5000 6000 7000 8000 9000 10000	16 12 10 8 6 4 3 1
Total	60

Solution:  We construct the table :

Salary (in Rs) (xi)	No. of workers  (fi)	fixi
3000 4000 5000 6000 7000 8000 9000 10000	16 12 10 8 6 4 3 1	48000 48000 50000 48000 42000 32000 27000 10000
Total	60	305000

 

We have ,

Mean

x=fixifi

=30500060=5083.33
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) The mean is an appropriate measure of central tendency when calculating the average score of a class in an exam. Add up all the individual scores of the students and then divide the sum by the total number of students to find the mean score.

(ii) The mean is not appropriate when measuring the central tendency of income in a city with a few extremely high earners, as it can be skewed by those outliers. In such cases, the median (middle value when data is arranged in ascending or descending order) is a better representation of the typical income.