SELINA Solution Class 9 Chapter 18 Statistics Exercise 18B

Question 1

Construct a frequency polygon for the following distribution:

Class-intervals	0-4	4 - 8	8 - 12	12 - 16	16 - 20	20 - 24
Frequency	4	7	10	15	11	6

Sol:

The frequency polygon is shown in the following figure

Steps:

(i) Drawing a histogram for the given data.

(ii) Marking the mid-point at the top of each rectangle of the histogram drawn.

(iii) Also, marking mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.

(iv) Joining the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

Question 2

Construct a combined histogram and frequency polygon for the following frequency distribution:

Class-Intervals	10 - 20	20 - 30	30 - 40	40 - 50	50 - 60
Frequency	3	5	6	4	2

Sol:

Steps:

1. Draw a histogram for the given data.

2. Mark the mid-point at the top of each rectangle of the histogram drawn.

3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.

4. Join the consecutive mid-point marked by straight lines to obtain the required frequency polygon.

5. The require combined histogram and frequency polygon are shown in the following figure:

Question 3

Construct a frequency polygon for the following data:

Class-Intervals	10 - 14	15 - 19	20 - 24	25 - 29	30 - 34
Frequency	5	8	12	9	4

Sol:

The class intervals are inclusive. We will first convert them into the exclusive form.

Class-Interval	Frequency
9.5 - 14.5	5
14.5 - 19.5	8
19.5 - 24.5	12
24.5 - 29.5	9
29.5 - 34.5	4

Steps:

1. Draw a histogram for the given data.
2. Mark the mid-point at the top of each rectangle of the histogram drawn.
3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon. 

The required frequency polygon is as follows:

Question 4

The daily wages in a factory are distributed as follows:

Daily wages (in Rs.)	125 - 175	175 - 225	225 - 275	275 - 325	325 - 375
Number of workers	4	20	22	10	6

Draw a frequency polygon for this distribution.

Sol:

Steps:

1. Draw a histogram for the given data.
2. Mark the mid-point at the top of each rectangle of the histogram drawn.
3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

The required frequency polygon is as follows:

Question 5.1

Draw frequency polygons for each of the following frequency distribution: 

(a) using histogram

(b) without using histogram

C.I	10 - 30	30 - 50	50 - 70	70 - 90	90 - 110	110 - 130	130 - 150
ƒ	4	7	5	9	5	6	4

Sol:

using histogram

C.I	ƒ
10 - 30	4
30 - 50	7
50 - 70	5
70 - 90	9
90 - 110	5
110 - 130	6
130 - 150	4

Steps:

1. 1. Draw a histogram for the given data.
   2. Mark the mid-point at the top of each rectangle of the histogram drawn.
   3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
   4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon. 

Without using Histogram:

Steps: 

1. Find the class-mark (mid-value) of each given class-interval.
   class mark = mid-value = ${\displaystyle \frac{\text{Upper limit + Lower limit}}{2}}$
2. On a graph paper, mark class-marks along X-axis and frequencies along Y-axis.

3. On this graph paper, mark points taking values of class-marks along the X-axis and the values of their corresponding frequencies along Y-axis.

4. Draw line segments joining the consecutive points marked in step (3) above.

   C.I.	Class-mark	f
   -10 - 10	0	0
   10 - 30	20	4
   30 - 50	40	7
   50 - 70	60	5
   70 - 90	80	9
   90 - 110	100	5
   110 - 130	120	6
   130 - 150	140	4
   150 - 170	160	0

5. 

Question 5.2

Draw frequency polygons for each of the following frequency distribution:
(a) using histogram
(b) without using histogram

C.I	5 -15	15 -25	25 -35	35 - 45	45-55	55-65
ƒ	8	16	18	14	8	2

Sol:

Using Histogram:

C.I.	f
5 - 15	8
15 - 25	16
25 - 35	18
35 - 45	14
45 - 55	8
55 - 65	2

Steps:

1. Draw a histogram for the given data.
2. Mark the mid-point at the top of each rectangle of the histogram drawn.
3. Also, mark the mid-point of the immediately lower class-interval and mid-point of the immediately higher class-interval.
4. Join the consecutive mid-points marked by straight lines to obtain the required frequency polygon.

Without using Histogram:
Steps:

1. Find the class-mark (mid-value) of each given class-interval.
2.  
3. On a graph paper, mark class-marks along X-axis and frequencies along Y-axis.
4. On this graph paper, mark points taking values of class-marks along the X-axis and the values of their corresponding frequencies along the Y-axis.
5.  
6. Draw line segments joining the consecutive points marked in step (3) above.
   

   C.I.	Class-mark	f
   -5 - 5	0	0
   5 - 15	10	8
   15 - 25	20	16
   25 - 35	30	18
   35 - 45	40	14
   45 - 55	50	8
   55 - 65	60	2
   65 - 75	70	0

   

SELINA Solution Class 9 Chapter 18 Statistics Exercise 18A

Question 1.1

State, the following variable is continuous or discrete:
number of children in your class.

Ans: Discrete Variable 

 Question 1.2

State, the following variable is continuous or discrete:
Distance travelled by car.

Ans: Continuous Variable 

Question 1.3

State, the following variable is continuous or discrete:
Sizes of shoes.

Ans: Discrete Variable 

Question 1.4

State, the following variable is continuous or discrete: Time.

Ans: Continuous Variable 

 

Question 1.5

State, the following variable is continuous or discrete: Number of patients in a hospital.

Ans: Discrete Variable 

 

Question 2

Given below are the marks obtained by 30 students in an examination:

08	17	33	41	47	23	20	34
09	18	42	14	30	19	29	11
36	48	40	24	22	02	16	21
15	32	47	44	33	01

Taking class intervals 1 - 10, 11 - 20, ....., 41 - 50;
make a frequency table for the above distribution.

Sol:

The frequency table for the given distribution is

Marks	Tally Marks	Frequency
1 - 10		4
11 - 20		8
21 - 30		6
31 - 40		6
41 - 50		6

Question 3

The marks of 24 candidates in the subject mathematics are given below:

45	48	15	23	30	35	40	11
29	0	3	12	48	50	18	30
15	30	11	42	23	2	3	44

The maximum marks are 50. Make a frequency distribution taking class intervals 0 - 10, 10-20, .......

Sol:

The frequency table for the given distribution is

Marks	Tally Marks	Frequency
0 - 10	||||	4
10 - 20		6
20 - 30	|||	3
30 - 40	||||	4
40 - 50		7

In this frequency distribution, the marks 30 are in the class of interval 30 - 40 and not in 20 - 30. Similarly, marks 40 are in the class of interval 40 - 50 and not in 30 - 40.

Question 4.1

Fill in the blank :
 A quantity which can very from one individual to another is called a ............. 

Sol:

A quantity which can very from one individual to another is called a Variable.

Question 4.2

Fill in the blank :
Sizes of shoes are ........... variables.

Sol:

Sizes of shoes are Discrete variables.

Question 4.3

Fill in the blank :
Daily temperatures is ........... variable.

Daily temperatures is Continuous  variable.

Sol:

Daily temperatures is Continuous  variable.

Question 4.4

Fill in the blank :
The range of data 7, 13, 6, 25, 18, 20, 16 is ............

Sol:

The range of the data 7, 13, 6, 25, 18, 20, 16 is The range is 25 - 6 = 19.

Question 4.5

Fill in the blank : 
In the class interval 35 - 46; the lower limit is .......... and the upper limit is .........

Sol:

In the class interval 35 - 46; the lower limit is 35 and the upper limit is 46.

Question 4.6

Fill in the blank :
The class mark of class interval 22 - 29 is ...........

Sol:

The class mark of class interval 22 - 29 is
The classmark is  22 - 29 = ${\displaystyle \frac{22+29}{2}=\frac{51}{2}}$ = 25.5.

Question 5

Find the actual lower class limits, upper-class limits and the mid-values of the classes:
10 - 19, 20 - 29, 30 - 39 and 40 - 49.

Sol:

In case of frequency 10 - 19 the lower class limit is 10, the upper class limit is 19 and mid-value is
${\displaystyle \frac{10+19}{2}}$ = 14.5

In case of frequency 20 - 29 the lower class limit is 20, the upper class limit is 29 and mid-value is
${\displaystyle \frac{20+29}{2}}$ = 24.5

In case of frequency 30 - 39 the lower class limit is 30, the upper class limit is 39 and mid-value is
${\displaystyle \frac{30+29}{2}}$ = 34.5

In case of frequency 40 - 49 the lower class limit is 40, the upper class limit is 49 and mid-value is
${\displaystyle \frac{40+49}{2}}$ = 44.5

Question 6

Find the actual lower and upper-class limits and also the class marks of the classes:
1.1 - 2.0, 2.1 -3.0 and 3.1 - 4.0.

Sol:

In the case of frequency 1.1 - 2.0 the lower class limit is 1.1, upper-class limit is 2.0 and class mark

is  ${\displaystyle \frac{1.1+2.0}{2}}$ = 1.55

In the case of frequency 2.1 - 3.0 the lower class limit is 2.1, upper-class limit is 3.0 and class mark

is  ${\displaystyle \frac{2.1+3.0}{2}}$ = 2.55

In the case of frequency 3.1 - 4.0 the lower class limit is 3.1, the upper class limit is 4.0 and class mark

is  ${\displaystyle \frac{3.1+4.0}{2}}$ = 3.55

Question 7

Use the table given below to find:
(a) The actual class limits of the fourth class.
(b) The class boundaries of the sixth class.
(c) The class mark of the third class.
(d) The upper and lower limits of the fifth class.
(e) The size of the third class.

Class Interval	Frequency
30 - 34	7
35 - 39	10
40 - 44	12
45 - 49	13
50 - 54	8
55 - 59	4

Sol:

(a) The actual class limit of the fourth class will be  44.5 - 49.5.

(b) The class boundaries of the sixth class will be 54.5 - 59.5

(c) The class mark of the third class will be the average of the lower bound and the upper bound of the interval. Therefore the class mark will be: ${\displaystyle \frac{40+44}{2}}$ = 42

(d) The upper and lower limit of the fifth class is 54 and 50 respectively.

(e) The size of the third class will be 44 - 40 + 1 = 5.

Question 8.1

Construct a cumulative frequency distribution table from the frequency table given below:

Class Interval	Frequency
0 -8	9
8 - 16	13
16 - 24	12
24 - 32	7
32 - 40	15

Sol:

 The cumulative frequency distribution table is

C.I	c.f
0 -8	9
8 - 16	22
16 - 24	34
24 - 32	41
32 - 40	56

Question 8.2

Construct a cumulative frequency distribution table from the frequency table given below:

Class Interval	Frequency
1 - 10	12
11 - 20	18
21 - 30	23
31 - 40	15
41 - 50	10

Sol:

The cumulative frequency distribution table is

C.I	c.f
1 - 10	12
11 - 20	30
21 - 30	53
31 - 40	68
41 - 50	78

Question 9.1

Construct a frequency distribution table from the following cumulative frequency distribution:

Class Interval	Cumulative Frequency
10 - 19	8
20 - 29	19
30- 39	23
40- 49	30

Sol:

The frequency distribution table is

C. I	c.f
10 - 19	8
20 - 29	11
30 - 39	4
40 - 49	7

Question 9.2

Construct a frequency distribution table from the following cumulative frequency distribution:

C.I	C.F
5 - 10	18
10 - 15	30
15 - 20	46
20 - 25	73
25 - 30	90

Sol:

The frequency distribution table is

C .I	c.f
5 -  10	18
10 - 15	12
15 - 20	16
20  - 25	27
25 - 30	17

Question 10

Construct a frequency table from the following data:

Marks	No. of students
less than 10	6
less than 20	15
less than 30	30
less than 40	39
less than 50	53
less than 60	70

Sol:

The frequency table is

C. I	c.f
0 - 10	6
10 - 20	9
20 - 30	15
30 - 40	9
40 - 50	14
50 - 60	17

Question 11

Construct the frequency distribution table from the following cumulative frequency table:

Ages	No. of students
Below 4	0
Below 7	85
Below 10	140
Below 13	243
Below 16	300

(i) State the number of students in the age group 10 - 13.
(ii) State the age-group which has the least number of students.

Sol:

The frequency distribution table is

C. I	c.f
4 -  7	85
7 - 10	55
10 -  13	103
13 - 16	57

(i)The number of students in the age group  is 10 -13 is 103

(ii)The age group which has the least number of students is  7 - 10

Question 12

Fill in the blank in the following table:

Class interval	Frequency	Cumulative Frequency
25 - 34	......	15
35 - 44	......	28
45 - 54	21	......
55 - 64	16	......
65 - 74	......	73
75 - 84	12	......

Sol: (To be added) 

Question 13

Fill in the blank in the following table:

Class interval	Frequency	Cumulative Frequency
25 - 34	......	15
35 - 44	......	28
45 - 54	21	......
55 - 64	16	......
65 - 74	......	73
75 - 84	12	......

Sol:

Fill in the blank in the following table:

Class interval	Frequency	Cumulative Frequency
25 - 34	......	15
35 - 44	......	28
45 - 54	21	......
55 - 64	16	......
65 - 74	......	73
75 - 84	12	......

S.chand Class 8 Maths Solution Chapter 18 Statistics Exercise 18

 Exercise 18

Question 1

1. A total of 20 patients admitted to a hospital have blood sugar levels as given below : $67,69,74,73,70,70,71,67,73,74,73,75,69,72,70,70,72,70,73,74$. Make a frequency table.

2. The marks obtained by 30 students of a class in a test out of 10 marks are as follows :

4,6,5,1,5,4,3,6,8,10,7,1,8,5,4,9,7,10,3,2,4,5,3,6,7,8,4,10,3,9

Make a frequency distribution table for the above data. Use the table to find :

(i) The number of students passed, if the minimum pass marks are $40 \%$.

(ii) How many students failed ?

(iii) How many students secured the highest marks?

(iv) How many students secured more than $60 \%$ marks?

[Hint. Pass marks $=40 \%$ of $10=4,60 \%$ of $10=6]$.

3. The weight (in $\mathrm{kg}$ ) of 30 students of a class are $50,49,45,49,49,50,50,54,55,44,44,42,44,56,57$, $49,49,42,41,50,50,50,57,45,45,45,50,54,43$ and 49 .

Prepare a frequency table for the above data and answer the following questions :

(i) What is the least weight?

(ii) Find the number of students having the least weight in the above data.

(iii) Find the number of students having the maximum weight in the above data.

(iv) Which weight do the maximum number of students have ?

4. The value of $\pi$ up to 50 decimal places is given below :

$3.141592653589793238462643383279502888419716939937510$

Write the frequencies of the following digits in the decimal part of the above number.

(i) 2

(ii) 3

(iii) 5

(iv) 6

(v) 9

(vi) 1

5. Fill in the blanks:

(i) The difference between the maximum and the minimum observations in a data is called the of the data.

(ii) The number of observations in a particular class interval is called the of the class interval.

(iii) The range of the data $15,13,14,17,19,16,14,15$ is

6. Fill in the blanks in the following table :

Weights in kg	10-20	20-30	30-40	40-50	50-60
Class Marks

[Hint. Class mark for Ist class interval $=\frac{10+20}{2}=\frac{30}{2}=15$. Similarly, the class marks of other class intervals are obtained]

7. For each set of data, make up a tally table, using the groupings suggested and complete the frequency column.

(a) The number of tomatoes picked from tomato plants.

$18,31,25,16,21,20,34,7 \quad$ Suggested grouping

$19,18,24,26,30,21,26,18 \quad 0-4,5-9,10-14,15-19, \ldots \ldots$

$28,31,11,25,33,23,17,24$

(b) The number of books on 18 of the shelves in a college library.

$35,42,43,31,27,39,30,45,37$ Suggested grouping

$33,36,26,30,29,38,36,34,43 \quad 1-25,26-30,31-35,36-40, \ldots . .$

8. The following are the monthly rents (in rupees) of 30 shops:

$42,49,37,82,37,75,62,54,79,84,75,63,44,74,36,69,54,48,74,39,48,45,61,71,47,38,80,51,31,43$

Using the class interval of equal width in which one class interval being $40-50$ (excluding 50 ), construct a frequency table for the above data.

9. Construct a frequency table for the following marks obtained by 45 students using equal class intervals, one of them being $16-24$ ( 24 not included).

$12,35,6,10,8,24,37,32,61,52,63,7,41,48,15,16,25,29,62,40,33,46,18,20,34,28,24,56,55,12,50,56,48,47,38,26,60,42,39,40,43,25,13,46,20$.

10. The following list shows the weights in $\mathrm{kg}$ of the 22 boys students in a class.

$\begin{array}{llllll}37.48 & 61.93 & 58.72 & 49.78 & 51.70 & 68.10 \\ 49.87 & 38.75 & 69.10 & 65.39 & 36.49 & 65.62 \\ 54.63 & 46.17 & 48.80 & 57.35 & 62.25 & \\ 38.50 & 62.82 & 59.73 & 56.60 & 50.15 & \end{array}$

[Hint. Since observations like $37.48,62.25,59.73$, do not fit in any intervals, the class intervals have to overlap in such a way that all values fit in. You may take the intervals as $35-40,40-45,45-50$, $65-70 .]$