Measures of Central Tendency

The sum of terms = 5.7 + 6.6 + 7.2 + 9.3 + 6.2 = 35

Number of terms = 5

Arithmetic mean (A.M.) = $\frac{\text{Sum of terms}}{\text{No. of terms}} = \frac{∑x_i}{n}$

$\therefore A.M. = \frac{35}{5} = 7.$

Hence, the mean of 5.7, 6.6, 7.2, 9.3, 6.2 is 7.

(i) The sum of marks of all students = 12 + 14 + 07 + 09 + 23 + 11 + 08 + 13 + 11 + 19 + 16 + 24 + 17 + 03 + 20 = 207.

$$\begin{aligned}\therefore \text{The mean of marks} = \frac{\text{Sum of marks of all students}}{\text{No. of students}} = \frac{∑x_i}{n} \\[1em] = \frac{207}{15} \\[1em] = 13.8\end{aligned}$$

Hence, the mean of marks of 15 students is 13.8

(ii) When the marks of each student are increased by 4, then the sum of their marks increases by 15 × 4 i.e. by 60.

∴ The new sum of marks of all students = 207 + 60 = 267.

$$\begin{aligned}\therefore \text{The new mean of marks} = \frac{\text{ New sum of marks}}{\text{No. of students}} \\[1em] = \frac{267}{15} \\[1em] = 17.8\end{aligned}$$

Hence, the mean of marks of 15 students, when the marks of each student are increased by 4 is 17.8

(iii) When the marks of each student is decreased by 2, then the sum of their marks decreases by 15 × 2 i.e. by 30.

∴ The new sum of marks of all students = 207 - 30 = 177.

$$\begin{aligned}\therefore \text{The new mean of marks} = \frac{\text{ New sum of marks}}{\text{No. of students}} \\[1em] = \frac{177}{15} \\[1em] = 11.8\end{aligned}$$

Hence, the mean of marks of 15 students, when the marks of each student is decreased by 2 is 11.8

(iv) When the marks of each student are doubled, then the sum of their marks will also be doubled.

∴ The new sum of marks of all students = 207 × 2 = 414.

$$\begin{aligned}\therefore \text{The new mean of marks} = \frac{\text{ New sum of marks}}{\text{No.of students}} \\[1em] = \frac{414}{15} \\[1em] = 27.6\end{aligned}$$

Hence, the mean of marks of 15 students, when the marks of each student is doubled is 27.6

Arithmetic mean (A.M.) = $\frac{\text{Sum of terms}}{\text{No. of terms}} = \frac{∑x_i}{n}$

$$\begin{aligned}\therefore 8 = \frac{6 + y + 7 + x + 14}{5} \\[1em] \Rightarrow 8 = \frac{x + y + 27}{5} \\[1em] \Rightarrow 40 = x + y + 27 \\[1em] \Rightarrow x + y = 13 \\[1em] \Rightarrow y = 13 - x.\end{aligned}$$

Hence, the value of y = 13 - x.

Let 9th variate be x.

Sum of terms = 7 + 12 + 9 + 14 + 21 + 3 + 8 + 15 + x = 89 + x

Number of terms = 9.

Arithmetic mean (A.M.) = $\frac{\text{Sum of terms}}{\text{No. of terms}}$

Given, A.M. = 11

$$\begin{aligned}\therefore 11 = \frac{89 + x}{9} \\[1em] \Rightarrow 99 = 89 + x \\[1em] \Rightarrow x = 99 - 89 \\[1em] \Rightarrow x = 10.\end{aligned}$$

Hence, the value of 9th variate is 10.

$$\begin{aligned}\text{Arithmetic mean (A.M.)} = \frac{\text{Sum of age of students}}{\text{No.of students}} \\[1em] \Rightarrow 13 = \frac{\text{Sum of age of students}}{33} \\[1em] \Rightarrow \text{Sum of age of students} = 33 \times 13 \\[1em] \Rightarrow \text{Sum of age of students} = 429.\end{aligned}$$

Let the age of girl that leaves the class be x. So, sum of age of students becomes 429 - x and total no of students = 32. Given, new mean = $12\frac{15}{16}$.

$$\begin{aligned}\therefore 12\frac{15}{16} = \frac{429 - x}{32} \\[1em] \frac{207}{16} = \frac{429 -x}{32} \\[1em] 207 \times 32 = 16(429 - x) \\[1em] 6624 = 6864 - 16x \\[1em] 16x = 6864 - 6624 \\[1em] 16x = 240 \\[1em] x = \frac{240}{16} = 15.\end{aligned}$$

Hence, the age of girl is 15 years.

$$\begin{aligned}\text{ Mean of marks} = \frac{\text{Incorrect sum of marks}}{\text{No. of students}} \\[1em] \Rightarrow 18.2 = \frac{\text{Incorrect sum of marks}}{40} \\[1em] \Rightarrow \text{Incorrect sum of marks} = 40 \times 18.2 \\[1em] \Rightarrow \text{Incorrect sum of marks} = 728.\end{aligned}$$

As the marks of one student was wrongly copied as 21 instead of 29, correct sum of marks = 728 - 21 + 29 = 736.

$$\begin{aligned}\therefore \text{Correct mean } = \frac{\text{Correct sum of marks}}{\text{No. of students}} \\[1em] = \frac{736}{40} = 18.4\end{aligned}$$

Hence, the correct mean is 18.4.

$$\begin{aligned}\text{Arithmetic mean (A.M.)} = \frac{\text{Sum of terms}}{\text{No. of terms}} \\[1em] \therefore \text{Sum of terms} = \text{A.M.} \times \text{No. of terms}.\end{aligned}$$

Given, mean of 10 numbers is 13.

∴ Sum of 10 terms = 13 × 10 = 130.

Given, mean of 15 numbers is 18.

∴ Sum of 15 terms = 15 × 18 = 270.

Sum of 25 terms = 130 + 270 = 400.

$\therefore A.M. = \frac{400}{25} = 16$

Hence, the mean of 25 numbers is 16.

We construct the following table:

xi	fi	fixi
5	1	5
10	2	20
15	5	75
20	6	120
25	3	75
30	2	60
35	1	35
Total	20	390

Mean = $\frac{∑f_ix_i}{∑f_i} = \frac{390}{20}$ = 19.5

Hence, the mean of the following distribution is 19.5

(i) We construct the following table:

xi	fi	fixi
35	6	210
36	10	360
37	18	666
38	25	950
39	21	819
40	12	480
41	8	328
Total	100	3813

Mean = $\frac{∑f_ix_i}{∑f_i} = \frac{3813}{100}$ = 38.1

Hence, the mean of number of matches per box is 38.1

(ii) Mean = $\frac{\text{No. of matches}}{\text{No. of boxes}}$

Let the no. of matches added to total contents of 100 boxes be x in order to bring mean to 39. So, total matches becomes 3813 + x.

$$\begin{aligned}\therefore 39 = \frac{3813 + x}{100} \\[1em] \Rightarrow 3900 = 3813 + x \\[1em] \Rightarrow x = 3900 - 3813 = 87.\end{aligned}$$

Hence, 87 extra matches need to added to bring the mean upto exactly 39 matches.

We construct the following table as under taking the assumed mean, a = 63.

xi	Cumulative frequency	fi	Deviation (di = xi - a)	fidi
60	8	8	-3	-24
61	18	18 - 8 = 10	-2	-20
62	33	33 - 18 = 15	-1	-15
63	40	40 - 33 = 7	0	0
64	49	49 - 40 = 9	1	9
65	55	55 - 49 = 6	2	12
66	60	60 - 55 = 5	3	15
Total		60		-23

Mean = $a + \frac{∑f_id_i}{∑f_i} = 63 + \frac{-23}{60} = 63- 0.38$ = 62.62.

Hence, the mean of the following distribution is 62.62.

(i) We construct the following table:

Category	Wages in ₹ per day (xi)	No. of workers (fi)	fixi
A	500	2	1000
B	600	4	2400
C	700	8	5600
D	800	12	9600
E	900	10	9000
F	1000	6	6000
G	1100	8	8800
Total		Σfi = 50	Σfixi = 42400

Mean = $\frac{Σf_ix_i}{Σf_i}$

= $\frac{42400}{50}$

= 848.

Hence, the mean wage is ₹ 848.

(ii) If the number of workers in each category is doubled then total wage will also be doubled.

New total wage = 42400 × 2 = 84800 and number of workers = 50 × 2 = 100.

Mean = $\frac{\text{New total wage}}{\text{No. of workers}}$

= $\frac{∑f_ix_i}{∑f_i}$

= $\frac{84800}{100}$​ = 848.

Hence, the new mean wage is also ₹ 848.

Marks (xi)	No. of students (fi)	fixi
5	3	15
10	7	70
15	f	15f
20	9	180
25	6	150
Total	Σfi = 25 + f	Σfixi = 415 + 15f

By formula; Mean = $\frac{Σf_ix_i}{Σf_i}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow 16 = \frac{415 + 15f}{25 + f}\\[1em] \Rightarrow 16(25 + f) = 415 + 15f\\[1em] \Rightarrow 400 + 16f = 415 + 15f\\[1em] \Rightarrow 16f - 15f = 415 - 400\\[1em] \Rightarrow f = 15.\end{aligned}$$

Hence, the value of f = 15.

We construct the following table:

Marks (xi)	No. of students (fi)	fixi
5	6	30
6	a	6a
7	16	112
8	13	104
9	b	9b
Total	35 + a + b	246 + 6a + 9b

Given, total no. of students = 40 and mean = 7.2

∴ 35 + a + b = 40
⇒ a + b = 5
⇒ a = 5 - b      .....(i)

$$\begin{aligned}\text{Mean} = \frac{∑f_ix_i}{∑f_i} \\[1em] \therefore 7.2 = \frac{246 + 6a + 9b}{40} \\[1em] \Rightarrow 288 = 246 + 6a + 9b \\[1em] \Rightarrow 288 - 246 = 6a + 9b \\[1em] \Rightarrow 6a + 9b = 42\end{aligned}$$

Putting value of a from Eq (i)

$$\begin{aligned}\Rightarrow 6(5 - b) + 9b = 42 \\[1em] \Rightarrow 30 - 6b + 9b = 42 \\[1em] \Rightarrow 30 + 3b = 42 \\[1em] \Rightarrow 3b = 12 \\[1em] \Rightarrow b = 4.\end{aligned}$$

a = 5 - b = 5 - 4 = 1.

Hence, the value of a = 1 and b = 4.

Class	Class-mark (yi)	Frequency (fi)	fiyi
5 - 15	10	2	20
15 - 25	20	6	120
25 - 35	30	4	120
35 - 45	40	8	320
45 - 55	50	4	200
Total		Σfi = 24	Σfiyi = 780

$$\begin{aligned}\text{Mean }= \frac{∑f_iy_i}{∑f_i}\\[1em] = \frac{780}{24}\\[1em] = 32.5\end{aligned}$$

Hence, mean of the following distribution is 32.5.

We construct the following table:

Classes	Class mark (yi)	Frequency (fi)	fiyi
0 - 10	5	8	40
10 - 20	15	5	75
20 - 30	25	12	300
30 - 40	35	35	1225
40 - 50	45	24	1080
50 - 60	55	16	880
Total		100	3600

∴ Mean = $\frac{∑f_iy_i}{∑f_i} = \frac{3600}{100}$ = 36.

Hence, mean of the following distribution is 36.

We construct the following table, taking assumed mean a = 25.
Here, c (width of each class) = 10.

Marks (Classes)	Class mark (yi)	$u_i = \frac{y_i - a}{c}$	No. of students (Frequency (fi))	fiui
0 - 10	5	-2	10	-20
10 - 20	15	-1	9	-9
20 - 30	25	0	25	0
30 - 40	35	1	30	30
40 - 50	45	2	16	32
50 - 60	55	3	10	30
Total			100	63

$$\begin{aligned}\therefore \text{ Mean} = a + c \times \frac{∑f_iu_i}{∑f_i} \\[1em] = 25 + 10 \times \frac{63}{100} \\[1em] = 25 + \frac{630}{100} \\[1em] = 25 + 6.3 \\[1em] = 31.3\end{aligned}$$

Hence, mean of the following distribution is 31.3

Construct table as under, taking assumed mean a = 45.

Number of patients (Classes)	Class mark (yi)	Deviation (di = yi - a)	Number of days (Frequency (fi))	fidi
10 - 20	15	-30	5	-150
20 - 30	25	-20	2	-40
30 - 40	35	-10	7	-70
40 - 50	45	0	9	0
50 - 60	55	10	2	20
60 - 70	65	20	5	100
Total			30	-140

$$\begin{aligned}\therefore \text{ Mean} = a + \frac{∑f_id_i}{∑f_i} \\[1em] = 45 + \frac{-140}{30} \\[1em] = 45 + (-4.67) \\[1em] = 40.33\end{aligned}$$

Hence, mean of the following distribution is 40.33

We construct the following table as under taking the assumed mean, a = 625.

Class	Class - mark (xi)	Frequency (fi)	Deviation (di = xi - a)	fidi
450 - 500	475	5	-150	-750
500 - 550	525	8	-100	-800
550 - 600	575	30	-50	-1500
600 - 650	625	25	0	0
650 - 700	675	14	50	700
700 - 750	725	12	100	1200
750 - 800	775	6	150	900
Total		Σfi = 100		Σfidi = -250

By formula,

$$\begin{aligned}\text{Mean }= a + \frac{∑f_id_i}{∑f_i}\\[1em] = 625 + \frac{-250}{100}\\[1em] = 625 - 2.5\\[1em] = 622.5\end{aligned}$$

Hence, mean of the following distribution is ₹622.5.

Construct the table as under, taking assumed mean as 45.5

Marks (Classes)	Class mark (yi)	Deviation (di = yi - a)	No. of students (Frequency (fi))	fidi
11 - 20	15.5	-30	2	-60
21 - 30	25.5	-20	6	-120
31 - 40	35.5	-10	10	-100
41 - 50	45.5	0	12	0
51 - 60	55.5	10	9	90
61 - 70	65.5	20	7	140
71 - 80	75.5	30	4	120
Total			50	70

$$\begin{aligned}\therefore \text{ Mean} = a + \frac{∑f_id_i}{∑f_i} \\[1em] = 45.5 + \frac{70}{50} \\[1em] = 45.5 + 1.4 \\[1em] = 46.9\end{aligned}$$

Hence, mean of the following distribution is 46.9 marks

We construct the following table :

No. of days (Classes)	Class mark (yi)	No. of students (Frequency (fi))	fiyi
0 - 6	3	11	33
6 - 10	8	10	80
10 - 14	12	7	84
14 - 20	17	4	68
20 - 28	24	4	96
28 - 38	33	3	99
38 - 40	39	1	39
Total		40	499

∴ Mean = $\frac{∑f_iy_i}{∑f_i} = \frac{499}{40}$ = 12.475

Hence, mean of the following distribution is 12.475

We construct the following table :

Marks (Classes)	Class mark (yi)	Number of students (Frequency (fi))	fiyi
0 - 10	5	7	35
10 - 20	15	a	15a
20 - 30	25	8	200
30 - 40	35	10	350
40 - 50	45	5	225
Total		30 + a	810 + 15a

$$\begin{aligned}\therefore \text{Mean} = \frac{∑f_iy_i}{∑f_i} \\[1em] \Rightarrow 24 = \frac{810 + 15a}{30 + a} \\[1em] \Rightarrow 24(30 + a) = 810 + 15a \\[1em] \Rightarrow 720 + 24a = 810 + 15a \\[1em] \Rightarrow 24a - 15a = 810 - 720 \\[1em] \Rightarrow 9a = 90 \\[1em] \Rightarrow a = 10.\end{aligned}$$

Hence, the value of a = 10.

Class Interval	Class mark (xi)	Frequency (fi)	fixi
0 - 20	10	6	60
20 - 40	30	f	30f
40 - 60	50	8	400
60 - 80	70	12	840
80 - 100	90	8	720
Total		Σfi = 34 + f	Σfixi = 2020 + 30f

By formula,

Mean = $\frac{∑f_ix_i}{∑f_i}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 50 = \frac{2020 + 30f}{34 + f}\\[1em] \Rightarrow 50(34 + f) = 2020 + 30f\\[1em] \Rightarrow 1700 + 50f = 2020 + 30f\\[1em] \Rightarrow 50f - 30f = 2020 - 1700\\[1em] \Rightarrow 20f = 320\\[1em] \Rightarrow f = \frac{320}{20}\\[1em] \Rightarrow f = 16.\end{aligned}$$

Hence, the value of f = 16.

We construct the following table :

Marks (Classes)	Class mark (yi)	Number of students (Frequency (fi))	fiyi
0 - 20	10	7	70
20 - 40	30	p	30p
40 - 60	50	12	600
60 - 80	70	q	70q
80 - 100	90	8	720
100 - 120	110	5	550
Total		32 + p + q	1940 + 30p + 70q

Given,

The sum of frequencies = 50.

∴ 32 + p + q = 50
⇒ p + q = 18
⇒ p = 18 - q      .....(i)

$$\begin{aligned}\therefore \text{Mean} = \frac{∑f_iy_i}{∑f_i} \\[1em] \Rightarrow 57.6 = \frac{1940 + 30p + 70q}{50} \\[1em] \Rightarrow 57.6 \times 50 = 1940 + 30p + 70q \\[1em] \Rightarrow 2880 = 1940 + 30p + 70q \\[1em] \Rightarrow 30p + 70q = 2880 - 1940 \\[1em] \Rightarrow 30p + 70q = 940\end{aligned}$$

Putting value of p from (i) in above equation,

$$\begin{aligned}\Rightarrow 30(18 - q) + 70q = 940 \\[1em] \Rightarrow 540 - 30q + 70q = 940 \\[1em] 540 + 40q = 940 \\[1em] 40q = 940 - 540 \\[1em] 40q = 400 \\[1em] q = 10\end{aligned}$$

Using (i),

⇒ p = 18 - q = 18 - 10 = 8.

Hence, the value of p = 8 and q = 10.

We construct the following table :

Lifetime in days (Classes)	Class mark (yi)	No. of tubes (Cumulative frequency)	Frequency (fi)	fiyi
0 - 50	25	8	8	200
50 - 100	75	23	23 - 8 = 15	1125
100 - 150	125	55	55 - 23 = 32	4000
150 - 200	175	81	81 - 55 = 26	4550
200 - 250	225	93	93 - 81 = 12	2700
250 - 300	275	100	100 - 93 = 7	1925
Total			100	14500

∴ Mean = $\frac{∑f_iy_i}{∑f_i} = \frac{14500}{100}$ = 145

Hence, mean of the following distribution is 145 days

In the given table i is the class interval which is equal to 10.

Class	Class mark (x)	d = (x - A)	u = d/i	Frequency (f)	fu
100-110	105	-30	-3	5	-15
110-120	115	-20	-2	10	-20
120-130	125	-10	-1	17	-17
130-140	A = 135	0	0	8	0
140-150	145	10	1	6	6
150-160	155	20	2	5	10
Total				Σf = 51	Σfu = -36

Mean = A + $\frac{Σfu}{Σf} \times i = 135 + \frac{-36}{51} \times 10$

= $135 + \frac{-360}{51}$

= 135 - 7.06

= 127.94 (approx)

Hence, mean duration = 127.94 minutes (approx).

In the given table,

Class size (i) = 4.

Monthly income	No.of employees (f)	Class mark	d = x - A	u = d/i	fu
0-4	55	2	-4	-1	-55
4-8	15	A = 6	0	0	0
8-12	06	10	4	1	06
12-16	08	14	8	2	16
16-20	12	18	12	3	36
20-24	4	22	16	4	16
Total	Σf = 100				Σfu = 19

By formula,

Mean = A + $\frac{Σfu}{Σf} \times i$

= 6 + $\frac{19}{100} \times 4$

= 6 + $\frac{76}{100}$

= 6 + 0.76

= 6.76

Hence, mean = 6.76

We construct the following table :

Class interval	Class mark (yi)	Frequency (fi)	fiyi
20 - 30	25	3	75
30 - 40	35	5	175
40 - 50	45	12	540
50 - 60	55	9	495
60 - 70	65	4	260
Total		33	1545

∴ Mean = $\frac{∑f_iy_i}{∑f_i} = \frac{1545}{33}$ = 46.8

Hence, mean of the following distribution is 46.8

On arranging the marks in ascending order, we get

1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8.

Here, n (no. of observations) = 11, which is odd.

$$\begin{aligned}\therefore \text{Median} = \frac{n + 1}{2} \text{th observation} \\[1em] = \frac{11 + 1}{2} \\[1em] = \frac{12}{2} \\[1em] = 6 \text{th observation}\end{aligned}$$

6th observation = 5.

Hence, the median marks are 5.

On arranging the numbers in ascending order we get,

3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81.

Here, n (no. of observations) = 12, which is even.

$$\begin{aligned}\therefore \text{Median} = \frac{\frac{n}{2} \text{th observation} + \big(\frac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\frac{12}{2} \text{th observation} + \big(\frac{12}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\text{ 6th observation + 7th observation}}{2} \\[1em] = \frac{15 + 17}{2} \\[1em] = \frac{32}{2} \\[1em] = 16.\end{aligned}$$

Hence, median = 16.

On arranging the numbers in ascending order we get,

0, 1, 1, 2, 2, 3, 3, 3, 4, 5.

Sum of terms = 0 + 1 + 1 + 2 + 2 + 3 + 3 + 3 + 4 + 5 = 24.

By definition,

$$\begin{aligned}\text{Mean} = \frac{\text{Sum of terms}}{\text{No. of terms}} \\[1em] = \frac{24}{10} \\[1em] = 2.4\end{aligned}$$

Here, n (no. of observations) = 10, which is even.

$$\begin{aligned}\therefore \text{Median} = \frac{\frac{n}{2} \text{th observation} + \big(\frac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\frac{10}{2} \text{th observation} + \big(\frac{10}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\text{ 5th observation + 6th observation}}{2} \\[1em] = \frac{2 + 3}{2} \\[1em] = \frac{5}{2} \\[1em] = 2.5\end{aligned}$$

Hence, mean = 2.4 and median = 2.5

Observations arranged in ascending order are :

11, 12, 14, (x - 2), (x + 4), (x + 9), 32, 38, 47.

Here, n (no. of observations) = 9, which is odd.

$$\begin{aligned}\therefore \text{Median} = \frac{n + 1}{2} \text{th observation} \\[1em] = \frac{9 + 1}{2} \\[1em] = \frac{10}{2} \\[1em] = 5 \text{th observation}\end{aligned}$$

Given, median = 24 = 5th observation = x + 4.

⇒ 24 = x + 4
⇒ x = 24 - 4 = 20.

Putting the value of x in observations :

11, 12, 14, 18, 24, 29, 32, 38, 47.

Sum of terms = 11 + 12 + 14 + 18 + 24 + 29 + 32 + 38 + 47 = 225.

By definition,

$$\begin{aligned}\text{Mean } = \frac{\text{Sum of terms}}{\text{No. of terms}} \\[1em] = \frac{225}{9} \\[1em] = 25.\end{aligned}$$

Hence, the value of x = 20 and mean = 25.

(i) Given, the mean of the numbers 1, 7, 5, 3, 4, 4 is m.

Sum of terms = 1 + 7 + 5 + 3 + 4 + 4 = 24.

By definition,

$$\begin{aligned}\text{Mean } = \frac{\text{Sum of terms}}{\text{No. of terms}} \\[1em] \therefore \text{m} = \frac{24}{6} = 4.\end{aligned}$$

Given, the mean of the numbers 3, 2, 4, 2, 3, 3, p is (m - 1) i.e. 3.

Sum of terms = 3 + 2 + 4 + 2 + 3 + 3 + p = 17 + p.

By definition,

$$\begin{aligned}\text{Mean } = \frac{\text{Sum of terms}}{\text{No. of terms}} \\[1em] \therefore 3 = \frac{17 + p}{7} \\[1em] \Rightarrow 21 = p + 17 \\[1em] \Rightarrow p = 21 - 17 = 4.\end{aligned}$$

Hence, the value of p is 4.

(ii) Given, the median of the numbers 3, 2, 4, 2, 3, 3, 4 is q.

Arranging the numbers in ascending order we get,

2, 2, 3, 3, 3, 4, 4.

Here, n (no. of observations ) = 7, which is odd.

$$\begin{aligned}\therefore \text{Median} = \frac{n + 1}{2} \text{th observation} \\[1em] = \frac{7 + 1}{2} \\[1em] = \frac{8}{2} \\[1em] = 4 \text{th observation}\end{aligned}$$

Given, median = q = 4th observation = 3.

⇒ q = 3.

Hence, the value of q is 3.

(iii) Mean of p and q i.e. 4 and 3 is,

$$\begin{aligned}\text{Mean } = \frac{\text{Sum of terms}}{\text{No. of terms}} \\[1em] = \frac{4 + 3}{2} \\[1em] = \frac{7}{2} \\[1em] = 3.5\end{aligned}$$

Hence, mean of p and q is 3.5

The given variates (wages) are already in ascending order. We construct the cumulative frequency table as under :

Wages per day (in ₹)	Frequency (No. of workers)	Cumulative frequency (fi)
380	14	14
450	8	22
480	7	29
550	10	39
620	6	45
680	2	47

Here, n (total no. of workers) = 47, which is odd.

$$\begin{aligned}\text{Median }= \Big(\frac{n + 1}{2}\Big)\text{th observation}\\[1em] = \Big(\frac{47 + 1}{2}\Big)\text{th observation}\\[1em] = \Big(\frac{48}{2}\Big)\text{th observation}\\[1em] = 24\text{th observation}\\[1em]\end{aligned}$$

All observations from 23rd to 29th are equal, each = 480.

Median = ₹480

Hence, median = ₹480.

On arranging the given variates (marks) in ascending order, we construct the cumulative frequency table as under :

Variate (Marks)	Frequency (No. of students)	Cumulative frequency
20	8	8
40	12	20
50	18	38
60	6	44
70	12	56
75	9	65
90	5	70

Here, n (total no. of students) = 70, which is even.

$$\begin{aligned}\therefore \text{Median} = \frac{\frac{n}{2} \text{th observation} + \big(\frac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\frac{70}{2} \text{th observation} + \big(\frac{70}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\text{ 35th observation + 36th observation}}{2} \\[1em]\end{aligned}$$

All observations from 21st to 38th are equal, each = 50.

Hence, median

$$\begin{aligned}= \frac{50 + 50}{2} \\[1em] = \frac{100}{2} \\[1em] = 50\end{aligned}$$

Hence, median marks = 50.

The given numbers are already in ascending order. We construct the cumulative frequency table as under :

Number (fi)	Frequency (xi)	Cumulative frequency	fixi
5	1	1	5
10	2	3	20
15	5	8	75
20	6	14	120
25	3	17	75
30	2	19	60
35	1	20	35
Total	20		390

Here, n (no. of observations) = 20, which is even.

$$\begin{aligned}\therefore \text{Median} = \frac{\frac{n}{2} \text{th observation} + \big(\frac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\frac{20}{2} \text{th observation} + \big(\frac{20}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\text{ 10th observation + 11th observation}}{2} \\[1em]\end{aligned}$$

All observations from 9th to 14th are equal, each = 20.

Hence, median

$$\begin{aligned}= \frac{20 + 20}{2} \\[1em] = \frac{40}{2} \\[1em] = 20.\end{aligned}$$

Now calculating mean,

$$\begin{aligned}\text{ Mean} = \frac{∑f_ix_i}{∑f_i} \\[1em] = \frac{390}{20} \\[1em] = 19.5\end{aligned}$$

Hence, mean = 19.5 and median = 20.

On arranging the given wages in ascending order we get,

21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53.

Here, n (no. of observations) = 19, which is odd.

(i) As n is odd,

$$\begin{aligned}\therefore \text{Median} = \frac{n + 1}{2} \text{th observation} \\[1em] = \frac{19 + 1}{2} \\[1em] = \frac{20}{2} \\[1em] = 10 \text{th observation}\end{aligned}$$

∴ Median = 10th observation = 31.

Hence, median = 31.

(ii) As n is odd,

$$\begin{aligned}\therefore \text{Lower quartile} (Q_1) = \frac{n + 1}{4} \text{th observation} \\[1em] = \frac{19 + 1}{4} \\[1em] = \frac{20}{4} \\[1em] = 5 \text{th observation}\end{aligned}$$

∴ Lower quartile (Q₁) = 5th observation = 27.

Hence, lower quartile = 27.

(iii) As n is odd,

$$\begin{aligned}\therefore \text{Upper quartile} (Q_3) = \frac{3(n + 1)}{4} \text{th observation} \\[1em] = \frac{3(19 + 1)}{4} \\[1em] = \frac{60}{4} \\[1em] = 15 \text{th observation}\end{aligned}$$

∴ Upper quartile (Q₃) = 15th observation = 41.

Hence, upper quartile = 41.

(iv) Inter quartile range = Q₃ - Q₁ = 41 - 27 = 14.

Hence, inter quartile range = 14.

The given variates are already in ascending order. We construct the cumulative frequency table as under

Variate	Frequency	Cumulative frequency
15	4	4
18	6	10
20	8	18
22	9	27
25	7	34
27	8	42
30	6	48

Here, n (no. of observations) = 48, which is even.

(i) As n is even,

$$\begin{aligned}\therefore \text{Median} = \frac{\frac{n}{2} \text{th observation} + \big(\frac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\frac{48}{2} \text{th observation} + \big(\frac{48}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\text{ 24th observation + 25th observation}}{2} \\[1em]\end{aligned}$$

All observations from 19th to 27th are equal, each = 22.

Hence, median

$$\begin{aligned}= \frac{22 + 22}{2} \\[1em] = \frac{44}{2} \\[1em] = 22.\end{aligned}$$

Hence, median = 22.

(ii) As n is even,

$$\begin{aligned}\therefore \text{Lower quartile} (Q_1) = \frac{n}{4} \text{th observation} \\[1em] = \frac{48}{4} \\[1em] = 12 \text{th observation}\end{aligned}$$

∴ Lower quartile (Q₁) = 12th observation = 20.

Hence, lower quartile = 20.

(iii) As n is even,

$$\begin{aligned}\therefore \text{Upper quartile} (Q_3) = \frac{3n}{4} \text{th observation} \\[1em] = \frac{3 × 48}{4} \\[1em] = \frac{144}{4} \\[1em] = 36 \text{th observation}\end{aligned}$$

∴ Upper quartile (Q₃) = 36th observation = 27.

Hence, upper quartile = 27.

(iv) Inter quartile range = Q₃ - Q₁ = 27 - 20 = 7.

Hence, inter quartile range = 7.

The given numbers are already in ascending order. We construct the cumulative frequency table as under

Variate	Frequency	Cumulative frequency
25	3	3
31	8	11
34	10	21
40	15	36
45	10	46
48	9	55
50	6	61
60	2	63

Here, n (no. of observations) = 63, which is odd.

(i) As n is odd,

$$\begin{aligned}\therefore \text{Median} = \frac{n + 1}{2} \text{th observation} \\[1em] = \frac{63 + 1}{2} \\[1em] = \frac{64}{2} \\[1em] = 32 \text{nd observation}\end{aligned}$$

∴ Median = 32nd observation = 40.

Hence, median = 40.

(ii) As n is odd,

$$\begin{aligned}\therefore \text{Lower quartile} (Q_1) = \frac{n + 1}{4} \text{th observation} \\[1em] = \frac{63 + 1}{4} \\[1em] = \frac{64}{4} \\[1em] = 16 \text{th observation}\end{aligned}$$

∴ Lower quartile (Q₁) = 16th observation = 34.

Hence, lower quartile = 34.

(iii) As n is odd,

$$\begin{aligned}\therefore \text{Upper quartile} (Q_3) = \frac{3(n + 1)}{4} \text{th observation} \\[1em] = \frac{3(63 + 1)}{4} \\[1em] = \frac{3 \times 64}{4} \\[1em] = 48 \text{th observation}\end{aligned}$$

∴ Upper quartile (Q₃) = 48th observation = 48.

Hence, upper quartile = 48.

(i) In the given data : 5, 7, 6, 8, 9, 0, 6, 8, 1, 8

8 is repeated more number of times than any other number,

∴ mode = 8.

(ii) In the given data : 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7

5 is repeated more number of times than any other number,

∴ mode = 5.

Arithmetic mean (A.M.) = $\frac{\text{Sum of terms}}{\text{No.of terms}} = \frac{∑x_i}{n}$

Sum of terms = 8 + 10 + 7 + 6 + 10 + 11 + 6 + 13 + 10 = 81.

$\therefore \text{A.M.} = \frac{81}{9} = 9.$

∴ Mean = 9.

On arranging the marks in ascending order, we get

6, 6, 7, 8, 10, 10, 10, 11, 13.

Here, n (no. of observations) = 9, which is odd.

$$\begin{aligned}\therefore \text{Median} = \frac{n + 1}{2} \text{th observation} \\[1em] = \frac{9 + 1}{2} \\[1em] = \frac{10}{2} \\[1em] = 5 \text{th observation}\end{aligned}$$

5th observation = 10.

∴ Median = 10.

In the given data : 8, 10, 7, 6, 10, 11, 6, 13, 10

10 is repeated more number of times than any other number,

∴ Mode = 10.

Hence, mean = 9, median = 10 and mode = 10.

Arithmetic mean (A.M.) = $\frac{\text{Sum of terms}}{\text{No.of terms}} = \frac{∑x_i}{n}$

Sum of terms = 3 + 1 + 5 + 6 + 3 + 4 + 5 + 3 + 7 + 2 = 39

$\therefore \text{A.M.} = \frac{39}{10} = 3.9$

∴ Mean = 3.9

On arranging the numbers in ascending order we get,

1, 2, 3, 3, 3, 4, 5, 5, 6, 7.

Here, n (no. of observations) = 10, which is even.

$$\begin{aligned}\therefore \text{Median} = \frac{\frac{n}{2} \text{th observation} + \big(\frac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\frac{10}{2} \text{th observation} + \big(\frac{10}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\text{5th observation + 6th observation}}{2} \\[1em] = \frac{3 + 4}{2} \\[1em] = \frac{7}{2} \\[1em] = 3.5\end{aligned}$$

∴ Median = 3.5

In the given data : 3, 1, 5, 6, 3, 4, 5, 3, 7, 2.

3 is repeated more number of times than any other number,

∴ Mode = 3.

Hence, mean = 3.9, median = 3.5 and mode = 3.

Here, n (no. of observations) = 10, which is even.

$$\begin{aligned}\therefore \text{Median} = \frac{\frac{n}{2} \text{th observation} + \big(\frac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\frac{10}{2} \text{th observation} + \big(\frac{10}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\text{5th observation + 6th observation}}{2} \\[1em] = \frac{x + (x + 4)}{2} \\[1em] = \frac{2x + 4}{2} \\[1em] = x + 2.\end{aligned}$$

Given, median marks = 48.

∴ x + 2 = 48
⇒ x = 46.

Putting value of x in data we get,

13, 35, 43, 46, 46, 50, 55, 61, 71, 80.

In the given data 46 is repeated more number of times than any other number.

Hence, the value of x = 46 and mode = 46.

The variates are already in ascending order. We construct the cumulative frequency table as under :

x	f	Cumulative frequency (C.F.)
10	1	1
11	4	5
12	7	12
13	5	17
14	9	26
15	3	29

Total number of observations = 29, which is odd.

$$\begin{aligned}\therefore \text{Median} = \frac{n + 1}{2} \text{th observation} \\[1em] = \frac{29 + 1}{2} \\[1em] = \frac{30}{2} \\[1em] = 15 \text{th observation}\end{aligned}$$

All observations from 13th to 17th are equal, each = 13, so median = 13.

As the variate 14 has maximum frequency 9, so mode = 14.

Hence, median = 13 and mode = 14.

(i) The variates are already in ascending order. We construct the cumulative frequency table as under :

Marks	Number of students	Cumulative frequency (C.F.)
1	1	1
2	2	3
3	3	6
4	3	9
5	6	15
6	10	25
7	5	30
8	4	34
9	3	37
10	3	40

Total number of observations = 40, which is even.

$$\begin{aligned}\therefore \text{Median} = \frac{\frac{n}{2} \text{th observation} + \big(\frac{n}{2} + 1\big)\text{th observation}}{2} \\[1em] = \frac{\frac{40}{2} \text{th observation} + \big(\frac{40}{2} + 1\big)\text{th observation}}{2} \\[1em] = \frac{\text{ 20th observation + 21st observation}}{2} \\[1em]\end{aligned}$$

All observations from 16th to 25th are equal, each = 6.

Hence, median

$$\begin{aligned}= \frac{6 + 6}{2} \\[1em] = \frac{12}{2} \\[1em] = 6.\end{aligned}$$

Hence, median = 6.

(ii) As the variate 6 has maximum frequency 10, so mode = 6.

Hence, mode = 6.

The variates (marks) are already in ascending order. We construct the cumulative frequency table as under :

Marks (xi)	No. of students (fi)	Cumulative frequency (C.F.)	fixi
0	1	1	0
1	3	4	3
2	6	10	12
3	10	20	30
4	5	25	20
5	5	30	25
Total	30		90

Mean = $\frac{∑f_ix_i}{∑f_i} = \frac{90}{30}$ = 3.

Total number of observations = 30, which is even.

$$\begin{aligned}\therefore \text{Median} = \frac{\frac{n}{2} \text{th observation} + \big(\frac{n}{2} + 1\big)\text{th observation}}{2} \\[1em] = \frac{\frac{30}{2} \text{th observation} + \big(\frac{30}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\text{ 15th observation + 16th observation}}{2} \\[1em]\end{aligned}$$

All observations from 11th to 20th are equal, each = 3.

Hence, median

$$\begin{aligned}= \frac{3 + 3}{2} \\[1em] = \frac{6}{2} \\[1em] = 3.\end{aligned}$$

As the variate 3 has maximum frequency 10, so mode = 3.

Hence, mean = 3, median = 3 and mode = 3.

The variates (marks) are already in ascending order. We construct the cumulative frequency table as under :

Marks (xi)	No. of students (fi)	Cumulative frequency (C.F.)	fixi
5	3	3	15
6	9	12	54
7	6	18	42
8	4	22	32
9	2	24	18
10	1	25	10
Total	25		171

Mean = $\frac{∑f_ix_i}{∑f_i} = \frac{171}{25}$ = 6.84.

Total number of observations = 25, which is odd.

$$\begin{aligned}\therefore \text{Median} = \frac{n + 1}{2} \text{th observation} \\[1em] = \frac{25 + 1}{2} \\[1em] = \frac{26}{2} \\[1em] = 13 \text{th observation}\end{aligned}$$

All observations from 13th to 18th are equal, each = 7, so median = 7.

As the variate 6 has maximum frequency 9, so mode = 6.

Hence, mean = 6.84, median = 7 and mode = 6.

(i) We construct the following table :

Weekly wages (xi)	No. of workers (fi)	Class mark (ui)	Cumulative frequency (C.F.)	fi.ui
500 - 550	5	525	5	2625
550 - 600	20	575	25	11500
600 - 650	10	625	35	6250
650 - 700	10	675	45	6750
700 - 750	9	725	54	6525
750 - 800	6	775	60	4650
800 - 850	12	825	72	9900
850 - 900	8	875	80	7000
Total	Σfi = 80			Σfiui = 55200

By formula,

$$\begin{aligned}\text{Mean }= \frac{∑f_iu_i}{∑f_i}\\[1em] = \frac{55200}{80}\\[1em] = 690.\end{aligned}$$

Hence, mean = ₹ 690.

(ii) The class 550 - 600 has maximum frequency 20.

Hence, modal class = 550-600.

(iii) From table,

The number of workers getting weekly wages below ₹800 = 60.

(iv) From table,

The number of workers getting ₹650 or more but less than ₹850 as weekly wages = 72 - 35 = 37.

Steps :

1. Take 1 cm along x-axis = 5 units and 1 cm along y-axis = 4 units.
2. Construct rectangles corresponding to the given data.
3. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.
4. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 14.

Hence, the required mode = 14.

Steps :

1. Take 1 cm along x-axis = 10 marks and 1 cm along y-axis = 4 (students).
2. Since, the scale on x-axis starts at 50, a break (zig-zag curve) is shown near the origin along x-axis to indicate that the graph is drawn to scale beginning at 50 and not at origin itself.
3. Construct rectangles corresponding to the given data.
4. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.
5. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 82.5.

Hence, the required mode = 82.5

Steps :

1. Take 1 cm along x-axis = 10 units and 1 cm along y-axis = 2 units.
2. Construct rectangles corresponding to the given data.
3. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.
4. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 23.

Hence, the required mode = 23.

Steps :

1. Take 1 cm along x-axis = 1000 runs and 1 cm along y-axis = 4 (batsman).
2. Construct rectangles corresponding to the given data.
3. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.
4. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 4600.

Hence, the required mode = 4600.

Steps :

1. Take 1 cm along x-axis = 1000 rupees and 1 cm along y-axis = 2 (No. of people).

2. Construct rectangles corresponding to the given data.

3. In highest rectangle, draw two st. lines AD and BC from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AD and BC.

4. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 5400.

Hence, mode = ₹ 5,400.

Steps :

1. Take 1 cm along x-axis = ₹5 and 1 cm along y-axis = 5 students.
2. Construct rectangles corresponding to the given data.
3. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.
4. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents ₹21.50.

Hence, the required mode = ₹21.50.

Steps :

1. The given frequency distribution is discontinuous, to convert it into continuous distribution,

     Adjustment factor = $\frac{45 - 44}{2} = \frac{1}{2}$ = 0.5

We construct the continuous frequency table for the given data :

Classes before adjustment	Classes after adjustment	No. of students
40 - 44	39.5 - 44.5	2
45 - 49	44.5 - 49.5	8
50 - 54	49.5 - 54.5	12
55 - 59	54.5 - 59.5	10
60 - 64	59.5 - 64.5	6
65 - 69	64.5 - 69.5	4

2. Take 2 cm along x-axis = 5 kg and 1 cm along y-axis = 2 (students).

3. Since, the scale on x-axis starts at 39.5, a break (zig-zag curve) is shown near the origin along x-axis to indicate that the graph is drawn to scale beginning at 39.5 and not at origin itself.

4. Construct rectangles corresponding to the given data.

5. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.

6. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 52.75 kg.

Hence, the required mode = 52.75 kg.

Size of each class = difference between two consecutive mid-values = 18 - 12 = 6.

Constructing the table as under :

Mid value	Class	Frequency
12	9 - 15	20
18	15 - 21	12
24	21 - 27	8
30	27 - 33	24
36	33 - 39	16
42	39 - 45	8
48	45 - 51	12

Steps :

1. Take 1 cm along x-axis = 6 units and 1 cm along y-axis = 4 units.

2. Since, the scale on x-axis starts at 9, a break (zig-zag curve) is shown near the origin along x-axis to indicate that the graph is drawn to scale beginning at 9 and not at origin itself.

3. Construct rectangles corresponding to the given data.

4. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.

5. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 30.5.

∴ Mode = 30.5

As the class 27 - 33 has the highest frequency, hence modal class = 27 - 33.

Hence, the required mode = 30.5 and modal class = 27 - 33.

Steps :

1. Take 2 cm along x-axis = 10 overs and 1 cm along y-axis = 10 runs.

2. Since, the scale on x-axis starts at 20, a break (zig-zag curve) is shown near the origin along x-axis to indicate that the graph is drawn to scale beginning at 20 and not at origin itself.

3. Construct rectangles corresponding to the given data.

4. In highest rectangle, draw two st. lines KN and LI from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let Z be the point of intersection of KN and LI.

5. Through Z, draw a vertical line to meet the x-axis at A. The abscissa of the point A represents 57.

Hence, mode = 57.

To draw an ogive :

1. The cumulative frequency table for the given continuous distribution is :

Height (in cm)	No. of students	Cumulative frequency
150 - 160	8	8
160 - 170	3	11
170 - 180	4	15
180 - 190	10	25
190 - 200	2	27

2. Since, the scale on x-axis starts at 150, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 150.

3. Take 2 cm along x-axis = 10 cm (height)

4. Take 1 cm along y-axis = 5 students

5. Plot the points (160, 8), (170, 11), (180, 15), (190, 25), (200, 27) representing upper class limits and the respective cumulative frequencies.
   Also plot the point representing lower limit of the first class i.e. 150 - 160.

6. Join these points by a freehand drawing.

The required ogive is shown in figure above.

To draw an ogive :

1. The given frequency distribution is discontinuous, to convert it into continuous distribution,
   Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2
   $= \frac{11 - 10}{2} = \frac{1}{2} = 0.5$

The cumulative frequency table for the given data is :

Classes before adjustment	Classes after adjustment	Frequency	Cumulative frequency
1 - 10	0.5 - 10.5	3	3
11 - 20	10.5 - 20.5	5	8
21 - 30	20.5 - 30.5	8	16
31 - 40	30.5 - 40.5	7	23
41 - 50	40.5 - 50.5	6	29
51 - 60	50.5 - 60.5	2	31

2. Since, the scale on x-axis starts at 0.5, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 0.5

3. Take 2 cm along x-axis = 10

4. Take 1 cm along y-axis = 5

5. Plot the points (10.5, 3), (20.5, 8), (30.5, 16), (40.5, 23), (50.5, 29), (60.5, 31) representing upper class limits and the respective cumulative frequencies.

Also plot the point representing lower limit of the first class i.e. 0.5 - 10.5

6. Join these points by a freehand drawing.

The required ogive is shown in figure above.

To draw an ogive :

1. The cumulative frequency table for the given continuous distribution is :

Marks obtained	No. of students	Cumulative frequency
24 - 29	1	1
29 - 34	2	3
34 - 39	5	8
39 - 44	6	14
44 - 49	4	18
49 - 54	3	21
54 - 59	2	23

2. Since, the scale on x-axis starts at 24, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 24.

3. Take 1 cm along x-axis = 5 marks

4. Take 1 cm along y-axis = 5 (students)

5. Plot the points (29, 1), (34, 3), (39, 8), (44, 14), (49, 18), (54, 21) and (59, 23) representing upper class limits and the respective cumulative frequencies.

Also plot the point representing lower limit of the first class i.e. 24 - 29.

6. Join these points by a freehand drawing.

The required cumulative frequency curve is shown in figure above.

1. The cumulative frequency table for the given continuous distribution is :

Weight (in kg)	No. of workers	Cumulative frequency
50 - 60	4	4
60 - 70	7	11
70 - 80	11	22
80 - 90	14	36
90 - 100	6	42
100 - 110	5	47
110 - 120	3	50

2. Take 1 cm along x-axis = 10 kg

3. Take 1 cm along y-axis = 5 (workers)

4. Since, scale on x-axis starts at 50, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 50.

5. Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42), (110, 47) and (120, 50) representing upper class limits and the respective cumulative frequencies.
   Also plot the point representing lower limit of the first class i.e. 50 - 60.

6. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) To find lower quartile :

Let A be the point on y-axis representing frequency = $\frac{n}{4} = \frac{50}{4}$ = 12.5

Through A, draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 71 kg.

To find upper quartile :

Let B be the point on y-axis representing frequency = $\frac{3n}{4} = \frac{150}{4}$ = 37.5

Through B, draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 93 kg.

Hence, lower quartile = 71 kg and upper quartile = 93 kg.

(ii) Let O be the point on x-axis representing 95 kg. Through O draw a vertical line to meet the ogive R. Through R, draw a horizontal line to meet the y-axis at point C. The ordinate of the point C represents 39.

So, the number of people whose weight is less than 95 kg = 39. So, overweight people = 50 - 39 = 11 kg.

Hence, the number of workers who are overweight are 11.

1. The cumulative frequency table for the given continuous distribution is :

Scores	No. of shooters	Cumulative frequency
0 - 10	9	9
10 - 20	13	22
20 - 30	20	42
30 - 40	26	68
40 - 50	30	98
50 - 60	22	120
60 - 70	15	135
70 - 80	10	145
80 - 90	8	153
90 - 100	7	160

2. Take 2 cm along x-axis = 10 scores

3. Take 2 cm along y-axis = 20 (shooters)

4. Plot the points (10, 9), (20, 22), (30, 42), (40, 68), (50, 98), (60, 120), (70, 135), (80, 145), (90, 153) and (100, 160) representing upper class limits and the respective cumulative frequencies.
   Also plot the point representing lower limit of the first class i.e. 0 - 10.

5. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 160.

To find the median :

Let A be the point on y-axis representing frequency = $\frac{n}{2} = \frac{160}{2}$ = 80.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 44.

Hence, the required median score = 44.

(ii) To find lower quartile :

Let B be the point on y-axis representing frequency = $\frac{n}{4} = \frac{160}{4}$ = 40

Through B, draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 29.

To find upper quartile :

Let C be the point on y-axis representing frequency = $\frac{3n}{4} = \frac{480}{4}$ = 120

Through C, draw a horizontal line to meet the ogive at R. Through R, draw a vertical line to meet the x-axis at O. The abscissa of the point O represents 60.

Inter quartile range = Upper quartile - Lower quartile = 60 - 29 = 31.

Hence, the inter quartile range = 31 scores.

(iii) Total marks = 100.

So, more than 85% marks mean more than 85 marks.

Let T be the point on x-axis representing marks = 85.

Through T, draw a vertical line to meet the ogive at S. Through S, draw a horizontal line to meet the y-axis at D. The ordinate of the point D represents 149.

Students who have scored less than 85% = 149.

So, students scoring more than 85% = Total students - Students who have scored less = 160 - 149 = 11.

Hence, there are 11 shooters who obtained a score of more than 85%.

1. The cumulative frequency table for the given continuous distribution is :

Wages (in ₹)	No. of workers	Cumulative frequency
400 - 450	2	2
450 - 500	6	8
500 - 550	12	20
550 - 600	18	38
600 - 650	24	62
650 - 700	13	75
700 - 750	5	80

2. Take 2 cm along x-axis = 50 rupees

3. Take 1 cm along y-axis = 10 workers

4. Since, scale on x-axis starts at 400, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 400.

5. Plot the points (450, 2), (500, 8), (550, 20), (600, 38), (650, 62), (700, 75) and (750, 80) representing upper class limits and the respective cumulative frequencies.
   Also plot the point representing lower limit of the first class i.e. 400 - 450.

6. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 80.

To find the median :

Let A be the point on y-axis representing frequency = $\frac{n}{2} = \frac{80}{2}$ = 40.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 604.

Hence, the required median wage = ₹604.

(ii) To find lower quartile :

Let B be the point on y-axis representing frequency = $\frac{n}{4} = \frac{80}{4}$ = 20

Through B, draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 550.

Hence, lower quartile wage = ₹550.

(iii) Let T be the point on x-axis representing wage = ₹625.

Through T, draw a vertical line to meet the ogive at S. Through S, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 51.

Workers who earn less than ₹625 = 51.

So, workers earning more than ₹625 = Total workers - workers who earn less than ₹625 = 80 - 51 = 29.

Hence, there are 29 workers earning more than ₹625 daily.

1. The cumulative frequency table for the given continuous distribution is :

Marks	No. of students	Cumulative frequency
0 - 10	5	5
10 - 20	11	16
20 - 30	10	26
30 - 40	20	46
40 - 50	28	74
50 - 60	37	111
60 - 70	40	151
70 - 80	29	180
80 - 90	14	194
90 - 100	6	200

2. Take 1 cm along x-axis = 10 scores

3. Take 1 cm along y-axis = 20 (students)

4. Plot the points (10, 5), (20, 16), (30, 26), (40, 46), (50, 74), (60, 111), (70, 151), (80, 180), (90, 194) and (100, 200) representing upper class limits and the respective cumulative frequencies.
   Also plot the point representing lower limit of the first class i.e. 0 - 10.

5. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 200.

To find the median :

Let A be the point on y-axis representing frequency = $\frac{n}{2} = \frac{200}{2}$ = 100.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 57.

Hence, the required median marks = 57.

(ii) Let N be the point on x-axis representing marks = 40.

Through N, draw a vertical line to meet the ogive at Q. Through Q, draw a horizontal line to meet the y-axis at B. The ordinate of the point B represents 46.

Students who scored less than 40 = 46.

Hence, 46 students failed in the examination.

(iii) Let O be the point on x-axis representing marks = 85.

Through O, draw a vertical line to meet the ogive at R. Through R, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 187.

Students who scored less than 85 = 187.

So, students scoring more than 85 = Total students - students scoring less than 85 = 200 - 187 = 13.

Hence, 13 students secured grade one in examination.

1. The cumulative frequency table for the given continuous distribution is :

Height (in cm)	No. of boys	Cumulative frequency
135 - 140	4	4
140 - 145	8	12
145 - 150	20	32
150 - 155	14	46
155 - 160	7	53
160 - 165	6	59
165 - 170	1	60

2. Take 2 cm along x-axis = 5 cm

3. Take 2 cm along y-axis = 10 (boys)

4. Since, scale on x-axis starts at 135, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 135.

5. Plot the points (140, 4), (145, 12), (150, 32), (155, 46), (160, 53), (165, 59) and (170, 60) representing upper class limits and the respective cumulative frequencies.
   Also plot the point representing lower limit of the first class i.e. 135 - 140.

6. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 60.

To find the median :

Let A be the point on y-axis representing frequency = $\frac{n}{2} = \frac{60}{2}$ = 30.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 149.5.

Hence, the median height = 149.5 cm.

(ii) To find lower quartile :

Let B be the point on y-axis representing frequency = $\frac{n}{4} = \frac{60}{4}$ = 15.

Through B, draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 146.

Hence, lower quartile = 146 cm.

(iii) Let O be the point on x-axis representing height = 158 cm.

Through O, draw a vertical line to meet the ogive at R. Through R, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 51.

No. of boys shorter than 158 cm = 51

So, no. of boys taller than 158 cm = Total boys - boys shorter than 158 cm = 60 - 51 = 9.

Hence, there are 9 tall boys in the class.

Cumulative frequency distribution table :

Distance in m	Frequency	Cumulative frequency
12 - 13	3	3
13 - 14	9	12
14 - 15	12	24
15 - 16	9	33
16 - 17	4	37
17 - 18	2	39
18 - 19	1	40

Steps of construction :

1. Since, the scale on x-axis starts at 12, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 12.

2. Take 2 cm along x-axis = 1 m.

3. Take 2 cm along y-axis = 5 students.

4. Plot the point (12, 0) as ogive starts from x-axis representing lower limit of first class.

5. Plot the points (13, 3), (14, 12), (15, 24), (16, 33), (17, 37), (18, 39) and (19, 40).

6. Join the points by a free hand curve.

(i) The total number of students is N = 40. The median position is found at $\frac{N}{2} = \frac{40}{2} = 20$.

Draw a line parallel to x-axis from point A (number of students) = 20, touching the graph at point B. From point B draw a line parallel to y-axis touching x-axis at point C.

From graph, C = 14.7

The median = 14.7 m.

(ii) Here, n = 40, which is even.

By formula,

Upper quartile = $\frac{3n}{4} = \frac{ 3 \times 40}{4} = \frac{120}{4}$ = 30.

Draw a line parallel to x-axis from point J (number of students) = 30, touching the graph at point K. From point K draw a line parallel to y-axis touching x-axis at point L.

From graph, L = 15.6

The upper quartile = 15.6 m

(iii) Draw a line parallel to y-axis from point D (Distance) = $16\frac{1}{2}$ m = 16.5 m, touching the graph at point E. From point E draw a line parallel to x-axis touching y-axis at point F.

From graph, F = 35.

It means that 35 students who cover a distance either less or equal to $16\frac{1}{2}$ m.

Number of student who cover a distance which is above $16\frac{1}{2}$ m = 40 - 35 = 5.

Number of students who cover a distance above $16\frac{1}{2}$ m = 5.

1. The cumulative frequency table for the given continuous distribution is :

Marks	No. of students	Cumulative frequency
0 - 10	3	3
10 - 20	7	10
20 - 30	12	22
30 - 40	17	39
40 - 50	23	62
50 - 60	14	76
60 - 70	9	85
70 - 80	6	91
80 - 90	5	96
90 - 100	4	100

2. Take 1 cm along x-axis = 10 (marks)

3. Take 1 cm along y-axis = 10 (students)

4. Plot the points (10, 3), (20, 10), (30, 22), (40, 39), (50, 62), (60, 76), (70, 85), (80, 91), (90, 96) and (100, 100) representing upper class limits and the respective cumulative frequencies. Also plot the point representing lower limit of the first class i.e. 0 - 10.

5. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 100.

To find the median :

Let A be the point on y-axis representing frequency = $\frac{n}{2} = \frac{100}{2}$ = 50.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 45

Hence, the median marks = 45.

(ii) To find lower quartile :

Let B be the point on y-axis representing frequency = $\frac{n}{4} = \frac{100}{4}$ = 25.

Through B, draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 32.

Hence, lower quartile = 32.

(iii) Total marks = 100.

85% marks = 85 numbers.

Let O be the point on x-axis representing marks = 85.

Through O draw a vertical line to meet the ogive at R. Through R, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 94.

Hence, 94 students score less than 85 so students scoring more than 85 = 100 - 94 = 6.

Hence, 6 students score more than 85% in the test.

(iv) 35% of 100 = 35.

Let T be the point on x-axis representing marks = 35.

Through T, draw a vertical line to meet the ogive at S. Through S, draw a horizontal line to meet the y-axis at D. The ordinate of the point D represents 30.

No. of students who scored less than 35 marks = 30.

Hence, 30 students were failed in the examination.

1. The cumulative frequency table for the given continuous distribution is :

Marks	No. of students	Cumulative frequency
0 - 10	5	5
10 - 20	9	14
20 - 30	16	30
30 - 40	22	52
40 - 50	26	78
50 - 60	18	96
60 - 70	11	107
70 - 80	6	113
80 - 90	4	117
90 - 100	3	120

2. Take 1 cm along x-axis = 10 marks

3. Take 1 cm along y-axis = 10 students

4. Plot the points (10, 5), (20, 14), (30, 30), (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117) and (100, 120) representing upper class limits and the respective cumulative frequencies.
   Also plot the point representing lower limit of the first class i.e. 0 - 10.

5. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 120.

To find the median :

Let A be the point on y-axis representing frequency = $\frac{n}{2} = \frac{120}{2}$ = 60.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents marks = 43.5

Hence, the median marks = 43.5.

(ii) Total marks = 100.

75% marks = 75 numbers.

Let O be the point on x-axis representing marks = 75.

Through O draw a vertical line to meet the ogive at R. Through R, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 110.

Hence, 110 students score less than 75 so students scoring more than 75 = 120 - 110 = 10.

Hence, 10 students score more than 75% in the test.

(iii) 40% of 100 = 40.

Let T be the point on x-axis representing marks = 40.

Through T, draw a vertical line to meet the ogive at S. Through S, draw a horizontal line to meet the y-axis at D. The ordinate of the point D represents 52.

No. of students who scored less than 40 marks = 52.

Hence, 52 students failed in the examination.

1. The cumulative frequency table for the given continuous distribution is :

Height (in cm)	No. of students	Cumulative frequency
140 - 145	12	12
145 - 150	20	32
150 - 155	30	62
155 - 160	38	100
160 - 165	24	124
165 - 170	16	140
170 - 175	12	152
175 - 180	8	160

2. Take 2 cm along x-axis = 5 cm (height)

3. Take 1 cm along y-axis = 20 (students)

4. Since, scale on x-axis starts at 140, a kink is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 140.

5. Plot the points (145, 12), (150, 32), (155, 62), (160, 100), (165, 124), (170, 140), (175, 152) and (180, 160) representing upper class limits and the respective cumulative frequencies. Also plot the point representing lower limit of the first class i.e. 140 - 145.

6. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 160.

To find the median :

Let A be the point on y-axis representing frequency = $\frac{n}{2} = \frac{160}{2}$ = 80.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents height = 157.5 cm.

Hence, the median height = 157.5 cm.

(ii) To find lower quartile :

Let B be the point on y-axis representing frequency = $\frac{n}{4} = \frac{160}{4}$ = 40

Through B, draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 151.5.

To find upper quartile :

Let C be the point on y-axis representing frequency = $\frac{3n}{4} = \frac{480}{4}$ = 120

Through C, draw a horizontal line to meet the ogive at R. Through R, draw a vertical line to meet the x-axis at O. The abscissa of the point O represents 164.

Inter quartile range = Upper quartile - Lower quartile = 164 - 151.5 = 12.5.

Hence, the inter quartile range = 12.5 cm.

(iii) Let T be the point on x-axis representing height = 172 cm.

Through T, draw a vertical line to meet the ogive at S. Through S, draw a horizontal line to meet the y-axis at D. The ordinate of the point D represents 144.

No. of students shorter than 172 cm = 144.

So, no. of students taller than 172 cm = Total students - No. of students shorter than 172 cm = 160 - 144 = 16.

Hence, there are 16 students taller than 172 cm.

(a) From graph,

The curve plotted is a cumulative frequency curve (ogive).

(b) From graph,

The total number of students = 40.

(c) From graph,

The median marks = 56.

(d) From graph,

No of students scoring below 80 = 37

No of students scoring below 50 = 12

∴ No. of students scoring between 50 and 80 = 37 - 12 = 25.

Hence, the no. of students scoring between 50 and 80 = 25.

On converting the class from discontinuous intervals into continuous intervals.

Adjustment factor = (Lower limt of one class - Upper limit of previous class) / 2

$$\begin{aligned}= \frac{11 - 10}{2} \\[1em] = \frac{1}{2} \\[1em] = 0.5\end{aligned}$$

Classes before adjustment	Classes after adjustment
1 - 10	0.5 - 10.5
11 - 20	10.5 - 20.5
21 - 30	20.5 - 30.5
31 - 40	30.5 - 40.5
41 - 50	40.5 - 50.5
51 - 60	50.5 - 60.5

Class interval = 10.5 - 0.5 = 10.

Hence, Option 2 is the correct option.

On converting the class from discontinuous intervals into continuous intervals.

Adjustment factor = (Lower limt of one class - Upper limit of previous class) / 2

$$\begin{aligned}= \frac{11 - 10}{2} \\[1em] = \frac{1}{2} \\[1em] = 0.5\end{aligned}$$

Classes before adjustment	Classes after adjustment
1 - 10	0.5 - 10.5
11 - 20	10.5 - 20.5
21 - 30	20.5 - 30.5
31 - 40	30.5 - 40.5
41 - 50	40.5 - 50.5
51 - 60	50.5 - 60.5
61 - 70	60.5 - 70.5

From table, the upper limit of the class 11 - 20 is 20.5

Hence, Option 4 is the correct option.

In a grouped frequency distribution, the mid-values of the classes are used to measure mean.

Hence, Option 3 is the correct option.

In the formula : x̄ = a + $\frac{Σ f_id_i}{Σ f_i}$ for finding the mean of the grouped data, d_i's are deviations from a (assumed mean) of mid-points of the classes.

Hence, Option 3 is the correct option.

Cumulative frequency distribution table is useful in determining the median.

Hence, Option 2 is the correct option.

The given class intervals are already in ascending order. We construct the cumulative frequency table as under :

Class Interval	Frequency	Cumulative frequency
0 - 10	2	2
10 - 20	4	6
20 - 30	3	9
30 - 40	5	14

Here, Cumulative frequency = 14, which is even.

By formula,

$$\begin{aligned}\text{Median }= \frac{\frac{n}{2}\text{th observation} + \Big(\frac{n}{2} + 1\Big)\text{th observation}}{2}\\[1em] = \frac{\frac{14}{2}\text{th observation} + \Big(\frac{14}{2} + 1\Big)\text{th observation}}{2}\\[1em] = \frac{7\text{th observation} + (7 + 1)\text{th observation}}{2}\\[1em] = \frac{7\text{th observation} + 8\text{th observation}}{2}\\[1em]\end{aligned}$$

All observations from 7th to 8th are equal, each lies in the interval 20 - 30.

So, median class = 20 - 30

Hence, option 3 is the correct option.

Converting the discontinuous interval into continuous interval.

Adjustment factor = (Lower limt of one class - Upper limit of previous class) / 2

$$\begin{aligned}= \frac{11 - 10}{2} \\[1em] = \frac{1}{2} \\[1em] = 0.5\end{aligned}$$

We construct the cumulative frequency distribution table as under :

Classes before adjustment	Classes after adjustment	Frequency	Cumulative frequency
0 - 5	0 - 5.5	13	13
6 - 11	5.5 - 11.5	10	23
12 - 17	11.5 - 17.5	15	38
18 - 23	17.5 - 23.5	8	46
24 - 29	23.5 - 29.5	11	57

Here n (total no. of observations) = 57.

As n is odd,

$$\begin{aligned}\therefore \text{Median } = \frac{n + 1}{2} \text{th observation} \\[1em] = \frac{57 + 1}{2} \\[1em] = \frac{58}{2} \\[1em] = 29 \text{th observation}\end{aligned}$$

As observation from 24^th to 38^th lies in the class 11.5 - 17.5

∴ Median class = 11.5 - 17.5, with upper limit = 17.5

Hence, Option 2 is the correct option.

We construct the cumulative frequency distribution table as under :

Class	Frequency	Cumulative frequency
0 - 5	10	10
5 - 10	15	25
10 - 15	12	37
15 - 20	20	57
20 - 25	9	66

Here n (total no. of observations) = 66.

As n is even,

∴ Median = ($\frac{n}{2}$th observation + ($\frac{n}{2} + 1$)th observation) / 2

$$\begin{aligned}= \frac{\frac{66}{2} + \Big(\frac{66}{2} + 1\Big)}{2} \\[1em] = \frac{33\text{th observation} + 34\text{th observation}}{2}\end{aligned}$$

As observation from 26^th to 37^th lie in the class 10 - 15,

∴ Median class = 10 - 15.

Since the class 15 - 20 has highest frequency i.e. 20.

∴ Modal class = 15 - 20.

Sum of lower limit of median and modal class = 10 + 15 = 25.

Hence, Option 2 is the correct option.

The "modal class" refers to the interval in a frequency distribution that has the highest number of observations, meaning it's the class with the most occurrences.

Hence, option 1 is the correct option.

An ogive curve is used to determine median.

Hence, Option 4 is the correct option.

No. of observations = 10, which is even.

Median = $\frac{n}{2}\text{th term} = \frac{10}{2}$ = 5th term.

Given,

Median = 64

∴ x + 4 = 64

⇒ x = 64 - 4 = 60.

Hence, Option 1 is the correct option.

The prime numbers between 5 and 20 are: 7, 11, 13, 17, 19 .

There are 5 such primes.

By formula,

Mean = $\frac{\text{sum of all observations}}{\text{number of observations}}$

∴ Reason (R) is true.

Mean = $\frac{7 + 11 + 13 + 17 + 19}{5} = \frac{67}{5}$ = 13.4

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Number of plants	Class mark(xi)	Number of houses(fi)	fi.xi
2 - 4	3	2	6
4 - 6	5	3	15
6 - 8	7	1	7
8 - 10	9	3	27
10 - 12	11	1	11
Total		Σfi = 10	Σfixi = 66

Mean = $\frac{∑f_ix_i}{∑f_i} = \frac{f_1x_1 + f_2x_2 + f_3x_3 + ........ + f_kx_k}{f_1 + f_2 + f_3 + ........ + f_k}$

∴ Reason (R) is true.

Mean = $\frac{66}{10}$​ = 6.6

∴ Assertion (A) is false.

∴ Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Given goals: 3, 1, 0, 7, 5, 3, 3, 4, 1, 2, 0, 2

Arrange the data in ascending order: 0, 0, 1, 1, 2, 2, 3, 3, 3, 4, 5, 7

There are 12 values (even).

By formula,

$$\begin{aligned}\text{Median }= \frac{\frac{n}{2}\text{th observation} + \Big(\frac{n}{2} + 1\Big)\text{th observation}}{2}\\[1em] = \frac{\frac{12}{2}\text{th observation} + \Big(\frac{12}{2} + 1\Big)\text{th observation}}{2}\\[1em] = \frac{6\text{th observation} + (6 + 1)\text{th observation}}{2}\\[1em] = \frac{6\text{th observation} + 7\text{th observation}}{2}\\[1em] = \frac{(2 + 3)}{2} = \frac{5}{2} \\[1em] = 2.5.\end{aligned}$$

∴ Assertion (A) is true.

The median is a measure of central tendency that splits your ordered data into two equal parts—half the observations lie below it, and half lie above it.

∴ Reason (R) is false.

∴ Assertion (A) is true, but Reason (R) is false.

Hence, option 1 is the correct option.

For the given 25 variables : x₁, x₂, x₃,........., x₂₅.

To compute the median, you must first arrange the data in either ascending or descending order. Only with ordered data, the middle value can be identified.

∴ Assertion (A) is true.

The median is a measure of central tendency that splits your ordered data into two equal parts—half the observations lie below it, and half lie above it.

Thus, the median is the central most term of the arranged data.

∴ Reason (R) is true.

∴ Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Mean = $\frac{\text{Sum of observations}}{\text{No. of observations}}$

∴ Sum of observations = Mean × No. of observations
⇒ Sum of observations = 18 × 20 = 360.

Since 3 is added to first ten numbers then total sum increases by 10 × 3 = 30.

New sum = 360 + 30 = 390.

Mean = $\frac{\text{Sum of observations}}{\text{No. of observations}} = \frac{390}{20}$ = 19.5

Hence, the mean of new set of 20 numbers is 19.5

Mean = $\frac{\text{Sum of observations}}{\text{No. of observations}}$

∴ Sum of observations = Mean × No. of observations
⇒ Sum of observations = 150 × 30 = 4500.

Since, one value of 165 cm was wrongly copied as 135 cm,

∴ Correct sum of observations = 4500 - 135 + 165 = 4530.

Correct mean = $\frac{\text{Sum of observations}}{\text{No. of observations}} = \frac{4530}{30}$ = 151.

Hence, the correct mean is 151 cm.

No. of girls = Total - No. of boys = 50 - 40 = 10.

Average weight of class = $\frac{\text{Total weight}}{\text{No. of students}}$

∴ Total weight = Average weight × No. of students
⇒ Total weight = 44 × 50 = 2200

Average weight of girls = $\frac{\text{Weight of girls}}{\text{No. of girls}}$

∴ Weight of girls = Average weight of girls × No. of girls
⇒ Weight of girls = 40 × 10 = 400.

Total weight of boys = Total weight - Weight of girls = 2200 - 400 = 1800.

Average weight of boys = $\frac{\text{Weight of boys}}{\text{No. of boys}} = \frac{1800}{40}$ = 45.

Hence, the average weight of boys is 45 kg.

We construct the table as under :

Height (xi)	No. of children (fi)	fixi
65	1	65
66	4	264
67	5	335
68	7	476
69	11	759
70	10	700
71	6	426
72	4	288
73	2	146
Total	50	3459

Mean = $\frac{Σf_ix_i}{Σf_i} = \frac{3459}{50}$ = 69.2 cm.

Hence, mean height = 69.2 cm.

We construct the table as under :

Variate (x)	Frequency (f)	fx
13	8	104
15	2	30
17	3	51
19	4	76
20 + p	5p	5p2 + 100p
23	6	138
Total	5p + 23	5p2 + 100p + 399

$$\begin{aligned}\text{Mean} = \frac{Σfx}{Σf} \\[1em] \therefore 18 = \frac{5p^2 + 100p + 399}{5p + 23} \\[1em] \Rightarrow 90p + 414= 5p^2 + 100p + 399 \\[1em] \Rightarrow 5p^2 + 100p - 90p + 399 - 414 = 0 \\[1em] \Rightarrow 5p^2 + 10p - 15 = 0 \\[1em] \Rightarrow 5p^2 + 15p - 5p - 15 = 0 \\[1em] \Rightarrow 5p(p + 3) - 5(p + 3) = 0 \\[1em] \Rightarrow (5p - 5)(p + 3) = 0 \\[1em]\end{aligned}$$

⇒ 5p - 5 = 0 or p + 3 = 0
⇒ 5p = 5 or p = -3
⇒ p = 1 or p = -3.

Since, frequency cannot be negative so p = 1.

Hence, the value of p = 1.

The above data is discontinuous converting in continuous data we get,

Adjustment factor = (Lower limt of one class - Upper limit of previous class) / 2

$$\begin{aligned}= \frac{30 - 29}{2} \\[1em] = \frac{1}{2} \\[1em] = 0.5\end{aligned}$$

Classes before adjustment	Classes after adjustment	Class mark (ui)	No. of persons (fi)	fiui
25 - 29	24.5 - 29.5	27	4	108
30 - 34	29.5 - 34.5	32	14	448
35 - 39	34.5 - 39.5	37	22	814
40 - 44	39.5 - 44.5	42	16	672
45 - 49	44.5 - 49.5	47	6	282
50 - 54	49.5 - 54.5	52	5	260
55 - 59	54.5 - 59.5	57	3	171
Total			70	2755

Mean = $\frac{Σf_iu_i}{Σf_i} = \frac{2755}{70}$ = 39.36 years

Hence, the mean age = 39.36 years.

We construct the following table :

Classes	Class mark (ui)	Frequency (fi)	fiui
0 - 20	10	5	50
20 - 40	30	8	240
40 - 60	50	p	50p
60 - 80	70	12	840
80 - 100	90	7	630
100 - 120	110	8	880
Total		40 + p	2640 + 50p

$$\begin{aligned}\text{Mean} = \frac{Σf_iu_i}{Σf_i} \\[1em] \Rightarrow 62.8 = \frac{2640 + 50p}{40 + p} \\[1em] \Rightarrow 62.8(40 + p) = 2640 + 50p \\[1em] \Rightarrow 2512 + 62.8p = 2640 + 50p \\[1em] \Rightarrow 62.8p - 50p = 2640 - 2512 \\[1em] \Rightarrow 12.8p = 128 \\[1em] \Rightarrow p = 10.\end{aligned}$$

Hence, the value of p = 10.

We construct the table as under :

Expenditure (in ₹)	Class mark (ui)	No. of families (fi)	fiui
140 - 160	150	5	750
160 - 180	170	25	4250
180 - 200	190	f1	190f1
200 - 220	210	f2	210f2
220 - 240	230	5	1150
Total		35 + f1 + f2	6150 + 190f1 + 210f2

As there are 100 families,

∴ 35 + f₁ + f₂ = 100
⇒ f₁ + f₂ = 100 - 35 = 65
⇒ f₁ = 65 - f₂      ......(i)

$$\begin{aligned}\text{Mean} = \frac{Σf_iu_i}{Σf_i} \\[1em] \Rightarrow 188 = \frac{6150 + 190f_1 + 210f_2}{35 + f_1 + f_2} \\[1em] \Rightarrow 188(35 + f_1 + f_2) = 6150 + 190f_1 + 210f_2 \\[1em] \Rightarrow 188(35 + 65 - f_2 + f_2) = 6150 + 190(65 - f_2) + 210f_2 \text{ .......(Using (i))} \\[1em] \Rightarrow 188(100) = 6150 + 12350 - 190f_2 + 210f_2 \\[1em] \Rightarrow 18800 = 18500 + 20f_2 \\[1em] \Rightarrow 18800 - 18500 = 20f_2 \\[1em] \Rightarrow 20f_2 = 300 \\[1em] \Rightarrow f_2 = 15.\end{aligned}$$

Using (i),

⇒ f₁ = 65 - f₂ = 65 - 15 = 50.

Hence, f₁ = 50 and f₂ = 15.

Here, n (no. of observations) = 10, which is even.

$$\begin{aligned}\therefore \text{Median} = \frac{\frac{n}{2} \text{th observation} + \big(\frac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\frac{10}{2} \text{th observation} + \big(\frac{10}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \frac{\text{ 5th observation + 6th observation}}{2} \\[1em] = \frac{x + 2 + x + 4}{2} \\[1em] = \frac{2x + 6}{2} \\[1em] = x + 3.\end{aligned}$$

Given, median = 25.

∴ x + 3 = 25
⇒ x = 22.

Hence, the value of x = 22.

Arranging the numbers in ascending order we get,

3, 4, 5, x, 8, 9, 11.

Here n (no. of observations) = 7

$$\begin{aligned}\therefore \text{Median} = \frac{n + 1}{2} \text{th observation} \\[1em] = \frac{7 + 1}{2} \\[1em] = \frac{8}{2} \\[1em] = 4 \text{th observation}.\end{aligned}$$

Given, median = 6.

∴ 6 = 4th observation = x.

Hence, the value of x = 6.

On arranging the numbers in ascending order we get,

1, 3, 5, 6, 8, 9, 10, 12, 13, 15, 17, 18, 20, 21, 21, 23.

(i) Here, n (no. of observations) = 16, which is even.

$$\begin{aligned}\therefore \text{Median} = \frac{\frac{n}{2} \text{th observation} + \big(\frac{n}{2} + 1\big)\text{th observation}}{2} \\[1em] = \frac{\frac{16}{2} \text{th observation} + \big(\frac{16}{2} + 1\big)\text{th observation}}{2} \\[1em] = \frac{\text{ 8th observation + 9th observation}}{2} \\[1em] = \frac{12 + 13}{2} \\[1em] = \frac{25}{2} \\[1em] = 12.5\end{aligned}$$

Hence, the median of following data = 12.5.

(ii) Here, n (no. of observations) = 16, which is even.

$$\begin{aligned}\text{Lower Quartile} = \frac{n}{4}\text{th observation} \\[1em] = \frac{16}{4} \\[1em] = 4\text{th observation} \\[1em] = 6.\end{aligned}$$

Hence, lower quartile = 6.

(iii) Here, n (no. of observations) = 16, which is even.

$$\begin{aligned}\text{Upper Quartile} = \frac{3n}{4}\text{th observation} \\[1em] = \frac{48}{4} \\[1em] = 12\text{th observation} \\[1em] = 18.\end{aligned}$$

Hence, upper quartile = 18.

(iv) Inter quartile range = Upper quartile - Lower quartile = 18 - 6 = 12.

Hence, inter quartile range = 12.

We construct the table as under :

Age in years (xi)	No. of students (fi)	Cumulative frequency	fixi
12	2	2	24
13	3	5	39
14	5	10	70
15	6	16	90
16	4	20	64
17	3	23	51
18	2	25	36
Total	25		374

Mean = $\frac{Σf_ix_i}{Σf_i} = \frac{374}{25}$ = 14.96

Here, n (no. of observations) = 25, which is odd

$$\begin{aligned}\therefore \text{Median} = \frac{n + 1}{2} \text{th observation} \\[1em] = \frac{25 + 1}{2} \\[1em] = \frac{26}{2} \\[1em] = 13 \text{th observation}.\end{aligned}$$

The age of observation from 11th to 16th = 15.

∴ Median = 15.

Highest no. of students are 15 years old.

∴ Mode = 15.

Hence, mean = 14.96, median = 15 and mode = 15.

Steps :

1. Take 2 cm along x-axis = 10 rupees and 1 cm along y-axis = 1 employee.

2. Construct rectangles corresponding to the given data.

3. In highest rectangle, draw two st. lines AC and BD from corners of the rectangles on either side of the highest rectangle to the opposite corners of the highest rectangle. Let P be the point of intersection of AC and BD.

4. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 23.

Hence, the required mode = ₹23.

1. The cumulative frequency table for the given continuous distribution is :

Marks obtained	No. of students	Cumulative frequency
0 - 10	8	8
10 - 20	10	18
20 - 30	22	40
30 - 40	40	80
40 - 50	20	100

2. Take 2 cm along x-axis = 10 marks

3. Take 1 cm along y-axis = 10 (students)

4. Plot the points (10, 8), (20, 18), (30, 40), (40, 80) and (50, 100) representing upper class limits and the respective cumulative frequencies.
   Also plot the point representing lower limit of the first class i.e. 0 - 10.

5. Join these points by a freehand drawing.

The required ogive is shown in figure above.

(i) Here, n (no. of students) = 100.

To find the median :

Let A be the point on y-axis representing frequency = $\frac{n}{2} = \frac{100}{2}$ = 50.

Through A draw a horizontal line to meet the ogive at P. Through P, draw a vertical line to meet the x-axis at M. The abscissa of the point M represents 32.5 marks.

(ii) Total no. of students = 100.

85% of students pass i.e. 85 students pass.

Remaining no. of students = 15.

Let B be the point on y-axis representing frequency 15.

Through B draw a horizontal line to meet the ogive at Q. Through Q, draw a vertical line to meet the x-axis at N. The abscissa of the point N represents 18 marks.

Hence, passing marks = 18 if 85% students pass.

(iii) Total no. of students = 100.

45% of students i.e. 45 students.

Remaining no. of students = 55.

Let C be the point on y-axis representing frequency 55.

Through C draw a horizontal line to meet the ogive at R. Through R, draw a vertical line to meet the x-axis at O. The abscissa of the point O represents 34 marks.

Hence, 45% of students exceed 34 marks.

(a) Frequency table :

Height (class)	Number of plants
30-40	4
40-50	2
50-60	8
60-70	12
70-80	6
80-90	3
90-100	4

(b) From graph,

The modal class is 60-70.

(c) From graph,

The mode = 64.

(d) From graph,

The number of plants whose height range is between 80 cm to 90 cm are 3.