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# Class 9 SELINA Solutions Maths Chapter 18 - Statistics

## Statistics Exercise Ex. 18

### Solution 1(a)

Correct option: (ii) sizes of shoes

A variable that takes distinct, countable values is called a discrete variable.

Size of shoes can be 4, 5, ….. and so on but cannot take any value between 4 and 5, 5 and 6, and so on.

### Solution 1(b)

Correct option: (ii) 54

Maximum value = 90

Minimum value = 36

Therefore, range = maximum value - minimum value = 90 - 36 = 54

### Solution 1(c)

Correct option: (iii) 40

### Solution 1(d)

Correct option: (iv) 10.5-20.5

Therefore, the class 11-20 after adjustment is

### Solution 1(e)

Correct option: (i) 12.5-17.5

Class size = 15 - 10 = 5

### Solution 2

(a)Discrete variable.

(b)Continuous variable.

(c)Discrete variable.

(d)Continuous variable.

(e)Discrete variable.

### Solution 3

The frequency table for the given distribution is

 Marks Tally Marks Frequency

### Solution 4

The frequency table for the given distribution is

 Marks Tally Marks Frequency

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.

### Solution 5

(a)Variable.

(b)Discrete variables.

(c)Continuous variable.

(d)The range is

(e)Lower limit is and upper limit is

(f)The class mark is

### Solution 6

In case of frequency 10 - 19 the lower class limit is 10, upper class limit is 19 and mid-value is

In case of frequency 20 - 29 the lower class limit is 20, upper class limit is 29 and mid-value is

In case of frequency 30 - 39 the lower class limit is 30, upper class limit is 39 and mid-value is

In case of frequency 40 - 49 the lower class limit is 40, upper class limit is 49 and mid-value is

### Solution 7

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

is

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

is

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

is

### Solution 8

(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 class mark will be:

(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.

### Solution 9

(i)The cumulative frequency distribution table is

 C.I c.f

(ii)The cumulative frequency distribution table is

 C.I c.f

### Solution 10

(i)The frequency distribution table is

 C.I c.f

(ii)The frequency distribution table is

 C.I c.f

### Solution 11

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.

### Solution 12

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 combined histogram and frequency polygon is shown in the following figure:

### Solution 13

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:

### Solution 14

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:

## Statistics Exercise Test Yourself

### Solution 1

The frequency table is

 C.I c.f

### Solution 2

The frequency distribution table is

 C.I c.f

(i)The number of students in the age group is

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

### Solution 3

 Class Interval Frequency Cumulative Frequency 65-74

### Solution 4

 X 0 1 2 3 4 5 6 7 8 9 F 2 5 5 8 4 5 4 4 5 8

Most occurring digits are 3 and 9. Least occurring digits are 0.

### Solution 5(i)

(a) Using Histogram:

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

(b) Without using Histogram:

Steps:

1. Find the class-mark (mid-value) of each given class-interval.

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 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

### Solution 5(ii)

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. 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 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 -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

### Solution 6

The frequency distribution table is as follows:

### Solution 7

The cumulative frequency distribution table is as follows:

 Class Interval Frequency Cumulative Frequency 20-29 18 18 30-39 23 18 + 23 = 41 40-49 36 41 + 36 = 77 50-59 42 77 + 42 = 119

### Solution 8

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 (10-20) and mid-point of the immediately higher class-interval (70-80).

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