Statistics

A discrete variable is a variable which is incapable of taking all possible numerical values.

From the given options, only shoe sizes are considered discrete variables, whereas daily temperature in your city, distance traveled by a person, and time are continuous variables.

Hence, option 2 is the correct option.

Range is defined as the difference between the highest and lowest values in the data.

From the given data, highest value = 90.

Lowest value = 36

Range = 90 - 36

= 54

Hence, option 2 is the correct option.

Class-mark is the value midway between its actual lower limit and actual upper limit.

$\text{Class-mark} = \frac{\text{Lower class limit + Upper class limit}}{2}$

Lower class limit = 35

Upper class limit = 45

$$\begin{aligned}= \frac{35 + 45}{2}\\[1em] = \frac{80}{2}\\[1em] = 40\end{aligned}$$

Hence, option 3 is the correct option.

The adjustment factor is $\frac{11-10}{2} = 0.5$

The adjusted class would then be as follows:

Class interval before adjustment	Class interval after adjustment
1 - 10	0.5 - 10.5
11 - 20	10.5 - 20.5
21 - 30	20.5 - 30.5

The class 11-20 after adjustment is 10.5-20.5.

Hence, option 4 is the correct option.

The given class marks are 10, 15, 20, 25, ........

The class width is the difference between any two consecutive class marks.

Class width = 15 - 10 = 5

If a is the class mark of a class interval of width h, then the lower and upper limits of the class interval are:

Lower limit = a - $\frac{h}{2}$ upper limit = a + $\frac{h}{2}$

So, a = 15 and h = 5

Class interval = $\Big(15 - \frac{5}{2}\Big) to \Big(15 + \frac{5}{2}\Big)$

= (15 - 2.5) to (15 + 2.5)

= 12.5 to 17.5

Hence, the class corresponding to the class mark 15 is 12.5-17.5.

Hence, option 1 is the correct option.

(a) Discrete variable

Reason

A discrete variable is a variable which is incapable of taking all possible numerical values. In this case, the number of children is a finite, whole number.

(b) Continuous variable

Reason

Continuous variable is a variable which can take any numerical value within a certain range. The distance can be measured in infinitely small increments.

(c) Discrete variable

Reason

A discrete variable is a variable which is incapable of taking all possible numerical values. Sizes of shoes are represented by specific values.

(d) Continuous variable

Reason

Continuous variable is a variable which can take any numerical value within a certain range. Time can take on an infinite number of values within a range.

(e) Discrete variable

Reason

A discrete variable is a variable which is incapable of taking all possible numerical values. The number of patients is countable and can only be whole numbers

The frequency table for the given distribution is :

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

The frequency table for the given distribution is :

Marks	Tally marks	Frequency
0 - 10	IIII	4
10 - 20	IIII I	6
20 - 30	III	3
30 - 40	IIII	4
40 - 50	IIII II	7

(i) A quantity which can vary from one individual to another is called a variable.

(ii) Sizes of shoes are distinct variables.

(iii) Daily temperature is continuous variable.

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

Range of the data is the difference between the highest and the lowest values.

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

(vi) The class mark of class interval 22 - 29 is 25.5.

Class limit	Exclusive class limit
10 - 19	9.5 - 19.5
20 - 29	19.5 - 29.5
30 - 39	29.5 - 39.5
40 - 49	39.5 - 49.5

In case of frequency 10-19 the lower class limit is 9.5, the upper class limit is 19.5 and the mid value is

$$\begin{aligned}= \frac{9.5 + 19.5}{2} \\[1em] = \frac{29}{2} \\[1em] = 14.5\end{aligned}$$

In case of frequency 20-29 the lower class limit is 19.5, the upper class limit is 29.5 and the mid value is

$$\begin{aligned}= \frac{19.5 + 29.5}{2} \\[1em] = \frac{49}{2} \\[1em] = 24.5\end{aligned}$$

In case of frequency 30-39 the lower class limit is 29.5, the upper class limit is 39.5 and the mid value is

$$\begin{aligned}= \frac{29.5 + 39.5}{2} \\[1em] = \frac{69}{2} \\[1em] = 34.5\end{aligned}$$

In case of frequency 40-49 the lower class limit is 39.5, the upper class limit is 49.5 and the mid value is

$$\begin{aligned}= \frac{39.5 + 49.5}{2} \\[1em] = \frac{89}{2} \\[1em] = 44.5\end{aligned}$$

Class limit	Exclusive class limit
1.1 - 2.0	1.05 - 2.05
2.1 - 3.0	2.05 - 3.05
3.1 - 4.0	3.05 - 4.05

In case of frequency 1.1-2.0 the lower class limit is 1.05, the upper class limit is 2.05 and the class mark is

$$\begin{aligned}= \frac{1.05 + 2.05}{2} \\[1em] = \frac{3.1}{2} \\[1em] = 1.55\end{aligned}$$

In case of frequency 2.1-3.0 the lower class limit is 2.05, the upper class limit is 3.05 and the class mark is

$$\begin{aligned}= \frac{2.05 + 3.05}{2} \\[1em] = \frac{5.1}{2} \\[1em] = 2.55\end{aligned}$$

In case of frequency 3.1-4.0 the lower class limit is 3.05, the upper class limit is 4.05 and the class mark is

$$\begin{aligned}= \frac{3.05 + 4.05}{2} \\[1em] = \frac{7.1}{2} \\[1em] = 3.55\end{aligned}$$

Class Interval	Exclusive class Interval	Frequency
30 - 34	29.5 - 34.5	7
35 - 39	34.5 - 39.5	10
40 - 44	39.5 - 44.5	12
45 - 49	44.5 - 49.5	13
50 - 54	49.5 - 54.5	8
55 - 59	54.5 - 59.5	4

(i) The actual class limits of the fourth class is 44.5-49.5.

(ii) The class boundaries of the sixth class is 54.5-59.5.

(iii) The class mark of the third class

$$\begin{aligned}= \frac{40 + 44}{2}\\[1em] = \frac{84}{2}\\[1em] = 42\end{aligned}$$

(iv) The upper and lower limits of the fifth class (50-54) is 54 and 50, respectively.

(v) Interval of third class = 40-44

The size of class = Upper limit - lower limit + 1

The size of third class = 44 - 40 + 1

= 4 + 1

= 5

Hence, the size of third class is 5.

(i) The cumulative frequency distribution table is:

Class Interval	Frequency	Cumulative frequency
0 - 8	9	9
8 - 16	13	9 + 13 = 22
16 - 24	12	9 + 13 + 12 = 34
24 - 32	7	9 + 13 + 12 + 7 = 41
32 - 40	15	9 + 13 + 12 + 7 + 15 = 56

(ii) The cumulative frequency distribution table is -

Class Interval	Frequency	Cumulative frequency
1 - 10	12	12
11 - 20	18	12 + 18 = 30
21 - 30	23	12 + 18 + 23 = 53
31 - 40	15	12 + 18 + 23 + 15 = 68
41 - 50	10	12 + 18 + 23 + 15 + 10 = 78

(i) The frequency distribution table is

Class Interval	Cumulative Frequency	Frequency
10 - 19	8	8
20 - 29	19	19 - 8 = 11
30 - 39	23	23 -19 = 4
40 - 49	30	30 -23 = 7

(ii) The frequency distribution table is

Class Interval	Cumulative Frequency	Frequency
5 - 10	18	18
10 - 15	30	30 - 18 = 12
15 - 20	46	46 - 30 = 16
20 - 25	73	73 - 46 = 27
25 - 30	90	90 -73 = 17

Class intervals	Frequency	Class mark
-4 - 0	0	$\frac{-4 + 0}{2} = -2$
0 - 4	4	$\frac{0 + 4}{2} = 2$
4 - 8	7	$\frac{4 + 8}{2} = 6$
8 - 12	12	$\frac{8 + 12}{2} = 10$
12 - 16	15	$\frac{12 + 16}{2} = 14$
16 - 20	11	$\frac{16 + 20}{2} = 18$
20 - 24	6	$\frac{20 + 24}{2} = 22$
24 - 28	0	$\frac{24 + 28}{2} = 26$

Steps:

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

   Class-mark = mid-value = $\frac{\text{Upper limit + Lower limit}}{2}$

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 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.

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 (in the given question, the immediately lower class-interval is 0 - 10) and mid-point of the immediately higher class-interval (in the given question the immediate upper class-interval is 60 - 70).

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

The given class interval are inclusive. For constructing polygon we will first convert them into the exclusive form.

Class - intervals	Frequency	Class marks
4.5 - 9.5	0	$\frac{4.5 + 9.5}{2} = 7$
9.5 - 14.5	5	$\frac{9.5 + 14.5}{2} = 12$
14.5 - 19.5	8	$\frac{14.5 + 19.5}{2} = 17$
19.5 - 24.5	12	$\frac{19.5 + 24.5}{2} = 22$
24.5 - 29.5	9	$\frac{24.5 + 29.5}{2} = 27$
29.5 - 34.5	4	$\frac{29.5 + 34.5}{2} = 32$
34.5 - 39.5	0	$\frac{34.5 + 39.5}{2} = 37$

Steps:

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

   Class-mark = mid-value = $\frac{\text{Upper limit + Lower limit}}{2}$

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 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.

Daily wages (in ₹)	No. of workers	Class marks
75 - 125	0	$\frac{75 + 125}{2} = 100$
125 - 175	4	$\frac{125 + 175}{2} = 150$
175 - 225	20	$\frac{175 + 225}{2} = 200$
225 - 275	22	$\frac{225 + 275}{2} = 250$
275 - 325	10	$\frac{275 + 325}{2} = 300$
325 - 375	6	$\frac{325 + 375}{2} = 350$
375 - 425	0	$\frac{375 + 425}{2} = 400$

Steps:

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

   Class-mark mid-value = $\frac{\text{Upper limit + Lower limit}}{2}$

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 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.

By formula, class-mark = $\frac{\text{Lower limit + Upper limit}}{2}$

Substituting the values, we get :

Class-mark of class 19 - 30 = $\frac{19 + 30}{2}$

Hence, option 1 is the correct option.

Lower limit of class = Mid value - $\frac{\text{Width}}{2}$

= 10 - $\frac{6}{2}$

= 10 - 3

= 7.

Hence, option 2 is the correct option.

Given, mid-value = m

Upper limit of a class = x

As we know,

Mid-value = $\frac{\text{lower class-limit + upper class-limit}}{2}$

So, statement 2 is true.

⇒ m = $\frac{\text{lower class-limit} + x}{2}$

⇒ 2m = lower class-limit + x

⇒ Lower class-limit = 2m - x

So, statement 1 is true.

∴ Both the statements are true.

Hence, option 1 is the correct option.

10 or more hours covers the 10 – 15 and 15 – 20 classes.

4 students are there in 10 - 15 class.

2 students are there in 15 - 20 class.

Total students = 4 + 2 = 6

∴ A is false, but R is true.

Hence, option 2 is the correct option.

`

The frequency table for the given distribution is :

Marks	Tally marks	No. of students
0 - 10	IIII I	6
10 - 20	IIII IIII	9
20 - 30	IIII IIII IIII	15
30 - 40	IIII IIII	9
40 - 50	IIII IIII IIII	14
50 - 60	IIII IIII IIII II	17

The frequency table for the given distribution is :

Ages	Cumulative frequency	No. of students
0 - 4	0	0
4 - 7	85	85 - 0 = 85
7 - 10	140	140 - 85 = 55
10 - 13	243	243 - 140 = 103
13 - 16	300	300 - 243 = 57

(i) There are 103 students in the age group 10-13.

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

(i) The frequency distribution table of the digits from 0 to 9 after the decimal place is :

Digit	Tally marks	Frequency
0	II	2
1	IIII	5
2	IIII	5
3	IIII III	8
4	IIII	4
5	IIII	5
6	IIII	4
7	IIII	4
8	IIII	5
9	IIII III	8
Total		50

(ii) From the frequency distribution table, the frequency of digit 9 and 3 is maximum (8) and the frequency of digit 0 is minimum (2).

Hence, the most occurring digits are 9 and 3 and the least occurring digit is 0.

(a) Using histogram

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 (in the given question, the immediately lower class-interval is -10 - 10) and mid-point of the immediately higher class-interval (in the given question the immediate upper class-interval is 150 - 170).

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.

   Class-mark = mid-value = $\frac{\text{Upper limit + Lower limit}}{2}$

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 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.	f	Class mark
-10 - 10	0	$\frac{-10 + 10}{2}$ = 0
10 - 30	4	$\frac{10 + 30}{2}$ = 20
30 - 50	7	$\frac{30 + 50}{2}$ = 40
50 - 70	5	$\frac{50 + 70}{2}$ = 60
70 - 90	9	$\frac{70 + 90}{2}$ = 80
90 - 110	5	$\frac{90 + 110}{2}$ = 100
110 - 130	6	$\frac{110 + 130}{2}$ = 120
130 - 150	4	$\frac{130 + 150}{2}$ = 140
150 - 170	0	$\frac{150 + 170}{2}$ = 160

(a) Using histogram

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 (in the given question, the immediately lower class-interval is 0 - 5) and mid-point of the immediately higher class-interval (in the given question the immediate upper class-interval is 65 - 75).

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.

   Class-mark = mid-value = $\frac{\text{Upper limit + Lower limit}}{2}$

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 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.	f	Class mark
-5 - 5	0	$\frac{-5 + 5}{2}$ = 0
5 - 15	8	$\frac{5 + 15}{2}$ = 10
15 - 25	16	$\frac{15 + 25}{2}$ = 20
25 - 35	18	$\frac{25 + 35}{2}$ = 30
35 - 45	14	$\frac{35 + 45}{2}$ = 40
45 - 55	8	$\frac{45 + 55}{2}$ = 50
55 - 65	2	$\frac{55 + 65}{2}$ = 60
65 - 75	0	$\frac{65 + 75}{2}$ = 70

The frequency table for the given distribution is :

Class intervals	Tally marks	Frequency
0 - 9	II	2
10 - 19	IIII	4
20 - 29	II	2
30 - 39	II	2
40 - 49	II	2

The cumulative frequency table for the given distribution is :

Marks	No. of students	Cumulative frequency
20 - 29	18	18
30 - 39	23	18 + 23 = 41
40 - 49	36	18 + 23 + 36 = 77
50 - 59	42	18 + 23 + 36 + 42 = 119

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 (in the given question, the immediately lower class-interval is 10 - 20) and mid-point of the immediately higher class-interval (in the given question the immediate upper class-interval is 70 - 80).

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