WBBSE Solutions For Class 9 Maths Chapter 11 Statistics

Class IX Maths Solutions WBBSE Chapter 11 Statistics Exercise 11.1

Question 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.

Solution: Frequency distribution table

Class Interval	Class Size	Length of Class	Tally Mark	Frequency
0-2	0-2	2	I	11
2-4	2-4	2	II	17
4-6	4-6	2	llll	9
6-8	8-Jun	2	III	3
			Total	40

 

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

34	27	45	21	30	40	11	47	1	15
3	40	12	47	48	18	30	24	25	28
32	31	25	22	27	41	12	13	2	44
43	7	9	49	13	19	32	39	24	3

 

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

Solution: Frequency distribution table

Class	Tally mark	Frequency
1-10	I	6
11-20	III	8
21-30	I	11
31-40	II	7
41-50	III	8
	Total	40

 

Question 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.

Solution: Maximum weight = 110 gm
Minimum weight = 30 gm
∴ Width = 110-30=80 gm
Size of class = 10

Class	Tally mark	Frequency	Cumulative Frequency
30-40	llll	4	4
40-50	I	6	4 + 6=10
50-60	III	3	10 + 3=13
60-70	llll	4	13 + 4=17
70-80	III	8	17 + 8 = 25
80-90	II	7	25 + 7 = 32
90-100	III	3	32 + 3 = 35
100-120	III	3	35 + 3 = 38
110-120	II	2	38 + 2 = 40
	Total	40

 

Note: If the same value occurs in two classes put the value in a higher class.

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

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, 98, 65, 110.

I construct a frequency distribution table for the above data.

Solution: Frequency distribution table

Class	Tally Mark	Frequency
50-60	II	2
60-70	I	6
70-80	llll	4
80-90	llll	4
90-100	II	7
100-110	II	7
110-120	I	6
120-130	II	7
130-140	II	2
	Total frequency	45

 

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

Ages (years)	10-20	20-30	30-40	40-50	50-60	60-70
The number of patients	80	40	50	70	40	20

 

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

Solution: More than type cumulative frequency distribution label

Age (years) Class	No. of patients Frequency	Cumulative frequency
10-20	80	220 + 80 = 300
20-30	40	180 + 40 = 220
30-40	50	130 + 50 = 180
40-50	70	60 + 70 = 130
50-60	40	20 + 40 = 60
60-70	20	20
Total	300

 

Question 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: Frequency distribution table

Class no. of students	Frequency	Cumulative frequency
10	17	17
10-20	22-17 = 5	22
20-30	29-22 = 7	29
30-40	37-29 =8	37
40-50	50-37 = 13	50
50-60	60-50 = 10	60

 

Question 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

 

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Solution: Frequency distribution table

Marks	Class	The no. of students	Frequency
More than 60	More than 60	0	0
More than 50	50-60	16	16-0 = 16
More than 40	40-50	40	40-16 = 24
More than 30	30-40	75	75-40 = 35
More than 20	20-30	87	87-75 = 12
More than 10	10-20	92	92-87 = 5
More than 0	0-10	100	100-92 = 8

 

Question 8.

1. Which one of the following is a graphical (pictorial) representation of statistical data?

1. Line graph
2. Raw data
3. Cumulative frequency
4. Frequency.

Solution: 1. Line graph

2. The range of the data 12, 25, 15, 18, 17, 20, 22, 26, 6, 16, 11, 8, 19, 10, 30, 20, 32 is

1. 10
2. 15
3. 18
4. 26

Solution: Greatest value = 32
Minimum value = 6
Range 326 = 26

∴ 4. 26

3. The class size of classes 1-5, 6-10 is

1. 4
2. 5
3. 4.5
4. 5.5

Solution: Class size=5-110-6=4
∴ 1. 4

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

1. 12.5-17.5
2. 17.5-22.5
3. 18.5 21.5
4. 19.5 20.5

Solution:  20-15-25-20-30-25=5
∴ Class size = 5
Class size of 1st class = 17.5-12.5 = 5

Mid value = \(\frac{12.5+17.5}{2}=\frac{30.0}{2}=15 \neq 20\)

In the class, 17.5-22.5, class size = 22.5 17.5 = 5

Mid value = \(\frac{12.5+17.5}{2}\) = 20

The mid value of the last two classes is 20, but the class length is not 5.

∴ 2. 17.5 22.5

5. In a frequency distribution table if the mid-point of a class is 10 and the class size of each class is 6; the lower limit of the class is

1. 6
2. 7
3. 8
4. 12

Solution: Let upper limit of = x and lower limit = y.

According to the problem, 1st conditions: \(\frac{x+y}{2}=10\)

or, x + y = 20 …..(1)

According to the problem, 2nd conditions:
or, x – y = 6 ….(2)

Subtracting (2) from (1), 2y = 14

or, \(y=\frac{14}{2}=7\)

∴ Lower limit of the class
∴ 2. 7

Question 9.

1. 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: Mid value of the class = m
Upper boundary of the class = u
∴ Mid value = m
∴ Lower class boundary= 2 x mid value =2 x m-u = 2m – u

2. In a continuous frequency distribution table, if the mid-point of a class is 42 and class size is 10, then let us write the upper and lower limits of the class.

Solution: Let the upper-class boundary be x and the lower boundary be y. According to the problem, 1st condition

According to the problem, 1st condition \(\frac{x+y}{2}=42\)

or, x + y = 84 ……(1)

According to the problem, 2nd condition 2x = 94

or, \(x=\frac{94}{2}=47\)

Putting the value of x in equation (1), 47 + y = 84
⇒ y= 84-47 = 37
∴ Upper limit = 47
Lower limit 37

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

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

Solution: 1st class = 70-74
Class size the 1st class = 74-70 = 4
Frequency of the 1st class = 3

Frequency dencing of the 1st class \(=\frac{\text { Frequency }}{\text { Classsize }}\)

\(=\frac{3}{4}\)= 0.75

4. Let us write the frequency density of the last class

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

 

Solution: Last class = 85-89
Frequency = 8
Total frequency=3+4+5+8=20

∴ Frequency of the class \(y=\frac{\text { Frequency of the class }}{\text { Total frequency }}\)

\(\begin{aligned}
& =\frac{8}{20} \\
& =\frac{2}{5}=0.4
\end{aligned}\)

 

5. Let us write from the following examples which one indicates attribute and which one indicates variable.

1. Population of the family.
2. Daily temperature.
3. Educational value.
4. Monthly income.
5. Grade obtained in Madhyamik Examination.

Solution:
1. Population of the family – Variable
2. Daily temperature – Variable
3. Educational – Attribute
4. Monthly income – Variable
5. Grade obtained in – Attribute

Class 9 Maths WBBSE Chapter 11 Statistics Exercise 11.2

 

Question 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: Along x-axis one side of the smallest square = 1 mark & along y-axis two sides of the smallest square 1 student. And putting the points and joining then ABCDEFGH frequency polygon is obtained.

 

 

Question 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:

Class	Mid value	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

 

 

1. Class along x-axis & frequency along y-axis.
2. Taking the points of mid-value & frequency points are plotted, B(2.5, 4) C(7.5, 10), D(12.5, 24), E(17.5, 12), F(22.5, 20), G(27.5, 8)
3. Joining A, B, C, D, E, F, G, H with a scale we get, ABCDEFGH a frequency polygon.

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

Daily profit (Rs.)	0-50	50-100	100-150	150-200	200-250
Number of shops	8	15	10	12	5

 

 

Solution: Along x-axis – Daily profit
Along y-axis No. of shops

One side of the smallest square along x-axis = Rs. 10 (Profit); two sides of the smallest square along y-axis = 1 shop.

I draw the histogram for the above data.

Question 4. By measuring, Mita wrote the heights of her 75 friends of their school in the table given below : I draw the histogram of the data collected by Mita.

Height (cm.)	136-142	142-148	148-154	154-160	160-166
Number of friends	12	18	26	14	5

 

Solution: Along x-axis – height (in cm) &
along y-axis No. of friends

unit:- 5 sides of the smallest square along x-axis = 6 cm & 1 side of the smallest square along y-axis = 1 friend. Putting the value, I draw the histogram for the given data

 

 

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

Age (in years)	10-15	16-21	22-27	28-33	34-39	40-45
Number of Hindi-speaking people	8	14	10	20	6	12

 

Solution: First make a frequency distribution table

Class (year)	Class boundary	Length of class	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	0.6	20
34-39	33.5 – 39.5	6	6
40-45	39.5 – 45.5	6	12

 

Along the x-axis – Age (year) & Along the y-axis – No. of Hindi-speaking people Unit 5 sides of the smallest square along the x-axis & 2 sides of the smallest square along the y-axis = 1 people

Putting the value, I draw the histogram.

 

 

Question 6. I draw the histogram of the frequency distribution table given below:

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

 

Solution: Frequency distribution table

Class	Class-boundary	Length of Class	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

 

Along x-axis-class boundary
& Along y axis – Frequency along x-axis
unit one side of the smallest square = 2 units
& Four side of the smallest square along y-axis = 1 unit
Putting the value, Histogram is obtained

 

 

Question 7. By drawing the 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	418	6

 

Solution: Frequency distribution table

Class (Rs.)	Mid value	Frequency (no. of members)
17.5-22.5	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

 

Along x-axis Amount of Subscription (Rs.)
& Along y-axis Number of members (Frequency)

Unit 2 sides of the smallest square along the x-axis = Re. 1; 2 sides of the smallest square along y-axis = 1 member.

Now I draw the histogram.

Now for the drawing of frequency polygon just before the first class interval I take class interval 12.5 17.5 and just after class interval, I take class interval 52.5 –
57.5. The frequencies of these two class intervals are 0.

(30,16), (35,10), (40,4), (45,18), (50-6) & (55,0) successively with straight lines, I have drawn the frequency polygon ABCDEFGHI.

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

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

 

Hints: At first, by the exclusive class method the statistical data will be constructed as a frequency distribution table with class boundaries given below:

Number of children	0-1	1-2	2-3	3-4	4-5	5-6
Number of families	120	85	50	25	15	5

 

Solution: Frequency distribution table

Class (No. of children)	Frequency (No. of families)
0-1	120
1-2	85
2-3	50
3-4	25
4-5	15
5-6	5

 

Along x-axis – No of children & Along y-axis – Number of families

Unit: 5 sides of the smallest square along x-axis = 1 child. 2 sides of the smallest square along y-axis = 5 families

Putting the values Histogram is obtained.

 

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

Ages (years)	25-31	31-37	37-43	43-49	49-55
Number of teachers	10	13	5	3	1

 

Now, we have to draw a histogram and frequency polygon with the given data graphically:

Solution: Frequency distribution table

Class Age (years)	Class boundary	Mid value	Length	Frequency(No. of teachers)
25-31	25-31	28	6	10
31-37	31 -37	34	6	13
37-43	37-43	40	6	5
43-49	43-49	-46	6	3
49-55	49-55	52	6	1

 

Along x-axis Age (years) & Along y-axis – The number of teachers

Unit: 1 side of the smallest square along x – axis = 1 year and 5 sides of the smallest square along the y-axis = 1 teacher.

Now I draw the histogram.

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Now for drawing frequency polygon just before 1st class interval, I take a class interval of 19-25, and just after the last class interval, I take another class interval of 55- 61. The frequencies of these two class intervals are 0 (Zero).

Then by joining the points (22,0), (28,10), (34,13), (40,5), (46,3), (52,1), and (58,0) successively with straight lines, frequency polygon ABCDEFG is obtained.

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

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

Solution: Frequency distribution table

Class	Mid value	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

 

Along x-axis Class & along y-axis Frequency

Unit: 2 sides of the smallest square along x-axis = 1 unit and 2 sides of the smallest square along the y-axis= 1 unit.

Now for drawing frequency polygon just before 1st class interval I take a class interval of 70-75 and just after the last class interval, I take another class interval of 105- 110. The frequencies of the two class intervals are 0 (Zero).

Then by doing the points (72.5,0) & (107.5,0) successively with a straight line the freguency polygon ABCDEFG is obtained.

Question 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	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	25.5	6
31-40	30.5 – 40.5	35.5	12
41-50	40.5 – 50.5	45.5	4

 

Along x-axis – Class & along y-axis – Frequency

Unit: 1 side of the smallest square along x-axis = 1 unit and 4 sides of the smallest square along y-axis = 1 unit

Now for drawing frequency polygon just before 1st class interval I take a class interval of -10,-0 and just after the last class interval, I take another class interval 51- 60.

The frequencies of these two class intervals are 0 (Zero). Then by joining the points (-4.5,0), (5.5,8), (15.5,3), (25.5,6), (35.5,12), (45.5,4), and (55.5,0) successively with straight lines we obtained ABCDEFG, frequency polygon.

 

 

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

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

 

Solution: To draw the frequency polygon

Frequency distribution table

Class Age (years)	Mid value	Frequency (No. of illiterates)
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

 

Along x-axis-Class (Age) & along y-axis is the Number of illiterates

Unit: 2 sides of the smallest square along x axis = 1 year & 1 side of the smallest square along the y-axis = 5 number of illiterates.

Now for drawing frequency polygon just before 1st class interval I take a class interval 5-10 and just after the last class interval, I take another ass interval of 35-40.

The frequencies of these two class intervals are 0 (Zero). Then by joining the points (7.5,0), (12.5,40), (17.5,90), (22.5,100), (27.5,60), (32.5,160), and (37.5,0) straight lines we get the required frequency polygon ABCDEFG.

 

Question 13. I have written in the following the frequency of the number of goals given by the teams in our Kolkata football league in the previous month. I draw the frequency polygon for the representation of the data.

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

 

Solution: Frequency distribution table

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

 

Along x-axis-score (goal) & along y-axis – Frequency

Unit 10 sides of the smallest square along x-axis = 1 unit and 2 sides of the smallest square along y-axis = 1 unit.

Now for drawing frequency polygon just before 1st class interval I take a class interval of -1,-0 and just after the last class interval, I take another class interval of 7-0.

The frequencies of these two class intervals are 0 (Zero). Then by joining the points (1,0), (0,15), (1,20), (2,12), (3,0), (4,6) and (5,3), (6,1) & (7,0) successively with straight lines we obtained frequency polygon ABCDEFGHI,

 

 

Question 14.  Let us discuss

1. Each of the area of each of the rectangle of a histogram is proportional to

1. The mid-point of that class
2. The class size of that class
3. The frequency of that class
4. The cumulative frequency of that class

Solution: 3. The frequency of that class

2. A frequency polygon is drawn

1. Upper limit of the class
2. Lower limit of the class
3. Mid-value of the class
4. Any value of the class

Solution 3. Mid-value of the class

3. To draw a histogram, the class

1. Along y-axis
2. Along x-axis
3. Along x-axis and y-axis both
4. In between x-axis and y-axis

Solution: 2. Along x-axis

4. In the case of drawing a histogram,

1. Frequency
2. Class boundary
3. Range.
4. Class size.

Solution: 4. Class size.

5. A histogram is the graphical representation of grouped data whose class- boundary and frequency are taken respectively,
1. Along the vertical axis and horizontal axis,
2. Only along the vertical axis,
3. Only along horizontal axis,
4. Along the horizontal axis and vertical axis.

Solution: 4. Along the horizontal axis and vertical axis.