# 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

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

or, a+b=20\\

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 \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.

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

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

ABCDEFG represents the frequency polygon.

Taking 5 unit = 5.5 Class, along X-axis and 1 unit = 1 frequency along Y-axis.

12. A special drive wall be taken for women literacy in total in our village. For this reason, we have collected following data:

 Age 10-15 15-20 20-25 25-30 30-35 Number of illiterates 40 90 100 60 160

Solution: Frequency Distribution table

 Class Mid. value of the classes Frequency 10-15 12.5 40 15-20 17.5 90 20-25 22.5 100 25-30 27.5 60 30-35 32.5 160

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 illiterate along Y-axis.

13. I have written in the following the frequency of the number of goals given by the teams in our Kolkata Football-League in previous month.

 Scores 0 1 2 3 4 5 6 Frequency 15 20 12 8 6 3 1

Solution: 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 illiterate along Y-axis.

## 14. M.C.Q.:

(i) Each of the area of each of the rectangles of a histogram is proportional to

(a) the mid-point of that class

(b) the class-size of that class

(c) the frequency of that class

(d) the cumulative frequency of that class

Ans: (c) the frequency of that class.

(ii) A frequency polygon is drawn by the frequency of the class and

(a) upper limit of the class

(b) lower limit of the class

(c) mid-value of the class

(d) any value of the class

Ans: (c) mid-value of the class

(iii) To draw a histogram, the class-boundaries are taken

(a) along Y-axis

(b) along X-axis

(c) along X-axis and Y-axis

(d) in between X-axis and Y-axis.

Ans: (b) along X-axis

(iv) In case of drawing a histogram, the base of the rectangle of each class is

(a) frequency

(b) class-boundary

(c) range

(d) class-size

Ans: (d) class-size

(v) A histogram is the graphical representation of a grouped data whose class-boundary and frequency are taken respectively,

(a) along vertical axis and horizontal axis

(b) only along vertical axis,

(c) only along horizontal axis

(d) along horizontal axis and vertical axis

Ans: (d) along horizontal axis and vertical axis

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