Chapter – 11 : Statistics | Chapter Solution Class 9

Book Name	: Ganit Prakash
Subject	: Mathematics (Maths)
Class	: 9 (Madhyamik/WB)
Publisher	: Prof. Nabanita Chatterjee
Chapter Name	: Statistics (11th Chapter)

Let us work out – 11.1

Mriganka has written the ages of 30 workers of their factory:

Age (Year)	21-23	23-25	25-27	27-29	29-31	31-33	33-35
No. of Workers	3	4	5	6	5	4	3

I prepare a cumulative frequency distribution table from the above data and from there, I find out the answers of the following question:

(i) Let me write the number of workers in the factory whose ages are less than 27 years.

(ii) Let me write the number of workers whose ages are 25 years or more than 25 years.

(iii) Let me write the ages of the workers whose ages are 25 years or more than 25 years but less than 33 years.

Solution: At first, I prepare the cumulative frequency distribution table.

Class Boundary (Age Year)	Less than type cumulative frequency
Less than 21	0
Less than 23	3
Less than 25	7
Less than 27	12
Less than 29	18
Less than 31	23
Less than 33	27
Less than 35	30

Class Boundary (Age Year)	Less than type cumulative frequency
35 or more than 35	0
33 or more than 33	3
31 or more than 31	7
29 or more than 29	12
27 or more than 27	18
25 or more than 25	23
23 or more than 23	27
21 or more than 21	30

(i) No. of workers in the factory whose ages are less than 27 years = 12

(ii) No. of workers in the factory whose ages are more than 25 = 23

(iii) No. of workers in the factory whose ages are 25 years or more than 25 years but less than 33 years = 23 – 3 = 20

Let us work out – 11.1

1. I have written the number of children belonging to each of 40 families in our locality below –

1 2 6 5 1 5 1 3 2 6

2 3 4 2 0 4 4 3 2 2

0 0 1 2 2 4 3 2 1 0

5 1 2 4 3 4 1 6 2 2

I prepare a frequency distribution table of the above-given data whose classes are 0 – 2, 2 – 4 ………. etc.

From this frequency distribution table, let me understand and write (i) class-1nterval, (ii) class-size, (iii) frequency of the class, (iv) class-limit

Solution:

Class interval	Class limit	length of class	frequency
0-2	0-2	2	11
2-4	2-4	2	17
4-6	4-6	2	9
4-6	4-6	2	3
Total frequency = 40

2. Given below are the marks obtained by 40 students in a test of school:

34 27 45 21 30 40 11 47 01 15

03 40 12 47 48 18 30 24 25 28

32 31 25 22 27 41 12 13 02 44

43 07 09 49 13 19 32 39 24 03

I construct a frequency distribution table of these mairks by taking the classes 1 – 10, 11 – 20, ……….. 41 – 50.

Class interval	Class limit	frequency
1-10	1-10	6
11-20	11-20	8
21-30	21-30	11
31-40	31-40	7
41-50	41-50	8
Total frequency = 40

3. There are many oranges in a basket. From this basket, by aimlessly taking 40 oranges I wrote below their weights (gm):

45, 35, 30, 55, 70, 100, 80, 110, 80, 75, 85, 70, 75, 85, 90, 75, 90, 30, 55, 45,

40, 65, 60, 50, 40, 100, 65, 60, 40, 100, 75, 110, 30, 45, 84, 70, 80, 95, 85, 70.

Now I construct a frequency distribution table and a less than type cumulative frequency distribution table for the above given data.

Class interval	tally mark	frequency
30-40	4	4
40-50	6	10
50-60	3	13
60-70	4	17
70-80	8	25
80-90	7	32
90-100	3	35
100-110	3	38
110-120	2	40
Total frequency = 40

4. Mitali and Mohidul wrote below the amount of money of electricity bills for this month of the 45 houses of their village.

116, 127, 100, 82, 80, 101, 91, 65, 95, 89, 75, 92, 129, 78, 87,

101, 65, 52, 59, 65, 95, 108, 115, 121, 128, 63, 76, 130, 116,

108, 118, 61, 129, 127, 91, 130, 125, 101, 116, 105, 92, 75, 18, 65, 110.

I construct a frequency distribution table for the above data.

Class interval	frequency
50-60	2
60-70	6
70-80	4
80-90	4
90-100	7
100-110	7
110-120	6
120-130	7
130-140	2
Total frequency = 40

5. Maria has written the ages of 300 patients of a hospital in the table given below:

I construct a more than type cumulative frequency distribution table for the above data.

Age (Year)	10-20	20-30	30-40	40-50	50-60	60-70
The number of patients	80	40	50	70	40	20

Solution:

Age (Year)	Number of patients frequency	Cumulative frequency (more than type)
10-20	80	300
20-30	40	220
30-40	50	180
40-50	70	130
50-60	40	60
60-70	20	20

6. Let us observe the following cumulative frequency distribution table and construct a frequency distribution table:

Classes	Below 10	Below 20	Below 30	Below 40	Below 50	Below 60
The number of Students	17	22	29	37	50	60

Solution:

Classes	Less than 10	10-20	20-30	30-40	40-50	50-60
The number of Students	17	5	7	8	13	10

7. Let us observe the following cumulative frequency distribution table and construct the frequency distribution table:

Marks obtained	The number of students
More than 60	0
More than 50	16
More than 40	40
More than 30	75
More than 20	87
More than 10	92
More than 0	100

Solution:

Obtained marks	0-10	10-20	20-30	30-40	40-50	50-60	More Than 0
Number of Students	8	5	12	35	24	16	0

8. M.C.Q.:

(i) Which one of the following is a graphical (Pictorial) representation of a statistical data?

(a) Line graph

(b) Raw Data

(c) Cumulative frequency

(d) Frequency

Solution: (a) Line graph.

(ii) The range of the data 12, 25, 15, 18, 17, 20, 22, 26, 16, 11, 8, 19, 10, 30, 20, 32 is

(a) 10

(b) 15

(c) 18

(d) 26

Solution: Range = Max – Min

= 32 – 6 = 26

$\therefore \quad$ (d) is correct option

(iii) The class size of the classes 1 – 5, 6 – 10 is

(a) 4

(b) 5

(c) 4.5

(d) 5.5

Solution: (a) is correct option

(iv) In a frequency distribution table, the mid-points of the classes are 15, 20, 25, 30 respectively. The class having mid-point as 20 is

(a) 12.5 – 17.5

(b) 17.5 – 22.5

(c) 18.5 – 21.5

(d) 19.5 – 20.5

Solution: (b) is correct option.

(v) In a frequency distribution table, the mid-point of a class is 10 and the class size of each class is 6; the lower limit of the class is

(a) 6

(b) 7

(c) 8

(d) 12

Solution: Let lower limit of the class = a. Upper limit of the class = b

$\therefore \quad b-a=6 \\$

b = 6 + a and $\frac{a+b}{2}=10\\$

or, $a+b=20\\$

or, $a+6+a=20 \quad[\because b=6+a]\\$

or, $2 a=20-6 \\$

or, $2 \mathrm{a}=14 \\$

$\therefore$ a = 14

$\therefore$ (b) is correct option.

9. Short answer type questions:

(a) In a continuous frequency distribution table if the mid-point of a class is m and the upper class- boundary is u, then let us find out the lower class-boundary.

Solution: We have,

$\text { Mid point }\text { of a class } = \frac{lower class boundary + upper class boundary }{2}$

or, $= \frac{lower class boundary + u }{2} = m$

$\therefore$ lower class boundary + u = 2

$\therefore$ lower class boundary = 2m – u

(b) In a continuous frequency distribution table, if the mid-point of a class is 42 and class size is 10, then let write the upper and lower limit of the class.

Solution: Let lower limit of the class = a

Upper limit of the class =b

We have, $\frac{a+b}{2}=42$

and b – a = 10, or, b = 10 + a

$\therefore \quad$ a + b = 84

$\text { or, } a+10+a=84 \quad[\because b=10+a] \\$

$\text { or, } 2 a=84-10 \\$

$\text { or, } 2 a=74$

$\therefore$ a = 37

$\therefore \quad b =10+a \\$

$=10+37 \quad=47 \\$

$\therefore \quad \text { lower limit of the class }=37 \\$

$\text { Upper limit of the class }=47$

$\therefore \quad$ lower limit of the class = 37

Upper limit of the class =47

(c)

Class Limit	70-74	75-79	80-84	85-89
Frequency	3	4	5	8

Let us write the frequency density, of the first class of the above frequency distribution table.

Solution:

We have,

$\text { Frequency density } =\frac{\text { Class frequency }}{\text { Class }- \text { size }} \\$

$=\frac{3}{74-70} \\$

$=\frac{3}{4}$

= 0.75 (Ans.)

(d) Let us write the frequency density of the last class of the question (c)

Solution:

$\text { Frequency density } =\frac{8}{89-85} \\$

$=\frac{8}{4}=2 \text { (Ans.) }\\$

(e) Let us write from the following examples which one indicates attribute and which one indicates variable.

(i) Population of the family

(ii) Daily temperature

(iii) Education value

(iv) Monthly income.

(v) Grade obtained in Madhyamik Examination.

Solution:

$\text { Attribute } \rightarrow \text { (iii) \& (v) } \\$

$\text { Variable } \rightarrow \text { (i), (ii) \& (iv) }\\$

Let us work out – 11.2

1. I construct the frequency polygon for the following marks obtained by 75 learners of Pritha’s school.

Marks obtained	30	40	50	60	70	80
Number of students	12	18	21	15	6	3

In the graph paper, taking suitable measures along horizontal and vertical lines, the points (20,0), (30,12), (40,18), (50,21), (60,15), (70,6), (80,3) and (90,0) are plotted on the graph paper and then I draw the frequency polygon by adding them.

Solution:

We take $\overrightarrow{\mathrm{OX}}$ as $\mathrm{X} -axis$ and $\overrightarrow{\mathrm{OY}}$ as $\mathrm{Y} -axis\\$

ABCDEFH represents the frequency polygon.

Taking the length lunit = 2 marks along x -axis & the length l unit = 1 student along Y-axis.

2. I draw the frequency polygon for the following frequency distribution table.

Classes	0-5	5-10	10-15	15-20	20-25	25-30
Frequency	4	10	24	12	20	8

Solution:

Classes	Mid-value of classes	Frequency
0-5	2.5	4
5-10	7.5	10
10-15	12.5	24
15-20	17.5	12
20-25	22.5	20
25-30	27.5	8

We take $\overrightarrow{\mathrm{OX}}$ as $\mathrm{X} -axis$ and $\overrightarrow{\mathrm{OY}}$ as $\mathrm{Y} -axis\\$

ABCDEFG represents the frequency polygon.

Taking the length 1 unit = 1 class along X-axis & the length 1 unit = 1 frequency along y-axis.

3. I write below in tabular form the daily profit of the 50 shops of the village Bakultala.

Daily Proft (Rs.)	0-50	50-100	100-155	150-200	200-250
Number of Shops	8	15	10	12	5

I draw the histogram for the above data.

Solution: We take XOX’ as X-axis and YOY’ as Y-axis respectively.

Taking the length 1 unit = 10 marks along x-axis and the length 1 unit = 1 profit along y-axis.

4. By measuring, Mita wrote the heights of her 75 friends of their school in the table given below.

Height	136-142	142-148	148-154	155-160	160-166
Number of friends	12	18	26	14	05

I draw the histogram of the data collected by Mita.

Solution: We take XOX’ as X-axis and YOY’ as Y-axis

Taking the length 5 units = 6cm. along X-axis and 1 unit = 1 friend along Y-axis.

5. In our locality, by collecting the number of Hindi-speaking people between ages of 10 years to 45 years, I write them in table given below:

Age(In year)	10-15	16-21	22-27	28-33	34-39	40-45
Number of Hindi speaking	8	14	10	20	5	12

I draw the histogram of the data collected by Mita.

Solution: We take XOX’ as X-axis and YOY’ as Y-axis

Class of age	Particular class with class boundary	Class-size	Frequency
10-15	9.5-15.5	6	8
16-21	15.5-21.5	6	14
22-27	21.5-27.5	6	10
28-33	27.5-33.5	6	20
34-39	33.5-39.5	6	6
40-45	39.5-45.5	6	12

Taking the length 5 units = 6 years along X-axis and the length

1 unit = 1 people along Y-axis.

6. I draw the historgram of the frequency distribution table below.

Classes	1-10	11-20	21-30	31-40	41-50	51-60
Frequency	8	3	6	12	2	7

Solution: We take XOX’ as X-axis and YOY’ as Y-axis

Taking the length 5 unit = 6 cm . along x -axis and 1 unit = 1 friend along Y-axis.

Class of age	Particular class with class boundary	Class-size	Frequency
1-10	0.5-10.5	10	8
11-20	10.5-20.5	10	3
21-30	20.5-30.5	10	6
31-40	30.5-40.5	10	12
41-50	40.5-50.5	10	2
51-60	50.5-60.5	10	7

We take XOX’ as X-axis and YOY’ as Y-axis

Taking the length 5 units =10 class along X-axis and 1 unit = 1 frequency along Y-axis.

7. By drawing histogram, I draw the frequency polygon of the frequency distribution table given below:

Amount of Subscriptions (Rs.)	20	25	30	35	40	45	50
Number of Members	20	26	16	10	4	18	6

Solution: To draw the histogram for the given data I have got the frequency distribution table:

Classes	Mid. value of the classes	Frequency
17.7-22.5	20	20
22.5-27.5	25	26
27.5-32.5	30	16
32.5-37.5	35	10
37.5-42.5	40	4
42.5-47.5	45	18
47.5-52.5	50	6

We take XOX’ as X-axis and YOY’ as Y-axis

ABCD represents the frequency polygon.

Taking 1 unit = 1 subscription along X-axis and 1 unit = 1 member along Y-axis.

8. I draw the histogram for the following frequency distribution table.

Number of children	0	1	2	3	4	5
Number of families	120	85	50	25	15	5

Solution: To draw the histogram for the given data I have got the frequency distribution table:

Classes	Mid. value of the classes	Frequency
0-1	0.5	120
1-2	1.5	85
2-3	2.5	50
3-4	3.5	25
4-5	4.5	15
5-6	5.5	5

We take XOX’ as X-axis and YOY’ as Y-axis

ABCDEFG represents the frequency polygon.

Taking 5 units = 1 children along X-axis and 1 unit = 1 family along Y-axis.

9. I have written the ages of 32 teachers of Primary School in the village Virsingha in a table given below :

Ages (Years)	25-31	31-37	37-43	43-49	49-55
Number of teachers	10	13	05	03	01

I represent the above-given data graphically by histogram and frequency polygon.

Solution: Frequency Distribution table

Classes	Mid. value of the classes	Frequency
25-31	28	10
31-37	34	13
37-43	40	05
43-49	46	03
49-55	52	01

We take XOX’ as X-axis and YOY’ as Y-axis

ABCDEFG represents the frequency polygon.

Taking 1 unit = 1 year, along X-axis and 1 unit = 1 teacher along Y-axis.

10. I draw the frequency polygon for the following frequency distribution table:

Class	75-80	80-85	85-90	90-100	100-105
Frequency	12	18	22	10	8

Solution: Frequency Distribution table

Classes	Mid. value of the classes	Frequency
75-80	77.5	12
80-85	82.5	18
85-90	87.5	22
90-100	95	10
100-105	102.5	8

We take XOX’ as X-axis and YOY’ as Y-axis

ABCDEFG represents the frequency polygon.

Taking 1 unit = 1 Class, along X-axis and 1 unit” frequency along Y-axis.

11. I draw the frequency polygon for the following frequency distribution table:

Class	1-10	11-20	21-30	31-40	41-50
Frequency	8	3	6	12	4

Solution: Frequency Distribution table

Class	Class Boundary	Mid. value of the classes	Frequency
1-10	0.5-10.5	5.5	8
11-20	10.5-20.5	15.5	3
21-30	20.5-30.5	14.5	6
31-40	30.5-40.5	35.5	12
41-50	40.5-50.5	45.5	4