Measures of Central Tendency

Let fifth variable be x.

By formula,

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

$$\begin{aligned}\therefore 20 = \frac{15 + 25 + 10 + 12 + x}{5} \\[1em] \Rightarrow 20 \times 5 = x + 62 \\[1em] \Rightarrow 100 = x + 62 \\[1em] \Rightarrow x = 100 - 62 = 38.\end{aligned}$$

Hence, Option 1 is the correct option.

By formula,

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

Given,

The mean height of 5 students is 140 cm.

$$\begin{aligned}\therefore \text{Mean height} = \frac{\text{Total height}}{\text{No. of students}} \\[1em] \Rightarrow 140 = \frac{\text{Total height}}{5} \\[1em] \Rightarrow \text{Total height} = 140 \times 5 \\[1em] \Rightarrow \text{Total height} = 700 \text{ cm}.\end{aligned}$$

Given,

Height of one student is 156 cm, so total height of remaining students = 700 - 156 = 544 cm.

Mean height of remaining (4) students = $\frac{\text{Tot. ht. of 4 students}}{4} = \frac{544}{4}$ = 136 cm.

Hence, Option 2 is the correct option.

By formula,

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

Given,

The mean age of eight boys is 16 years.

$$\begin{aligned}\therefore \text{Mean age} = \frac{\text{Total age}}{\text{No. of boys}} \\[1em] \Rightarrow 16 = \frac{\text{Total age}}{8} \\[1em] \Rightarrow \text{Total age} = 16 \times 8 \\[1em] \Rightarrow \text{Total age} = 128 \text{ years}.\end{aligned}$$

Given,

Two more boys with ages 18 years and 14 years join the group.

Total age now = 128 + 18 + 14 = 160.

New mean age = $\frac{\text{New total age}}{\text{No. of boys}} = \frac{160}{10}$ = 16 years.

Hence, Option 3 is the correct option.

By formula,

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

Given, the mean value of 15 numbers is 20.

$$\begin{aligned}\therefore 20 = \frac{\text{Sum of observations}}{15} \\[1em] \Rightarrow \text{Sum of observations} = 20 \times 15 = 300.\end{aligned}$$

Given,

One of these numbers is wrongly taken as 45 instead of 15.

Correct sum of observations = 300 - 45 + 15 = 270.

$$\begin{aligned}\Rightarrow \text{Correct mean} = \frac{\text{Correct sum of observations}}{\text{No. of observations}} \\[1em] \Rightarrow \text{Correct mean} = \frac{270}{15} = 18.\end{aligned}$$

Hence, Option 4 is the correct option.

x	f	fx
30	15	450
20	5	100
Total	Σf = 20	Σfx = 550

By formula,

Mean = $\frac{Σfx}{Σf} = \frac{550}{20}$ = 27.5

Hence, Option 3 is the correct option.

(a) Sum of observations = 60 + 67 + 52 + 76 + 50 + 51 + 74 + 45 + 56 = 531.

By formula,

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

= $\frac{531}{9}$ = 59.

Hence, mean = 59.

(b) We know that,

If each given number is increased or decreased by the same quantity, the value of mean is also increased or decreased by same quantity.

∴ If marks of each student is increased by 4, then mean will also increase by 4.

New mean = 59 + 4 = 63.

Hence, new mean = 63.

(a) Sum of observations = 7 + 11 + 6 + 5 + 6 = 35.

By formula,

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

= $\frac{35}{5}$ = 7.

Hence, mean = 7.

(b) We know that,

If each given number is increased or decreased by the same quantity, the value of mean is also increased or decreased by same quantity.

∴ If each number is diminished by 2, then mean will also decrease by 2.

New mean = 7 - 2 = 5.

Hence, new mean = 5.

Sum of observations = 6 + 4 + 7 + a + 10 = a + 27.

By formula,

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

⇒ 8 = $\frac{a + 27}{5}$

⇒ 40 = a + 27

⇒ a = 13.

Hence, the value of a = 13.

Sum of observations = 6 + y + 7 + x + 14 = x + y + 27.

By formula,

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

⇒ 8 = $\frac{x + y + 27}{5}$

⇒ 40 = x + y + 27

⇒ x + y = 40 - 27

⇒ x + y = 13

⇒ y = 13 - x.

Hence, y = 13 - x.

Age (x)	Frequency (f)	fx
12	2	24
13	4	52
14	6	84
15	9	135
16	8	128
17	7	119
18	4	72
Total	Σf = 40	Σfx = 614

We know that,

n = Σf = 40.

By formula,

Mean = $\frac{Σfx}{n} = \frac{614}{40}$ = 15.35

Hence, arithmetic mean = 15.35

Sum of observations : 72 + 70 + x + 62 + 50 + 71 + 90 + 64 + 58 + 82 = 619 + x.

By formula,

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

⇒ 69.5 = $\frac{619 + x}{10}$

⇒ 695 = x + 619

⇒ x = 695 - 619

⇒ x = 76.

Hence, x = 76.

Height (x)	No. of plants (f)	fx
50	2	100
55	4	220
58	10	580
60	f	60f
65	5	325
70	4	280
71	3	213
Total	Σf = 28 + f	Σfx = 1718 + 60f

We know that,

n = Σf = 28 + f

By formula,

⇒ Mean = $\frac{Σfx}{n}$

⇒ 60.95 = $\frac{1718 + 60f}{28 + f}$

⇒ 60.95(28 + f) = 1718 + 60f

⇒ 1706.6 + 60.95f = 1718 + 60f

⇒ 60.95f - 60f = 1718 - 1706.6

⇒ 0.95f = 11.4

⇒ f = $\frac{11.4}{0.95}$ = 12.

Hence, f = 12.

Category	Wages in ₹/day (x)	No. of workers (f)	fx
A	50	2	100
B	60	4	240
C	70	8	560
D	80	12	960
E	90	10	900
F	100	6	600
Total		42	3360

Mean = $\frac{Σfx}{Σf} = \frac{3360}{42}$ = 80.

(i) Original mean = $\frac{Σfx}{Σf}$

If no. of workers is doubled, then

New mean = $\frac{2Σfx}{2Σf} = \frac{Σfx}{Σf}$ = original mean.

∴ If the numbers of workers in each category is doubled, then new mean wage will remain same.

Hence, mean = 80.

(ii) If the wages per day in each category are increased by 60% then new mean wage also increases by 60%.

New mean = 80 + $\frac{60}{100} \times 80$

= 80 + 48 = 128.

Hence, new mean = 128.

(iii) There will be no change in mean due to change in number of workers.

If wages is reduced by 40% then, mean will also reduce by 40%.

New mean = 80 - $\frac{40}{100} \times 80$

= 80 - 32

= 48.

Hence, new mean = 48.

(i)

No. of matches (x)	No. of boxes (f)	fx
35	6	210
36	10	360
37	18	666
38	25	950
39	21	819
40	12	480
41	8	328
Total	100	3813

By formula,

Mean = $\frac{Σfx}{Σf} = \frac{3813}{100}$ = 38.13 ≈ 38.1

Hence, mean = 38.1

(ii) By formula,

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

If mean = 39.

$$\begin{aligned}\Rightarrow 39 = \frac{\text{No. of matches}}{100} \\[1em] \Rightarrow \text{No. of matches} = 3900\end{aligned}$$

No. of matches to add = 3900 - 3813 = 87.

Hence, 87 extra matches need to be added.

x	f	fx
1	9	9
2	6	12
3	9	27
5	3	15
p + 4	6	6p + 24
Total	Σf = 33	Σfx = 6p + 87

By formula,

Mean = $\frac{Σfx}{Σf}$

Substituting values we get,

⇒ 3 = $\frac{6p + 87}{33}$

⇒ 99 = 6p + 87

⇒ 12 = 6p

⇒ p = $\frac{12}{6}$

⇒ p = 2.

Hence, p = 2.

x	f	fx
0	46	0
50	f1	50f1
100	f2	100f2
150	25	3750
200	10	2000
250	5	1250
Total	86 + f1 + f2	7000 + 50f1 + 100f2

Given,

⇒ Σf = 200

⇒ 86 + f₁ + f₂ = 200

⇒ f₁ + f₂ = 200 - 86

⇒ f₁ + f₂ = 114

⇒ f₁ = 114 - f₂ ........(1)

Given, mean = 73

$$\begin{aligned}\Rightarrow \frac{Σfx}{Σf} = 73 \\[1em] \Rightarrow \frac{7000 + 50f_1 + 100f_2}{86 + f_1 + f_2} = 73 \\[1em] \Rightarrow \frac{7000 + 50f_1 + 100f_2}{86 + 114} = 73 \\[1em] \Rightarrow 7000 + 50f_1 + 100f_2 = 200 \times 73 \\[1em] \Rightarrow 7000 + 50(f_1 + 2f_2) = 14600 \\[1em] \Rightarrow 50(f_1 + 2f_2) = 14600 - 7000 \\[1em] \Rightarrow 50(f_1 + 2f_2) = 7600 \\[1em] \Rightarrow f_1 + 2f_2 = \frac{7600}{50} \\[1em] \Rightarrow f_1 + 2f_2 = 152 \\[1em] \Rightarrow 114 - f_2 + 2f_2 = 152 \text{ (From 1)}\\[1em] \Rightarrow 114 + f_2 = 152 \\[1em] \Rightarrow f_2 = 152 - 114 \\[1em] \Rightarrow f_2 = 38.\end{aligned}$$

Substituting value of f₂ in (1), we get :

⇒ f₁ = 114 - f₂ = 114 - 38 = 76.

Hence, f₁ = 76 and f₂ = 38.

Let assumed mean (A) be 25.

x	f	d = x - A	t = (x - A)/i	ft
5	20	-20	-4	-80
10	43	-15	-3	-129
15	75	-10	-2	-150
20	67	-5	-1	-67
A = 25	72	0	0	0
30	45	5	1	45
35	39	10	2	78
40	9	15	3	27
45	8	20	4	32
50	6	25	5	30
Total	Σf = 384			Σft = -214

By formula,

Mean = A + $\frac{Σft}{Σf} \times i = 25 + \frac{-214}{384} \times 5$

= 25 - $\frac{1070}{384}$

= 25 - 2.786

= 22.21 ≈ 22.

Hence, mean = 22.

Let the assumed mean (A) be 45

x	f	d = x - A	fd
40	14	40 - 45 = -5	-70
41	28	41 - 45 = -4	-112
43	38	43 - 45 = -2	-76
A = 45	50	45 - 45 = 0	0
46	40	46 - 45 = 1	40
49	20	49 - 45 = 4	80
50	10	50 - 45 = 5	50
Total	Σf = 200		Σfd = -88

By formula,

Mean = A + $\frac{Σfd}{Σf} = 45 + \frac{-88}{200}$

= 45 - 0.44

= 44.56 ≈ 44.6

Hence, mean = 44.6

C.I.	f	Mean value (x)	fx
30-40	2	35	70
40-50	3	45	135
Total	Σf = 5		Σfx = 205

By formula,

Mean = $\frac{Σfx}{Σf} = \frac{205}{5}$ = 41.

Hence, Option 1 is the correct option.

In the table,

x is the mean value or class mark,

i = class-size = 10,

C.I.	f	x	u = (x - A)/i	f × u
0-10	5	5	(5 - 15)/10 = -10/10 = -1	-5
10-20	10	A = 15	(15 - 15)/10 = 0/10 = 0	0
20-30	10	25	(25 - 15)/10 = 10/10 = 1	10
Total	Σf = 25			Σfu = 5

By formula,

Mean = A + $\frac{Σfu}{Σf} \times i$

Substituting values we get :

Mean = 15 + $\frac{5}{25} \times 10 = 15 + \frac{50}{25}$ = 15 + 2 = 17.

Hence, Option 1 is the correct option.

x	f	fx
10	10	100
20	a	20a
30	30	900
Total	Σf = 40 + a	Σfx = 1000 + 20a

By formula,

Mean = $\frac{Σfx}{Σf}$

Substituting values we get :

$\Rightarrow 24 = \frac{1000 + 20a}{40 + a}$

⇒ 24(40 + a) = 1000 + 20a

⇒ 960 + 24a = 1000 + 20a

⇒ 24a - 20a = 1000 - 960

⇒ 4a = 40

⇒ a = 10.

Hence, Option 3 is the correct option.

x	f	fx
10	x1	10x1
20	20	400
30	x2	30x2
Total	Σf = 20 + x1 + x2	Σfx = 400 + 10x1 + 30x2

By formula,

Mean = $\frac{Σfx}{Σf}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 20 = \frac{400 + 10x_1 + 30x_2}{20 + x_1 + x_2} \\[1em] \Rightarrow 20(20 + x_1 + x_2) = 400 + 10x_1 + 30x_2 \\[1em] \Rightarrow 400 + 20x_1 + 20x_2 = 400 + 10x_1 + 30x_2 \\[1em] \Rightarrow 20x_1 - 10x_1 + 20x_2 - 30x_2 = 400 - 400 \\[1em] \Rightarrow 10x_1 - 10x_2 = 0 \\[1em] \Rightarrow 10(x_1 - x_2) = 0 \\[1em] \Rightarrow x_1 - x_2 = 0.\end{aligned}$$

Hence, Option 3 is the correct option.

Table from the given graph is :

C.I.	Class mark (x)	Frequency (f)	fx
40-50	45	40	1800
50-60	55	10	550
Total		Σf = 50	Σfx = 2350

By formula,

Mean = $\frac{Σfx}{Σf} = \frac{2350}{50}$ = 47.

Hence, Option 2 is the correct option.

Age (years)	Mid value (x)	No. of students (f)	fx
16 - 18	17	2	34
18 - 20	19	7	133
20 - 22	21	21	441
22 - 24	23	17	391
24 - 26	25	3	75
Total		Σf = 50	Σfx = 1074

By formula,

Mean = $\frac{Σfx}{Σf} = \frac{1074}{50}$ = 21.48

Hence, mean = 21.48

(i) Let assumed mean (A) be 65.

Marks	Mid value (x)	No. of boys (f)	d = x - A	fd
30 - 40	35	10	35 - 65 = -30	-300
40 - 50	45	12	45 - 65 = -20	-240
50 - 60	55	14	55 - 65 = -10	-140
60 - 70	65	12	65 - 65 = 0	0
70 - 80	75	9	75 - 65 = 10	90
80 - 90	85	7	85 - 65 = 20	140
90 - 100	95	6	95 - 65 = 30	180
Total		Σf = 70		Σfx = -270

n = Σf = 70.

By formula,

Mean = A + $\frac{Σfd}{n} = 65 + \frac{-270}{70}$

= 65 - 3.86 = 61.14

Hence, mean = 61.14

(ii) We get mean values from part (i) and assumed mean (A) = 65. Let i = 10.

Marks	Mid value (x)	No. of boys (f)	d = x - A	t = (x - A)/i	ft
30 - 40	35	10	35 - 65 = -30	-3	-30
40 - 50	45	12	45 - 65 = -20	-2	-24
50 - 60	55	14	55 - 65 = -10	-1	-14
60 - 70	65	12	65 - 65 = 0	0	0
70 - 80	75	9	75 - 65 = 10	1	9
80 - 90	85	7	85 - 65 = 20	2	14
90 - 100	95	6	95 - 65 = 30	3	18
Total		Σf = 70			Σft = -27

n = Σf = 70.

By formula,

Mean = A + $\frac{Σft}{n} \times i$

= 65 + $\frac{-27}{70} \times 10$

= 65 - $\frac{27}{7}$

= 65 - 3.86

= 61.14

Hence, mean = 61.14

Let assumed mean (A) be 87.5 and i = 7.

C.I.	Class mark (x)	Frequency (f)	d = x - A	t = (x - a)/i	ft
63 - 70	66.5	9	66.5 - 87.5 = -21	-3	-27
70 - 77	73.5	13	73.5 - 87.5 = -14	-2	-26
77 - 84	80.5	27	80.5 - 87.5 = -7	-1	-27
84 - 91	87.5	38	87.5 - 87.5 = 0	0	0
91 - 98	94.5	32	94.5 - 87.5 = 7	1	32
98 - 105	101.5	16	101.5 - 87.5 = 14	2	32
105 - 112	108.5	15	108.5 - 87.5 = 21	3	45
Total		Σf = 160			Σft = 29

n = Σf = 160.

By formula,

Mean = A + $\frac{Σft}{n} \times i$

= 87.5 + $\frac{29}{160} \times 7$

= 87.5 + $\frac{203}{160}$

= 87.5 + 1.3

= 88.8

Hence, mean = 88.8

Class interval	Class mark (x)	Frequency (f)	fx
0 - 10	5	8	40
10 - 20	15	22	330
20 - 30	25	31	775
30 - 40	35	f	35f
40 - 50	45	2	90
Total		Σf = 63 + f	Σfx = 1235 + 35f

By formula,

Mean = $\frac{Σfx}{Σf}$

$$\begin{aligned}\Rightarrow 21\frac{1}{7} = \frac{1235 + 35f}{63 + f} \\[1em] \Rightarrow \frac{148}{7} = \frac{1235 + 35f}{63 + f} \\[1em] \Rightarrow 148(63 + f) = 7(1235 + 35f) \\[1em] \Rightarrow 9324 + 148f = 8645 + 245f \\[1em] \Rightarrow 245f - 148f = 9324 - 8645 \\[1em] \Rightarrow 97f = 679 \\[1em] \Rightarrow f = \frac{679}{97} = 7.\end{aligned}$$

Hence, f = 7.

The table for the adjoining histogram is :

Class interval	Class mean (x)	Frequency (f)	fx
15 - 25	20	10	200
25 - 35	30	20	600
35 - 45	40	25	1000
45 - 55	50	15	750
55 - 65	60	5	300
Total		Σf = 75	Σfx = 2850

By formula,

Mean = $\frac{Σfx}{Σf} = \frac{2850}{75}$ = 38.

Hence, mean = 38.

By formula,

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

Class	Class mark (x)	Frequency (f)	fx
0 - 20	10	7	70
20 - 40	30	p	30p
40 - 60	50	10	500
60 - 80	70	9	630
80 - 100	90	13	1170
Total		Σf = 39 + p	2370 + 30p

By formula,

Mean = $\frac{Σfx}{Σf}$

⇒ 54 = $\frac{2370 + 30p}{39 + p}$

⇒ 54(39 + p) = 2370 + 30p

⇒ 2106 + 54p = 2370 + 30p

⇒ 54p - 30p = 2370 - 2106

⇒ 24p = 264

⇒ p = $\frac{264}{24}$

⇒ p = 11.

Hence, p = 11.

By formula,

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

Class	Class mark (x)	Frequency (f)	fx
0 - 20	10	5	50
20 - 40	30	f1	30 f1
40 - 60	50	10	500
60 - 80	70	f2	70f2
80 - 100	90	7	630
100 - 120	110	8	880
Total		Σf = f1 + f2 + 30	2060 + 30f1 + 70f2

Given,

Sum of frequencies = 50

⇒ f₁ + f₂ + 30 = 50

⇒ f₁ + f₂ = 20

⇒ f₁ = 20 - f₂ ........(1)

By formula,

Mean = $\frac{Σfx}{Σf}$

⇒ 62.8 = $\frac{2060 + 30f_1 + 70f_2}{50}$

⇒ 2060 + 30f₁ + 70f₂ = 3140

⇒ 30f₁ + 70f₂ = 1080

Substituting value of f₁ in above equation from (1), we get :

⇒ 30(20 - f₂) + 70f₂ = 1080

⇒ 600 - 30f₂ + 70f₂ = 1080

⇒ 40f₂ = 480

⇒ f₂ = $\frac{480}{40}$ = 12.

⇒ f₁ = 20 - f₂ = 20 - 12 = 8.

Hence, f₁ = 8 and f₂ = 12.

The above distribution is discontinuous, converting into continuous distribution, we get :

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

= $\frac{21 - 20}{2} = \frac{1}{2}$

= 0.5

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Let assumed mean (A) be 45.5

Marks (Classes before adjustment)	Marks (Classes after adjustment)	Class mean (x)	d = x - A	No. of students (frequency)	fd
11 - 20	10.5 - 20.5	15.5	-30	2	-60
21 - 30	20.5 - 30.5	25.5	-20	6	-120
31 - 40	30.5 - 40.5	35.5	-10	10	-100
41 - 50	40.5 - 50.5	45.5	0	12	0
51 - 60	50.5 - 60.5	55.5	10	9	90
61 - 70	60.5 - 70.5	65.5	20	7	140
71 - 80	70.5 - 80.5	75.5	30	4	120
Total				50	70

n = Σf = 50

Mean = A + $\frac{Σfd}{n}$

= $45.5 + \frac{70}{50}$

= 45.5 + 1.4

= 46.9

Hence, mean = 46.9

Arranging the numbers in ascending order, we get :

14, 15, 18, 21, 29.

The number of terms (n) are 5, which is odd.

∴ Median = $\frac{n + 1}{2}$ th term

Substituting value we get :

Median = $\frac{5 + 1}{2} = \frac{6}{2}$ = 3rd term = 18.

Hence, Option 2 is the correct option.

Arranging the numbers in ascending order, we get :

0, 2, 3, 4, 6, 8, 11, 16.

The number of terms (n) are 8, which is even.

∴ Median = $\frac{\Big(\frac{n}{2}\Big) \text{ th term} + \Big(\frac{n}{2} + 1\Big)\text{ th term}}{2}$

Substituting values we get :

$$\begin{aligned}\text{Median } = \frac{\Big(\frac{8}{2}\Big) \text{ th term} + \Big(\frac{8}{2} + 1\Big) \text{ th term}}{2} \\[1em] = \frac{\text{4 th term + 5 th term}}{2} \\[1em] = \frac{4 + 6}{2} \\[1em] = \frac{10}{2} \\[1em] = 5.\end{aligned}$$

Hence, Option 3 is the correct option.

Given, numbers :

5, 15, 20, x, 28, 30, 35

The number of terms (n) are 7, which is odd.

∴ Median = $\frac{n + 1}{2}$ th term

Substituting value we get :

$$\begin{aligned}\Rightarrow 23 = \frac{7 + 1}{2} \text{ th term} \\[1em] \Rightarrow 23 = \frac{8}{2} \text{ th term} \\[1em] \Rightarrow \text{4 th term} = 23 \\[1em] \Rightarrow x = 23.\end{aligned}$$

Hence, Option 4 is the correct option.

Numbers in ascending order : 10, 20, 30, 40, 50, 60, 70 and 80.

The number of terms (n) are 8, which is even.

∴ Lower quartile = $\frac{n}{4} = \frac{8}{4}$ = 2^nd term = 20.

∴ Upper quartile = $\frac{3n}{4} = \frac{3 \times 8}{4} = 3 \times 2$ = 6^th term = 60.

Inter-quartile = Upper quartile - Lower quartile = 60 - 20.

Hence, Option 3 is the correct option.

From figure,

Class 50-60 has the highest frequency.

Hence, Option 3 is the correct option.

Arranging the given data in ascending order :

0, 1, 3, 3, 4, 5, 6, 7, 8

Here, n = 9. Since, n is odd.

By formula,

Median = $\frac{n + 1}{2}$ th term.

Substituting values we get,

Median = $\frac{9 + 1}{2} = \frac{10}{2}$ = 5th term.

Here, 5th term = 4.

Hence, median = 4.

Arranging the given data in ascending order:

20.5, 21, 21, 22, 24, 24, 25.5, 27.5, 28, 28.5

Here, n = 10, which is even.

By formula,

Median = $\frac{\frac{n}{2} \text{ th term} + \Big(\frac{n}{2} + 1\Big) \text{ th term}}{2}$

$$\begin{aligned}= \frac{\frac{10}{2} \text{th term} + \Big(\frac{10}{2} + 1\Big)\text{ th term}}{2} \\[1em] = \frac{5\text{ th term} + 6\text{ th term}}{2} \\[1em] = \frac{24 + 24}{2} \\[1em] = \frac{48}{2} \\[1em] = 24.\end{aligned}$$

Hence, median = 24.

Arranging in ascending order:

22, 24, 25, 26, 26, 27, 27, 28, 28, 29, 21, 32, 32, 33, 35, 35, 36, 36, 37

(i) Here, n = 19, which is odd.

By formula,

Median = $\frac{n + 1}{2}$ th term.

= $\frac{19 + 1}{2} = \frac{20}{2}$ = 10th term

= 29.

Hence, median = 29.

(ii) Since, n = 19, which is odd.

By formula,

Lower quartile = $\frac{n + 1}{4}\text{ th term} = \frac{19 + 1}{4} = \frac{20}{4}$

= 5th term = 26.

Hence, lower quartile = 26.

(iii) Since, n = 19, which is odd.

By formula,

Upper quartile = $\frac{3(n + 1)}{4}\text{ th term} = \frac{3(19 + 1)}{4} = \frac{3 \times 20}{4}$

= 15th term = 35.

Hence, upper quartile = 35.

(iv) Inter quartile range = Upper quartile - Lower quartile

= 35 - 26

= 9.

Hence, inter quartile range = 9.

Cumulative frequency distribution table :

Weight (kg)	No. of boys (f)	Cumulative frequency
37	10	10
38	14	24 (10 + 14)
39	18	42 (24 + 18)
40	12	54 (42 + 12)
41	6	60 (6 + 54)

(i) Here, n = 60, which is even.

By formula,

Median = $\frac{\frac{n}{2} \text{ th term} + \Big(\frac{n}{2} + 1\Big) \text{ th term}}{2}$

Substituting values we get :

Median = $\frac{\frac{60}{2} \text{ th term} + \Big(\frac{60}{2} + 1\Big)\text{ th term}}{2} = \frac{\text{30th term + 31st term}}{2}$

From table,

The weight of each boy from 25th to 42nd is 39 kg.

∴ 30th term and 31st term = 39

Substituting value to get median :

Median = $\frac{39 + 39}{2} = \frac{78}{2}$ = 39 kg.

(ii) Here, n = 60, which is even.

By formula,

Lower quartile = $\Big(\frac{n}{4}\Big)$ th term

= $\Big(\frac{60}{4}\Big)$ = 15th term.

From table,

The weight of each boy from 11th to 24th term is 38 kg.

Hence, lower quartile = 38.

(iii) Here, n = 60, which is even.

By formula,

Upper quartile = $\Big(\frac{3n}{4}\Big)$ th term

= $\Big(\frac{3 \times 60}{4}\Big)$ = 45th term.

From table,

The weight of each boy from 43rd to 54th is 40 kg.

Hence, upper quartile = 40.

(iv) Inter quartile range = Upper quartile - Lower quartile

= 40 - 38

= 2.

Hence, inter-quartile range = 2.

Cumulative frequency distribution table :

Marks	Cumulative frequency
0 - 10	5
10 - 20	24
20 - 30	37
30 - 40	40
40 - 50	42
50 - 60	48
60 - 70	70
70 - 80	77
80 - 90	79
90 - 100	80

Here, n = 80, which is even.

By formula,

Median = $\frac{n}{2}$ th term

= $\frac{80}{2}$ = 40th term.

Lower quartile = $\frac{n}{4}$ th term

= $\frac{80}{4}$ = 20th term.

Upper quartile = $\frac{3n}{4}$ th term

= $\frac{3 \times 80}{4}$ = 60th term.

Steps of construction of ogive :

1. Take 1 cm = 10 units on x-axis.

2. Take 1 cm = 10 units on y-axis.

3. Plot the point (0, 0), as ogive always starts on x-axis representing the lower limit of the first class.

4. Plot the points (10, 5), (20, 24), (30, 37), (40, 40), (50, 42), (60, 48), (70, 70), (80, 77), (90, 79) and (100, 80).

5. Join the points by a free hand curve.

6. Draw a line parallel to x-axis from point L (frequency) = 40, touching the graph at point T. From point T draw a line parallel to y-axis touching x-axis at point M.

7. Draw a line parallel to x-axis from point N (frequency) = 20, touching the graph at point O. From point O draw a line parallel to y-axis touching x-axis at point P.

8. Draw a line parallel to x-axis from point Q (frequency) = 60, touching the graph at point R. From point R draw a line parallel to y-axis touching x-axis at point S.

(i) From graph, M = 40

Hence, median = 40.

(ii) From graph, P = 18

Hence, lower quartile = 18.

(iii) From graph, S = 66

Hence, upper quartile = 66.

The above distribution is discontinuous, converting into continuous distribution, we get :

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

= $\frac{131 - 130}{2} = \frac{1}{2}$

= 0.5

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Classes before adjustment	Classes after adjustment	No. of pupils	Cumulative frequency
121 - 130	120.5 - 130.5	12	12
131 - 140	130.5 - 140.5	16	28 (12 + 16)
141 - 150	140.5 - 150.5	30	58 (28 + 30)
151 - 160	150.5 - 160.5	20	78 (58 + 20)
161 - 170	160.5 - 170.5	14	92 (78 + 14)
171 - 180	170.5 - 180.5	8	100 (92 + 8)

Here, n = 100 which is even.

By formula,

Median = $\frac{n}{2}\text{ th term} = \frac{100}{2}$ = 50th term.

Steps of construction of ogive :

1. Since, the scale on x-axis starts at 120.5, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 120.5.

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

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

4. Plot the point (120.5, 0), as ogive always starts on x-axis representing the lower limit of the first class.

5. Plot the points (130.5, 12), (140.5, 28), (150.5, 58), (160.5, 78), (170.5, 92) and (180.5, 100).

6. Join the points by a free hand curve.

7. Draw a line parallel to x-axis from point A (frequency) = 50, 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 = 148 cm

Hence, median = 148 cm.

Steps :

1. Draw a histogram of the given distribution.

2. Inside the highest rectangle, which represents the maximum frequency (or modal class), draw two lines AC and BD diagonally from the upper corners C and D of adjacent rectangles.

3. Through point K (the point of intersection of diagonals AC and BD), draw KL perpendicular to the horizontal axis.

4. The value of point L on the horizontal axis represents the value of mode.

∴ Mode = 24.

Hence, mode = 24.

Steps :

1. Draw a histogram of the given distribution.

2. Inside the highest rectangle, which represents the maximum frequency (or modal class), draw two lines MJ and LO diagonally from the upper corners M and L of adjacent rectangles.

3. Through point Z (the point of intersection of diagonals MJ and LO), draw ZP perpendicular to the horizontal axis.

4. The value of point P on the horizontal axis represents the value of mode.

∴ Mode = 34.

Hence, mode = 34.

(i) From above data,

16 occurs for the maximum time.

Hence, mode = 16.

(ii) Arranging the numbers in ascending order :

7, 10, 12, 12, 14, 15, 16, 16, 16, 17, 19.

Here, n = 11, which is odd.

By formula,

Median = $\frac{n + 1}{2}$ th term

= $\frac{11 + 1}{2} = \frac{12}{2}$

= 6th term

= 15.

Hence, median = 15.

(iii) Total marks = 7 + 10 + 12 + 12 + 14 + 15 + 16 + 16 + 16 + 17 + 19

= 154.

Hence, total marks = 154.

(iv) Mean = $\frac{\text{Total marks}}{\text{No. of tests}}$

= $\frac{154}{11}$

= 14.

Hence, mean = 14.

(i) From above table,

The score 4 has the maximum frequency.

Hence, modal score = 4.

(ii) Cumulative frequency distribution table :

Score (x)	No. of shots (f)	Cumulative frequency	fx
0	0	0	0
1	3	3 (0 + 3)	3
2	6	9 (3 + 6)	12
3	4	13 (9 + 4)	12
4	7	20 (13 + 7)	28
5	5	25 (20 + 5)	25
Total	25	80

Here, n = 25, which is odd.

By formula,

Median = $\frac{n + 1}{2}$ th term = $\frac{25 + 1}{2} = \frac{26}{2}$ = 13th term

From table, score of 10th to 13th term is 3.

Hence, median = 3.

(iii) From cumulative frequency distribution table we get,

Total score = 80.

Hence, total score = 80.

(iv) By formula,

Mean = $\frac{Σfx}{Σf} = \frac{80}{25}$ = 3.2

Hence, mean = 3.2

Given,

A.P. = 2, 4, 6, 8, ……, 40

First term (a) = 2

Common difference (d) = 4 - 2 = 2

Last term (l) = 40

Let no. of terms in A.P. be n.

⇒ 40 = a + (n - 1)d

⇒ 40 = 2 + 2(n - 1)

⇒ 40 = 2 + 2n - 2

⇒ 2n = 40

⇒ n = $\frac{40}{2}$ = 20.

By formula,

Sum of first n terms of an A.P. = $\frac{n}{2}(a + l) = \frac{20}{2}(2 + 40) = 420$.

Mean = $\frac{\text{Sum of observations}}{\text{No. of observations}} = \frac{420}{20}$ = 21.

Solving,

$\Rightarrow \frac{40 + 2}{2} = \frac{42}{2}$ = 21.

Hence, option 1 is the correct option.

Numbers in ascending order :

8, 8, 9, 10, 12, 12, 12, 13, 15.

Number of terms (n) = 9

∴ Median = $\frac{n + 1}{2} = \frac{9 + 1}{2} = \frac{10}{2}$ = 5^th term = 12.

Hence, Option 1 is the correct option.

Numbers in ascending order :

10, 12, 14, 16, 17 and x.

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

Substituting values we get :

$$\begin{aligned}\text{Mean } = \frac{10 + 12 + 14 + 16 + 17 + x}{6} \\[1em] = \frac{69 + x}{6}.\end{aligned}$$

No. of terms (n) = 6, which is even.

Median = $\frac{\Big(\frac{n}{2}\Big) \text{ th term} + \Big(\frac{n}{2} + 1\Big) \text{ th term}}{2}$

Substituting values we get :

$$\begin{aligned}\text{Median } = \frac{\Big(\frac{6}{2}\Big) \text{ th term} + \Big(\frac{6}{2} + 1\Big) \text{ th term}}{2} \\[1em] = \frac{\text{3rd term + 4th term}}{2} \\[1em] = \frac{14 + 16}{2} \\[1em] = \frac{30}{2} \\[1em] = 15.\end{aligned}$$

Given,

Mean = Median

$$\begin{aligned}\therefore \frac{69 + x}{6} = 15 \\[1em] \Rightarrow 69 + x = 15 \times 6 \\[1em] \Rightarrow 69 + x = 90 \\[1em] \Rightarrow x = 90 - 69 = 21.\end{aligned}$$

Hence, Option 4 is the correct option.

Prime numbers : 2, 3, 5, 7, 11, 13.

No. of terms (n) = 6, which is even.

Median = $\frac{\Big(\frac{n}{2}\Big) \text{ th term} + \Big(\frac{n}{2} + 1\Big) \text{ th term}}{2}$

Substituting values we get :

$$\begin{aligned}\text{Median } = \frac{\Big(\frac{6}{2}\Big) \text{ th term} + \Big(\frac{6}{2} + 1\Big) \text{ th term}}{2} \\[1em] = \frac{\text{3rd term + 4th term}}{2} \\[1em] = \frac{5 + 7}{2} \\[1em] = \frac{12}{2} \\[1em] = 6.\end{aligned}$$

Hence, Option 3 is the correct option.

No. of terms (N) = 80

Lower quartile (Q₁) = $\frac{N}{4}$ th term

= $\frac{80}{4}$ = 20 th term = 10.

Upper quartile (Q₃) = $\frac{3N}{4}$ th term

= $\frac{3 \times 80}{4} = \frac{240}{4}$ = 60 th term.

1. From point A = 60. Draw a line parallel to x-axis touching the graph at B.

2. From point B, draw a line parallel to y-axis touching graph at C.

From graph, point C = 52.

∴ Upper quartile (Q₃) = 52

By formula,

Inter quartile range = Upper quartile - Lower quartile

= Q₃ - Q₁

= 52 - 10 = 42.

Hence, Option 1 is the correct option.

Given, the mean age of 9 boys is 28 years.

when one new boy joins, the mean age increases by 1 year, making the new mean 29 years

By formula,

Mean = $\frac{\text{Sum of all observations}}{\text{Number of all observations}}$

The mean age of 9 boys is 28 years.

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{\text{Sum of ages of 9 boys}}{\text{Number of boys}}\\[1em] \Rightarrow 28 = \frac{\text{Sum of ages of 9 boys}}{9}\\[1em] \Rightarrow \text{Sum of ages of the boys} = 28 \times 9\\[1em]\end{aligned}$$

After the new boy joins, the mean age becomes 29 years for 10 boys.

$$\begin{aligned}\Rightarrow 29 = \frac{\text{Sum of ages of 10 boys}}{10}\\[1em] \Rightarrow \text{Sum of ages of 10 boys} = 29 \times 10 \\[1em]\end{aligned}$$

The age of the new boy is the difference between the total age of 10 boys and the total age of 9 boys = 29 x 10 - 28 x 9

∴ A is true, R is false.

Hence, option 1 is the correct option.

Given, data = 37, 41, 56, 62, 70, 74, 81, 89, 95 and 90.

Arrange the data in ascending order: 37, 41, 56, 62, 70, 74, 81, 89, 90, 95

Number of data = 10

If the number of data points (n) is odd, the median is the $\Big(\frac{n+1}{2}\Big)^{th}$ term

If the number of data points (n) is even, the median is the $\Big(\frac{\Big(\frac{n}{2}\Big)^{th} + \Big(\frac{n}{2} + 1\Big)^{th}}{2}\Big)$

Here, n = 10

$$\begin{aligned}\text{Median }= \Big(\frac{\Big(\frac{10}{2}\Big)^{th} + \Big(\frac{10}{2} + 1\Big)^{th}}{2}\Big)\\[1em] = \Big(\frac{5^{th} + (5 + 1)^{th}}{2}\Big)\\[1em] = \Big(\frac{5^{th} + 6^{th}}{2}\Big)\\[1em] = \Big(\frac{70 + 74}{2}\Big)\\[1em] = \Big(\frac{144}{2}\Big)\\[1em] = 72.\end{aligned}$$

∴ Both A and R are true and R is incorrect reason for A.

Hence, option 4 is the correct option.

Cumulative frequency represents the running total of frequencies up to a certain class interval. For instance, the cumulative frequency for the class interval 10–20 includes all frequencies from the previous intervals as well. Therefore, the cumulative frequency for the 10–20 interval is the sum of the frequencies for the 0–10 and 10–20 intervals.

The cumulative frequency for the 10–20 interval is the sum of the frequencies for the 0–10 and 10–20 intervals:

a = 15 + 25 = 40

The cumulative frequency for the 20–30 interval is 50, which includes all previous frequencies. Therefore, the frequency for the 20–30 interval is:

b = 50 - a = 50 - 40 = 10.

∴ A is true, R is false.

Hence, option 1 is the correct option.

Given data : 9, 11, 15, 19, 17, 13 and 7

Arrange the data in ascending order : 7, 9, 11, 13, 15, 17, 19.

For data with n terms, the lower quantile is $\Big(\frac{n + 1}{4}\Big)^{th}$ term, if n is odd

Here, n = 7

$$\begin{aligned}\text{The lower quantile } = \Big(\frac{n + 1}{4}\Big)^{th} \\[1em] = \Big(\frac{7 + 1}{4}\Big)^{th} \\[1em] = \Big(\frac{8}{4}\Big)^{th} \\[1em] = 2^{\text{nd term}} \\[1em] = 9.\end{aligned}$$

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Given:

C.I.	f	x(Midpoint) = (Lower limit + upper limit)/2
0-20	20	10
20-40	x	30
40-60	10	50

By formula; Mean = $\frac{∑f.x}{∑f}$

Substituting the values, we get

$$\begin{aligned}\Rightarrow 26 = \frac{20 \times 10 + x \times 30 + 10 \times 50}{x + 30}\\[1em] \text{So, statement 2 is true.}\\[1em] \Rightarrow 26 = \frac{200 + 30x + 500}{30 + x}\\[1em] \Rightarrow 26(30 + x) = 200 + 30x + 500\\[1em] \Rightarrow 780 + 26x = 700 + 30x\\[1em] \Rightarrow 30x - 26x = 780 - 700\\[1em] \Rightarrow 4x = 80 \\[1em] \Rightarrow x = \frac{80}{4} \\[1em] \Rightarrow x = 20\end{aligned}$$

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Given, mean = 14 and median = 15

By formula,

Mode = 3 Median - 2 Mean

So, statement 2 is true.

Substituting the values, we get :

⇒ Mode = 3 x 15 - 2 x 14

⇒ Mode = 45 - 28

⇒ Mode = 17.

So, statement 1 is true.

∴ Both the statement are true.

Hence, option 1 is the correct option.

Given,

Mean of 1, 7, 5, 3, 4 and 4 is m.

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

Mean (m) = $\frac{\text{Sum of observations}}{\text{No. of observations}} = \frac{24}{6}$ = 4.

Given,

Numbers 3, 2, 4, 2, 3, 3 and p have mean m - 1 or mean = 3.

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

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

⇒ 3 = $\frac{17 + p}{7}$

⇒ 21 = 17 + p

⇒ p = 4.

Observations in ascending order are = 2, 2, 3, 3, 3, 4, 4.

Here, n = 7, which is odd.

By formula,

Median = $\frac{n + 1}{2}$ th term

= $\frac{7 + 1}{2} = \frac{8}{2}$ = 4th term = 3.

∴ q = 3.

Hence, p = 4 and q = 3.

Cumulative frequency distribution table :

Date (July)	Number (frequency)	Cumulative frequency
1	5	5
2	12	17 (5 + 12)
3	20	37 (17 + 20)
4	27	64 (37 + 27)
5	46	110 (64 + 46)
6	30	140 (110 + 30)
7	31	171 (140 + 31)
8	18	189 (171 + 18)
9	11	200 (189 + 11)
10	5	205 (200 + 5)
11	0	205 (205 + 0)
12	1	206 (205 + 1)

Here, n = 206, which is even

Lower quartile = $\frac{n}{4}$ th term

= $\frac{206}{4}$ = 51.5 th term

From table,

It is observed the date of 38th term to 64th term is 4th july.

Upper quartile = $\frac{3n}{4}$ th term

= $\frac{3 \times 206}{4}$ = 154.5 th term

From table,

It is observed the date of 141st term to 171st term is 7th july.

From table,

5th july has the highest no. of cases diagnosed.

Hence, mode = 5th july, upper quartile = 7th july and lower quartile = 4th july.

Cumulative frequency distribution table :

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

(i) Steps of construction of ogive :

1. Take 1 cm = 10 marks on x-axis.

2. Take 1 cm = 20 students on y-axis.

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

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).

5. Join the points by a free hand curve.

6. Draw a line parallel to x-axis from point A (no. of students) = 60, 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 = 43

Hence, median = 43.

(ii) Total marks = 100.

75% of 100 marks = $\frac{75}{100} \times 100$ = 75.

Draw a line parallel to y-axis from point D (marks) = 75, 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 = 110.

It means that 110 students score either less or equal to 75% marks.

No. of students left = 120 - 110 = 10.

Hence, no. of students scoring more than 75% marks = 10.

(iii) Total marks = 100.

40% of 100 marks = $\frac{40}{100} \times 100$ = 40.

Draw a line parallel to y-axis from point G (marks) = 40, touching the graph at point H. From point H draw a line parallel to x-axis touching y-axis at point I.

From graph, I = 52.

Hence, no. of failed students = 52.

(iv) Here, n = 120, which is even.

By formula,

Lower quartile = $\frac{n}{4} = \frac{120}{4}$ = 30th term.

Draw a line parallel to x-axis from point J (no. 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 = 30

Hence, lower quartile = 30.

(i) Cumulative frequency distribution table :

Weight	Frequency	Cumulative frequency
40 - 45	5	5
45 - 50	17	22 (5 + 17)
50 - 55	22	44 (22 + 22)
55 - 60	45	89 (44 + 45)
60 - 65	51	140 (89 + 51)
65 - 70	31	171 (140 + 31)
70 - 75	20	191 (171 + 20)
75 - 80	9	200 (191 + 9)

Steps of construction :

1. Since, the scale on x-axis starts at 40, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 40.

2. Take 1 cm along x-axis = 5 kg.

3. Take 1 cm along y-axis = 20 units.

4. Plot the point (40, 0) as ogive starts from x-axis representing lower limit of first class.

5. Plot the points (45, 5), (50, 22), (55, 44), (60, 89), (65, 140), (70, 171), (75, 191) and (80, 200).

6. Join the points by a free hand curve.

7. Draw a line parallel to y-axis from point J (weight) = 55, touching the graph at point Q. From point Q draw a line parallel to x-axis touching y-axis at point K.

From graph, K = 44.

Hence, 44 students weight 55 kg or less.

Students weighing more than 55 kg = 200 - 44 = 156.

Percentage of students weighing more than 55 kg = $\frac{156}{200} \times 100$ = 78%.

Hence, percentage of students weighing more than 55 kg = 78%.

(ii) 30% of students = $\frac{30}{100} \times 200$ = 60.

Total students = 200

No. of Students not in heaviest 30% = 200 - 60 = 140.

Draw a line parallel to x-axis from point O (no. of students) = 140, touching the graph at point R. From point R draw a line parallel to y-axis touching x-axis at point P.

From graph, P = 65

Hence, above 65 kg the heaviest 30% of the students fall.

(iii) Draw a line parallel to y-axis from point L (weight) = 55.70 kg, touching the graph at point M. From point M draw a line parallel to x-axis touching y-axis at point N.

(a) From graph,

N = 46.

∴ 46 students have weight less than 55.70 kg

Hence, 46 students are underweight.

(b) Since, 46 students have weight less than 55.70 kg

∴ 154 (200 - 46) students have weight more than 55.70 kg

Hence, 154 students are overweight.

Cumulative frequency distribution table :

Marks obtained (x)	No. of students (f)	Cumulative frequency	fx
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	Σf = 25		Σfx = 171

By formula,

Mean = $\frac{Σfx}{Σf} = \frac{171}{25}$ = 6.84

Here, n = 25, which is odd.

Median = $\frac{n + 1}{2}$ th term

= $\frac{25 + 1}{2} = \frac{26}{2}$ = 13th term.

From table,

Marks obtained by 13th to 18th student = 7.

Median = 7.

From table,

6 marks has highest frequency.

Mode = 6.

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

(i) Cumulative frequency distribution table :

Monthly income	No. of employees	Cumulative frequency
6 - 7	20	20
7 - 8	45	65 (20 + 45)
8 - 9	65	130 (65 + 65)
9 - 10	95	225 (130 + 95)
10 - 11	60	285 (225 + 60)
11 - 12	30	315 (285 + 30)
12 - 13	5	320 (315 + 5)

Here, n = 320, which is even.

By formula,

Median = $\frac{n}{2}$ th term

= $\frac{320}{2}$ = 160th term.

Steps of construction :

1. Since, the scale on x-axis starts at 6, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 6.

2. Take 2 cm along x-axis = 1 thousand rupees.

3. Take 1 cm along y-axis = 40 employees.

4. Plot the point (6, 0) as ogive starts from x-axis representing lower limit of first class.

5. Plot the points (7, 20), (8, 65), (9, 130), (10, 225), (11, 285), (12, 315) and (13, 320).

6. Join the points by a free hand curve.

7. Draw a line parallel to x-axis from point M (no. of employees) = 160, touching the graph at point N. From point N draw a line parallel to y-axis touching x-axis at point O.

From graph, O = 9.2 (thousands)

Hence, median wage = ₹ 9200.

(ii) Draw a line parallel to y-axis from point P (income) = ₹ 8.5 (thousands), touching the graph at point Q. From point Q draw a line parallel to x-axis touching y-axis at point R.

From graph, R = 95.

Hence, 95 employees have income less than ₹ 8500.

Sum of observations = 45 + 52 + 60 + x + 69 + 70 + 26 + 81 + 94 = 497 + x

No. of observations (n) = 9

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

68 = $\frac{497 + x}{9}$

612 = 497 + x

x = 612 - 497 = 115.

Here, n = 9, which is odd.

Median = $\frac{n + 1}{2}$ th term

= $\frac{9 + 1}{2} = \frac{10}{2}$

= 5th term

= 69.

Hence, mean = 115 and median = 69.

Here, n = 10, which is even.

By formula,

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

Given,

Median = 48

⇒ x + 2 = 48

⇒ x = 46.

Set of observations : 13, 35, 42, 46, 46, 50, 55, 61, 71, 80.

Here, 46 has the maximum frequency.

∴ Mode = 46.

Hence, x = 46 and mode = 46.

(i) Frequency distribution table :

Marks (Class)	No. of students (frequency)
0 - 10	2
10 - 20	5
20 - 30	8
30 - 40	4
40 - 50	6

(ii) Mean

Marks (Class)	No. of students (frequency)	Class mean (x)	fx
0 - 10	2	5	10
10 - 20	5	15	75
20 - 30	8	25	200
30 - 40	4	35	140
40 - 50	6	45	270
Total	Σf = 25		Σfx = 695

By formula,

Mean = $\frac{Σfx}{Σf}$

= $\frac{695}{25}$ = 27.8

Hence, mean = 27.8

(iii) From table,

Class 20 - 30 has the highest frequency.

Hence, modal class = 20 - 30.

(a) Since the difference between two consecutive class-marks is 10, subtract $\frac{10}{2}$ = 5 from each class-mark to get the lower limit and add 5 to get the upper limit.

Class-mark (x)	Class interval	Frequency (f)	fx	Cumulative frequency
15	10 - 20	10	150	10
25	20 - 30	12	300	22
35	30 - 40	14	490	36
45	40 - 50	16	720	52
55	50 - 60	8	440	60
Total	Σf = 60	Σfx = 2100

(i) By formula,

Mean = $\frac{Σfx}{Σf} = \frac{2100}{60}$ = 35.

Hence, mean = 35.

(ii) Calculating mode :

Steps of construction :

1. Draw a histogram of the given distribution.

2. Inside the highest rectangle, which represents the maximum frequency (or modal class) draw two lines AC and BD diagonally from the upper corners C and D of adjacent rectangles.

3. Through the point K (the point of intersection of diagonals AC and BD), draw KM perpendicular to the horizontal axis.

4. The value of point M on the horizontal axis represents the value of mode.

∴ Mode = 42 (approx).

Hence, mode = 42 (approx).

(iii) n = Σf = 60

Median = $\Big(\frac{n}{2}\Big)^{th}$ term = $\frac{60}{2}$ = 30th term.

Steps of construction of ogive :

1. Take 1 cm = 10 units along x-axis.

2. Take 1 cm = 10 units along y-axis.

3. Ogive always starts from a point on x-axis representing the lower limit of the first class. Mark point (10, 0).

4. Take upper class limits along x-axis and corresponding cumulative frequencies along y-axis, mark the points (20, 10), (30, 22), (40, 36), (50, 52) and (60, 60).

5. Join the points marked by a free hand curve.

6. Mark 30 on y-axis, draw a horizontal line which meets the curve at point A.

7. Through point A, on the curve, draw a vertical line which meets x-axis at point B.

From graph,

B = 36 (approx)

Hence, median = 36 (approx).

(b) Since the difference between two consecutive class-marks is 5, subtract $\frac{5}{2}$ = 2.5 from each class-mark to get the lower limit and add 2.5 to get the upper limit.

Class-mark (x)	Class interval	Frequency (f)	fx	Cumulative frequency
35	32.5 - 37.5	8	280	8
40	37.5 - 42.5	10	400	18
45	42.5 - 47.5	12	540	30
50	47.5 - 52.5	15	750	45
55	52.5 - 57.5	20	1100	65
60	57.5 - 62.5	15	900	80
Total	Σf = 80	Σfx = 3970

(i) By formula,

Mean = $\frac{Σfx}{Σf} = \frac{3970}{80}$ = 49.625

Hence, mean = 49.625

(ii) Calculating mode :

Steps of construction :

1. Draw a histogram of the given distribution.

2. Inside the highest rectangle, which represents the maximum frequency (or modal class) draw two lines AC and BD diagonally from the upper corners C and D of adjacent rectangles.

3. Through the point K (the point of intersection of diagonals AC and BD), draw KM perpendicular to the horizontal axis.

4. The value of point M on the horizontal axis represents the value of mode.

∴ Mode = 55 (approx).

Hence, mode = 55 (approx).

(iii) n = Σf = 80

Median = $\Big(\frac{n}{2}\Big)^{th}$ term = $\frac{80}{2}$ = 40th term.

Steps of construction of ogive :

1. Take 1 cm = 5 units along x-axis.

2. Take 1 cm = 10 units along y-axis.

3. Ogive always starts from a point on x-axis representing the lower limit of the first class. Mark point (32.5, 0).

4. Take upper class limits along x-axis and corresponding cumulative frequencies along y-axis, mark the points (37.5, 8), (42.5, 18), (47.5, 30), (52.5, 45), (57.5, 65) and (62.5, 80).

5. Join the points marked by a free hand curve.

6. Mark 40 on y-axis, draw a horizontal line which meets the curve at point A.

7. Through point A, on the curve, draw a vertical line which meets x-axis at point B.

From graph,

B = 49.5 (approx)

Hence, median = 49.5 (approx).

(i) The average age for which maximum cases occur is the mode of the distribution.

The maximum frequency is 25, so the modal class is 30 - 40.

Here, l = 30, f₁ = 25, f₀ = 23, f₂ = 14 and h = 10.

By formula,

$$\begin{aligned}\text{Mode} = l + \frac{f_1 - f_0}{2f_1 - f_0 - f_2} \times h \\[1em] = 30 + \frac{25 - 23}{2(25) - 23 - 14} \times 10 \\[1em] = 30 + \frac{2}{50 - 37} \times 10 \\[1em] = 30 + \frac{2}{13} \times 10 \\[1em] = 30 + \frac{20}{13} = 30 + 1.54 = 31.54.\end{aligned}$$

Hence, the average age for which maximum cases occur is 31.54 years (approx.).

(ii) The modal class is 30 - 40.

Hence, the upper limit of the modal class is 40.

(iii) Mean of the data :

Age (in years)	Class-mark (x)	No. of patients (f)	fx
10 - 20	15	13	195
20 - 30	25	23	575
30 - 40	35	25	875
40 - 50	45	14	630
50 - 60	55	5	275
Total		Σf = 80	Σfx = 2550

By formula,

Mean = $\frac{Σfx}{Σf} = \frac{2550}{80}$ = 31.875 ≈ 31.88.

Hence, the mean of the given data is 31.88 years (approx.).

(i) Mean monthly consumption :

Monthly consumption (in units)	Class-mark (x)	No. of families (f)	fx
0 - 100	50	160	8000
100 - 200	150	120	18000
200 - 300	250	180	45000
300 - 400	350	60	21000
400 - 500	450	40	18000
Total		Σf = 560	Σfx = 110000

By formula,

Mean = $\frac{Σfx}{Σf} = \frac{110000}{560}$ = 196.43 (approx.).

Hence, the mean monthly consumption is 196.43 units (approx.).

(ii) The maximum frequency is 180, which corresponds to the class 200 - 300.

Hence, the modal monthly class is 200 - 300.