Median, Quartiles & Mode

(i) By arranging data in ascending order, we get:

6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25

Number of observations, n = 11, which is odd.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\text{n} + 1}{2} \text{ th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \frac{11 + 1}{2} \text{ th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \frac{12}{2} \text{ th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 6 \text{ th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 16.\end{aligned}$$

Hence, median = 16.

(ii) By arranging data in ascending order, we get:

0, 13, 15, 18, 32, 35, 37, 39, 41, 51, 56

Number of observations, n = 11, which is odd.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\text{n} + 1}{2} \text{ th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \frac{11 + 1}{2} \text{ th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \frac{12}{2} \text{ th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 6 \text{ th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 35\end{aligned}$$

Hence, median = 35.

(iii) By arranging data in ascending order, we get:

25, 27, 28, 31, 35, 36, 38, 40

Number of observations, n = 8, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\Big(\frac{\text{n}}{2}\Big) \text{th} \text{ term} + \Big(\frac{\text{n}}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\Big(\frac{8}{2}\Big) \text{th} \text{ term} + \Big(\frac{8}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{4\text{th} \text{ term} + (4 + 1)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{4\text{ th} \text{ term} + 5\text{ th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{31 + 35}{2} \\[1em] \Rightarrow \text{Median} = \frac{66}{2} \\[1em] \Rightarrow \text{Median} = 33\end{aligned}$$

Hence, median = 33.

(iv) By arranging data in ascending order, we get:

35, 42, 51, 56, 66, 69, 72, 81, 85, 92

Number of observations, n = 10, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\Big(\frac{\text{n}}{2}\Big) \text{th} \text{ term} + \Big(\frac{\text{n}}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\Big(\frac{10}{2}\Big) \text{th} \text{ term} + \Big(\frac{10}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{5\text{th} \text{ term} + (5 + 1)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{5\text{th} \text{ term} + 6\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{66 + 69}{2} \\[1em] \Rightarrow \text{Median} = \frac{135}{2} \\[1em] \Rightarrow \text{Median} = 67.5\end{aligned}$$

Hence, median = 67.5.

(v) By arranging data in ascending order, we get:

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

Number of observations, n = 12, which is even.

By formula,

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

Hence, median = 16.

By arranging data in ascending order, we get:

17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40

Number of observations, n = 15, which is odd.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\text{n} + 1}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \frac{15 + 1}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \frac{16}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 8 \text{ th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 23\end{aligned}$$

Hence, median score = 23.

By arranging data in ascending order, we get:

143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1

Number of observations, n = 9, which is odd.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\text{n} + 1}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \frac{9 + 1}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \frac{10}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 5 \text{ th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 147.3\end{aligned}$$

Hence, median height = 147.3 cm.

By arranging data in ascending order, we get:

9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2

Number of observations, n = 8, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\Big(\frac{\text{n}}{2}\Big) \text{th} \text{ term} + \Big(\frac{\text{n}}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\Big(\frac{8}{2}\Big) \text{th} \text{ term} + \Big(\frac{8}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{4\text{th} \text{ term} + (4 + 1)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{4\text{ th} \text{ term} + 5\text{ th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{13.4 + 14.3}{2} \\[1em] \Rightarrow \text{Median} = \frac{27.7}{2} \\[1em] \Rightarrow \text{Median} = 13.85\end{aligned}$$

Hence, median weight = 13.85 kg.

(i) Set of numbers are arranged in ascending order,

11, 12, 14, 18, (x + 4), 30, 32, 35, 41

Given,

Median = 24.

Here,

Number of observations (n) = 9, which is odd.

$$\begin{aligned}\Rightarrow \text{Median} = \frac{9 + 1}{2} \text{th} \text{ observation} \\[1em] \Rightarrow 24 = \frac{10}{2} \text{th} \text{ observation} \\[1em] \Rightarrow 24 = 5 \text{ th} \text{ observation} \\[1em] \Rightarrow 24 = \text{x} + 4 \\[1em] \Rightarrow \text{x} = 24 - 4 \\[1em] \Rightarrow \text{x} = 20\end{aligned}$$

Hence, the value of x = 20.

(ii) Set of numbers are arranged in ascending order,

10, 13, 15, 18, x + 1, x + 3, 31, 36, 38, 42

Given,

Median = 28.

Here,

Number of observations (n) = 10, which is even.

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\Big(\frac{\text{n}}{2}\Big) \text{th} \text{ term} + \Big(\frac{\text{n}}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow 28 = \frac{\Big(\frac{10}{2}\Big) \text{th} \text{ term} + \Big(\frac{10}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow 28 = \frac{5 \text{th} \text{ term} + (5 + 1)\text{th} \text{ term}}{2} \\[1em] \Rightarrow 28 = \frac{5 \text{th} \text{ term} + 6\text{th} \text{ term}}{2} \\[1em] \Rightarrow 28 \times 2 = (\text{x} + 1) + (\text{x} + 3) \\[1em] \Rightarrow 56 = 4 + 2\text{x} \\[1em] \Rightarrow 2\text{x} = 56 - 4 \\[1em] \Rightarrow 2\text{x} = 52 \\[1em] \Rightarrow \text{x} = \frac{52}{2} \\[1em] \Rightarrow \text{x} = 26.\end{aligned}$$

Hence, the value of x = 26.

Cumulative frequency distribution table :

Weight (in nearest kg)	Number of students	Cumulative frequency
45	8	8
46	5	13 (8 + 5)
48	6	19 (13 + 6)
50	9	28 (19 + 9)
52	7	35 (28 + 7)
54	4	39 (35 + 4)
55	2	41 (39 + 2)

Here number of observations, n = 41, which is odd.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{41 + 1}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \frac{42}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 21 \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \text{Weight of 21 st student}\end{aligned}$$

From the above table, weight of each student from 20th to 28th are 50.

∴ Weight of 21st student = 50.

Hence, median weight = 50 kg.

The given varieties are arranged in ascending order.

Cumulative frequency distribution table :

Variate	Frequency	Cumulative frequency
15	3	3
17	5	8 (3 + 5)
20	9	17 (8 + 9)
22	4	21 (17 + 4)
25	6	27 (21 + 6)
30	10	37 (27 + 10)

Here number of observations, n = 37, which is odd.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{37 + 1}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \frac{38}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 19 \text{ th} \text{ observation}\end{aligned}$$

From the above table, variate corresponding to a cumulative frequency from 18th to 21st are 22.

∴ 19th observation = 22.

Hence, median = 22.

Diameter of heart (in mm) (x)	Number of patients (f)	fx	Cumulative frequency
120	5	600	5
121	8	968	13 (5 + 8)
122	12	1464	25 (13 + 12)
123	9	1107	34 (25 + 9)
124	6	744	40 (34 + 6)
125	10	1250	50 (40 + 10)
Total	Σf = 50	Σfx = 6133

(i) We know that,

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{\text{Σfx}}{\text{Σf}} \\[1em] \Rightarrow \text{Mean} = \frac{6133}{50} \\[1em] \Rightarrow \text{Mean} = 122.66 \text{ mm}.\end{aligned}$$

Hence, mean diameter of heart = 122.66 mm.

(ii) Here,

Number of observations (n) = 50, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\Big(\frac{\text{n}}{2}\Big) \text{th} \text{ term} + \Big(\frac{\text{n}}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\Big(\frac{50}{2}\Big) \text{th} \text{ term} + \Big(\frac{50}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{25 \text{th} \text{ term} + (25 + 1)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{25 \text{ th} \text{ term} + 26\text{ th} \text{ term}}{2}\end{aligned}$$

From table,

Diameter of heart corresponding to 25th term is 122

Diameter of heart corresponding to 26th term is 123

$$\begin{aligned}\Rightarrow \text{Median} = \frac{25 \text{th} \text{ term} + 26\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{122 + 123}{2} \\[1em] \Rightarrow \text{Median} = \frac{245}{2} \\[1em] \Rightarrow \text{Median} = 122.5 \text{ mm}.\end{aligned}$$

Hence, median diameter of heart = 122.5 mm.

By arranging data in ascending order, we get:

0, 5, 6, 8, 14, 16, 17 17, 19, 20, 20, 25, 25, 27, 30

Number of observations, n = 15, which is odd.

(i) By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\text{n} + 1}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \frac{15 + 1}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \frac{16}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 8 \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 17\end{aligned}$$

Hence, median = 17.

(ii) By formula,

Lower Quartile = $\Big(\frac{\text{n} + 1}{4}\Big)$ th term

= $\Big(\frac{15 + 1}{4}\Big) = \frac{16}{4}$ th term

= 4th term

= 8

Hence, lower quartile (Q₁) = 8.

(iii) By formula,

Upper Quartile (Q₃) = $\Big(\frac{3(\text{n} + 1)}{4}\Big)$ th term

= $\Big(\frac{3 \times (15 + 1)}{4}\Big)$ th term

= $\Big(\frac{3 \times 16}{4}\Big) = \frac{48}{4}$ th term

= 12th term

= 25.

Hence, Upper Quartile (Q₃) = 25.

(iv) By formula,

Inter quartile range = Upper quartile - Lower quartile

= 25 - 8

= 17

Hence, the inter-quartile range is 17.

(v) By formula,

Semi-interquartile range = $\frac{1}{2}$ × Inter quartile range

= $\frac{1}{2} \times 17$

= 8.5

Hence, semi-interquartile range = 8.5.

By arranging data in ascending order, we get:

0, 2, 5, 8, 9, 13, 14, 16, 18, 19, 23, 26

Number of observations, n = 12, which is even.

(i) By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\Big(\frac{\text{n}}{2}\Big) \text{th} \text{ term} + \Big(\frac{\text{n}}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\Big(\frac{12}{2}\Big) \text{th} \text{ term} + \Big(\frac{12}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{6 \text{th} \text{ term} + (6 + 1)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{6 \text{th} \text{ term} + 7\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{13 + 14}{2} \\[1em] \Rightarrow \text{Median} = \frac{27}{2} \\[1em] \Rightarrow \text{Median} = 13.5\end{aligned}$$

Hence, Median = 13.5.

(ii) By formula,

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

= $\Big(\frac{12}{4}\Big)$ th term

= 3 rd term

= 5.

Hence, lower quartile (Q₁) = 5.

(iii) By formula,

Upper Quartile (Q₃) = $\Big(\frac{3\text{n}}{4}\Big)$ th term

= $\Big(\frac{3 \times 12}{4}\Big)$ th term

= $\Big(\frac{36}{4}\Big)$ th term

= 9 th term

= 18.

Hence, Upper Quartile (Q₃) = 18.

(iv) By formula,

Inter quartile range = Upper quartile - Lower quartile

= 18 - 5

= 13.

Hence, the inter-quartile range is 13.

(v) By formula,

Semi-interquartile range = $\frac{1}{2}$ × Inter quartile range

= $\frac{1}{2} \times 13$

= 6.5

Hence, semi-interquartile range = 6.5.

The given varieties are arranged in ascending order.

Cumulative frequency distribution table :

Variate	Frequency	Cumulative frequency
13	6	6
15	4	10 (6 + 4)
18	11	21 (10 + 11)
20	9	30 (21 + 9)
22	16	46 (30 + 16)
24	12	58 (46 + 12)
25	2	60 (58 + 2)

Here number of observations, n = 60, which is even.

(i) By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\Big(\frac{\text{n}}{2}\Big) \text{th} \text{ term} + \Big(\frac{\text{n}}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\Big(\frac{60}{2}\Big) \text{th} \text{ term} + \Big(\frac{60}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{30 \text{th} \text{ term} + (30 + 1)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{30 \text{th} \text{ term} + 31\text{th} \text{ term}}{2}\end{aligned}$$

From table,

30th term is 20

31st term is 22 (All observations from 31st to 46th term = 22)

$$\begin{aligned}\Rightarrow \text{Median} = \frac{30 \text{th} \text{ term} + 31 \text{st} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{20 + 22}{2} \\[1em] \Rightarrow \text{Median} = \frac{42}{2} \\[1em] \Rightarrow \text{Median} = 21\end{aligned}$$

Hence, median = 21.

(ii) By formula,

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

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

= 15 th term

= 18.

Hence, lower quartile = 18.

(iii) By formula,

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

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

= $\Big(\frac{180}{4}\Big)$ th term

= 45 th term

= 22

Hence, Upper Quartile = 22.

(iv) By formula,

Semi-interquartile range = $\frac{1}{2}$ × (Upper quartile - Lower quartile)

= $\frac{1}{2} \times (22 - 18) = \frac{1}{2} \times 4$

= 2

Hence, semi-interquartile range = 2.

The given varieties are arranged in ascending order.

Cumulative frequency distribution table :

Height (in cm)	Number of students	Cumulative frequency
150	9	9
151	12	21 (9 + 12)
152	10	31 (21 + 10)
153	8	39 (31 + 8)
154	11	50 (39 + 11)
155	7	57 (50 + 7)
156	6	63 (57 + 6)

Here number of observations, n = 63, which is odd.

(i) By formula,

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

From table,

32 nd term is 153.

Hence, median = 153.

(ii) By formula,

Lower Quartile = $\Big(\frac{\text{n} + 1}{4}\Big)$ th term

= $\Big(\frac{63 + 1}{4}\Big) = \frac{64}{4}$ th term

= 16 th term

From table,

16 th term is 151.

Hence, lower quartile = 151.

(iii) By formula,

Upper Quartile = $\Big(\frac{3(\text{n} + 1)}{4}\Big)$ th term

= $\Big(\frac{3 \times (63 + 1)}{4}\Big)$ th term

= $\Big(\frac{3 \times 64}{4}\Big) = \frac{192}{4}$ th term

= 48th term

From table,

48 th term is 154.

Hence, Upper Quartile = 154.

(iv) By formula,

Inter quartile range = Upper quartile - Lower quartile

= 154 - 151

= 3.

Hence, the inter-quartile range is 3.

The given varieties are arranged in ascending order.

Cumulative frequency distribution table :

Variate	Frequency	Cumulative frequency
16	9	9
18	8	17 (9 + 8)
20	13	30 (17 + 13)
23	4	34 (30 + 4)
25	4	38 (34 + 4)
26	6	44 (38 + 6)
28	11	55 (44 + 11)
30	5	60 (55 + 5)

Here number of observations, n = 60, which is even.

(i) By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\Big(\frac{\text{n}}{2}\Big) \text{th} \text{ term} + \Big(\frac{\text{n}}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{\Big(\frac{60}{2}\Big) \text{th} \text{ term} + \Big(\frac{60}{2} + 1\Big)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{30 \text{th} \text{ term} + (30 + 1)\text{th} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{30 \text{th} \text{ term} + 31\text{th} \text{ term}}{2}\end{aligned}$$

From table,

30th term = 20

31st term = 23

$$\begin{aligned}\Rightarrow \text{Median} = \frac{30 \text{ th} \text{ term} + 31 \text { st} \text{ term}}{2} \\[1em] \Rightarrow \text{Median} = \frac{20 + 23}{2} \\[1em] \Rightarrow \text{Median} = \frac{43}{2} \\[1em] \Rightarrow \text{Median} = 21.5\end{aligned}$$

Hence, median = 21.5.

(ii) By formula,

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

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

= 15 th term

= 18.

Hence, lower quartile = 18.

(iii) By formula,

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

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

= $\Big(\frac{180}{4}\Big)$ th term

= 45 th term

= 28.

Hence, Upper Quartile = 28.

(iv) By formula,

Interquartile range = Upper quartile - Lower quartile

= 28 - 18

= 10

Hence, interquartile range = 10.

Steps of construction :

1. Take 2 cm = 5 years on x-axis.

2. Take 1 cm = 10 policy holders on y-axis.

3. Plot the points (20, 0), (25, 2), (30, 6), (35, 18), (40, 38), (45, 66), (50, 88), (55, 96) and (60, 100).

4. Join the points by a free-hand curve.

Here, n = 100

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

(a) Through J = 50 draw a horizontal line to meet the ogive at K. Through K, draw a vertical line to meet the x-axis at L. The abscissa of the point L represents 42.

Hence, the median age = 42 years.

(b) Through M = 52 draw a vertical line to meet the ogive at N. Through N, draw a horizontal line to meet the y-axis at O. The ordinate of the point O represents 91.

Hence, 91 people have their age less than or equal to 52.

∴ No. of people whose age is greater than 52 = 100 - 91 = 9.

Hence, number of policy holders whose age is above 52 years equal to 9.

Marks %	No. of students (f)	CF
0-10	5	5
10-20	7	12
20-30	11	23
30-40	20	43
40-50	40	83
50-60	52	135
60-70	36	171
70-80	15	186
80-90	9	195
90-100	5	200

Steps :

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

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

3. Plot the points (10, 5), (20, 12), (30, 23), (40, 43), (50, 83), (60, 135), (70, 171), (80, 186), (90, 195) and (100, 200).

4. Join the points by free hand curve.

(a) n = 200, which is even

Median = $\frac{n}{2} = \frac{200}{2}$ = 100^th term.

Through point L = 100 draw a horizontal line parallel to x-axis touching the graph at point M, through M draw a vertical line parallel to y-axis touching x-axis at point N = 53.

Hence, median = 53.

(b) Total marks = 100

65% of 100 = 65

Through point O = 65 draw a vertical line parallel to y-axis touching the graph at point P, through P draw a horizontal line parallel to x-axis touching y-axis at point Q = 154.

∴ 154 students score less than or equal to 65%.

∴ 46 (200 - 154) students score more than 65%.

Hence, 46 students score more than 65%.

(c) Total marks = 100

35% of 100 = 35

Through point R = 35 draw a vertical line parallel to y-axis touching the graph at point S, through S draw a horizontal line parallel to x-axis touching y-axis at point T = 33.

Hence, 33 students did not pass the exam.

Cumulative frequency distribution table :

Marks	Number of students	Cumulative frequency
0 - 10	5	5
10 - 20	10	15 (10 + 5)
20 - 30	14	29 (15 + 14)
30 - 40	21	50 (29 + 21)
40 - 50	25	75 (50 + 25)
50 - 60	34	109 (75 + 34)
60 - 70	36	145 (109 + 36)
70 - 80	27	172 (145 + 27)
80 - 90	16	188 (172 + 16)
90 - 100	12	200 (188 + 12)

Here, n = 200, which is even.

Steps of construction:

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

2. Take 2 cm along y-axis = 20 students

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, 15), (30, 29), (40, 50), (50, 75), (60, 109), (70, 145), (80, 172), (90, 188), (100, 200).

5. Joint the points by a free hand curve.

(i) To find the median :

Let A be the point on y-axis representing frequency = $\frac{\text{n}}{2} = \frac{200}{2}$ = 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 points M represents 57.5.

Hence, the median is 57.5.

(ii) By formula,

$$\begin{aligned}\text{Upper quartile} = \Big(\frac{3\text{n}}{4}\Big)^ \text{th}\text{ term} \\[1em] \Rightarrow \text{Upper quartile} = \Big(\frac{(3 \times 200)}{4}\Big)^\text{th}\text{ term} \\[1em] \Rightarrow \text{Upper quartile} = \frac{600}{4}^\text{th}\text{ term} \\[1em] \Rightarrow \text{Upper quartile} = 150^\text{ th}\text{ term} \\[1em]\end{aligned}$$

Draw a line parallel to x-axis from point R (Number of students) = 150, touching the graph at point Q. From point Q draw a line parallel to y-axis touching x-axis at point N.

From graph,

N = 72

Hence, the upper quartile is 72.

(iii) Total marks = 100.

Let E be the point on x-axis representing marks = 65.

Through E draw a vertical line to meet the ogive at B. Through B, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 128.

Hence, 128 students score less than or equal to 65, so, students scoring more than 65 = 200 - 128 = 72.

Hence, the number of students who scored more than 65 marks is 72.

(iv) Given, 10 students qualify for merit-scholarship. This corresponds to 190th student. Since, 200 - 10 = 190

Draw a line parallel to x-axis from point F = 190, touching the graph at point G. From point G draw a line parallel to y-axis touching x-axis at point S.

From graph,

S = 91

Hence, the minimum marks required to qualify is 91.

Cumulative frequency distribution table :

Scores obtained	Number of shooters	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)

Here, n = 120, which is even.

(i) Steps of construction:

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

2. Take 2 cm along y-axis = 20 shooters

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), (100, 120)

5. Joint the points by a free hand curve.

To find the median :

Let A be the point on y-axis representing frequency = $\frac{\text{n}}{2} = \frac{120}{2}$ = 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 points M represents 43.

Hence, the median is 43.

(ii) To find lower quartile:

Let B be the point on y-axis representing frequency = $\frac{\text{n}}{4} = \frac{120}{4}$ = 30.

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 30.

To find upper quartile:

Let C be the point on y-axis representing frequency = $\frac{3\text{n}}{4} = \frac{3 \times 120}{4} = \frac{360}{4}$ = 90.

Through C, draw a horizontal line to meet the ogive at R. Through R, draw a vertical line to meet the x-axis at S. The abscissa of the point S represents 56.

Inter-quartile range = Upper quartile - Lower quartile = 56 - 30 = 26

Hence, the inter quartile range is = 26.

(iii) Total score = 100

So, more than 75% score mean more than 75 score.

Let T be the point on x-axis representing scores = 75

Through T, draw a vertical line to meet the ogive at G. Through G, draw a horizontal line to meet the y-axis at D. The ordinate of the point D represents 110

Shooters who have scored less than 75% = 110

So, students scoring more than 75% = Total students - Students who have scored less = 120 - 110 = 10

Hence, there are 10 number of shooters who obtained more than 75% score.

Cumulative frequency distribution table :

Wages (in ₹)	Number of workers	Cumulative frequency
400 - 450	2	2
450 - 500	6	8 (2 + 6)
500 - 550	12	20 (8 + 12)
550 - 600	18	38 (20 + 18)
600 - 650	24	62 (38 + 24)
650 - 700	13	75 (62 + 13)
700 - 750	5	80 (75 + 5)

Here, n = 80, which is even.

Steps of construction:

1. Take 2 cm along x-axis = ₹ 50

2. Take 2 cm along y-axis = 10 workers

3. 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.

4. Plot the points (450, 2), (500, 8), (550, 20), (600, 38), (650, 62), (700, 75), (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.

5. Joint the points by a free hand curve.

(i) Here, n = 80

To find the median :

Let A be the point on y-axis representing frequency = $\frac{\text{n}}{2} = \frac{80}{2}$ = 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 points M represents 600.5.

Hence, the median is ₹ 600.5.

(ii) To find lower quartile:

Let B be the point on y-axis representing frequency = $\frac{\text{n}}{4} = \frac{80}{4}$ = 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. The ordinate of 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.

Cumulative frequency distribution table :

Weight (in kg)	No. of students	Cumulative frequency
40 - 45	5	5
45 - 50	17	22 (17 + 5)
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)

Here, n = 200, which is even.

Steps of construction:

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

2. Take 2 cm along y-axis = 20 units.

3. Since, 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.

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 I (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 = 50.

∴ 50 students have weight less than 55.70 kg

Hence, 50 students are underweight.

(b) Since, 50 students have weight less than 55.70 kg

∴ 150 (200 - 50) students have weight more than 55.70 kg.

Hence, 150 students are overweight.

Cumulative frequency distribution table :

Marks	Number of students	Cumulative frequency
0 - 10	5	5
10 - 20	10	15 (10 + 5)
20 - 30	20	35 (15 + 20)
30 - 40	25	60 (35 + 25)
40 - 50	15	75 (60 + 15)
50 - 60	12	87 (75 + 12)
60 - 70	9	96 (87 + 9)
70 - 80	4	100 (96 + 4)

Here, n = 100, which is even.

(i) Steps of construction:

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

2. Take 2 cm along y-axis = 20 students

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, 15), (30, 35), (40, 60), (50, 75), (60, 87), (70, 96), (80, 100).

5. Joint the points by a free hand curve.

To find the median :

Let A be the point on y-axis representing frequency = $\frac{\text{n}}{2} = \frac{100}{2}$ = 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 points M represents 36.

Hence, the median is 36.

(ii) Total marks = 100.

Let E be the point on x-axis representing marks = 65.

Through E draw a vertical line to meet the ogive at B. Through B, draw a horizontal line to meet the y-axis at C. The ordinate of the point C represents 93.

Hence, 93 students score less than or equal to 65, so, students scoring more than 65 = 100 - 93 = 7.

Hence, the number of students who scored more than 65 marks is 7.

Cumulative frequency distribution table :

Scores obtained	Number of shooters	Cumulative frequency
0 - 10	9	9
10 - 20	13	22 (13 + 9)
20 - 30	20	42 (22 + 20)
30 - 40	26	68 (42 + 26)
40 - 50	30	98 (68 + 30)
50 - 60	22	120 (98 + 22)
60 - 70	15	135 (120 + 15)
70 - 80	10	145 (135 + 10)
80 - 90	8	153 (145 + 8)
90 - 100	7	160 (153 + 7)

Here, n = 160, which is even.

(i) Steps of construction:

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

2. Take 2 cm along y-axis = 20 shooters

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

4. Plot the points (10, 9), (20, 22), (30, 42), (40, 68), (50, 98), (60, 120), (70, 135), (80, 145), (90, 153), (100, 160)

5. Joint the points by a free hand curve.

To find the median :

Let A be the point on y-axis representing frequency = $\frac{\text{n}}{2} = \frac{160}{2}$ = 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 points M represents 44.

Hence, the median score is 44.

(ii) To find lower quartile:

Let B be the point on y-axis representing frequency = $\frac{\text{n}}{4} = \frac{160}{4}$ = 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 = $\frac{3\text{n}}{4} = \frac{3 \times 160}{4} = \frac{480}{4}$ = 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 S. The abscissa of the point S represents 60.

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

Hence, the inter quartile range is = 31.

(iii) Total score = 100

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

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

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

Shooters 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 number of shooters who obtained more than 85% score.

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

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

2. Take 2 cm along y-axis = 10 (No. of boys)

3. 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.

4. 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 (135, 0) representing lower limit of the first class i.e. 135 - 140.

5. 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 = $\frac{\text{n}}{2} = \frac{60}{2}$ = 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.

Hence, the median height = 149 cm.

(ii) To find lower quartile :

Let B be the point on y-axis representing frequency = $\frac{\text{n}}{4} = \frac{60}{4}$ = 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 S be the point on x-axis representing height = 158 cm.

Through S, 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	No. of students	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 $\frac{\text{N}}{2} = \frac{40}{2}$ = 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 = 14.6

Hence, the median = 14.6 m.

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

By formula,

Upper quartile = $\frac{3\text{N}}{4} = \frac{3 \times 40}{4} = \frac{120}{4}$ = 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

Hence, 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 there are 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.

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

Cumulative frequency distribution table :

Weight (in kg)	Number of students (f)	Cumulative frequencies (c.f.)
28-30	2	2
30-32	4	6
32-34	10	16
34-36	13	29
36-38	15	44
38-40	9	53
40-42	5	58
42-44	2	60
Total	Σf = 60

Here, n = 60, which is even.

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

Steps of construction :

1. Take 2 cm = 2 kg on x-axis.

2. Take 2 cm = 10 students on y-axis.

3. Since, x axis starts at 28 hence, a kink is drawn at the starting of x-axis. Plot the point (28, 0) as ogive starts on x-axis representing lower limit of first class.

4. Plot the points (30, 2), (32, 6), (34, 16), (36, 29), (38, 44), (40, 53), (42, 58) and (44, 60).

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

6. Draw a line parallel to x-axis from point A (no. of students) = 30, 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 = 36.2

Hence, median = 36.2 kg

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

By formula,

Upper quartile = $\frac{3\text{n}}{4} = \frac{3 \times 60}{4} = \frac{180}{4}$ = 45 th term.

Draw a line parallel to x-axis from point D (no. of students) = 45, touching the graph at point E. From point E draw a line parallel to y-axis touching x-axis at point F.

From graph, F = 38.2 kg

Hence, upper quartile = 38.2 kg.

(iii) Draw a line parallel to y-axis from point G (weight) = 37 kg, 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 = 36.

∴ 36 students have weight less than or equal to 36 kg.

No. of students whose weight is more than 36 kg = 60 - 36 = 24.

Hence, no. of students whose weight is more than 36 kg = 24.

(i) Arranging given observations in ascending order:

11, 17, 19, 19, 23

Sum of observations = 17 + 19 + 11 + 23 + 19 = 89.

By formula,

Mean = $\frac{\text{Sum of observation}}{\text{No. of observation}} = \frac{89}{5}$ = 17.8

Here n = 5, which is odd.

By formula,

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

$$\begin{aligned}= \frac{5 + 1}{2} \text{ th observation} \\[1em] = \frac{6}{2} \text{ th observation} \\[1em] = 3 \text{ rd observation} \\[1em] = 19.\end{aligned}$$

From set of observations, we see that:

19 has the maximum frequency.

Hence, mean = 17.8, median = 19, mode = 19.

(ii) Arranging given observations in ascending order:

7, 8, 8, 8, 9, 9, 11, 12

Sum of observations = 7 + 8 + 8 + 8 + 9 + 9 + 11 + 12 = 72

By formula,

Mean = $\frac{\text{Sum of observation}}{\text{No. of observation}} = \frac{72}{8}$ = 9

Here n = 8, which is even.

By formula,

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

$$\begin{aligned}= \frac{\frac{8}{2} \text{ th observation} + \Big(\frac{8}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \frac{4 \text{ th observation} + (4 + 1) \text{ th observation}}{2} \\[1em] = \frac{4 \text{ th observation} + 5 \text{ th observation}}{2} \\[1em] = \frac{8 + 9}{2} \\[1em] = \frac{17}{2} \\[1em] = 8.5.\end{aligned}$$

From set of observations, we see that :

8 has the maximum frequency.

Hence, mean = 9, median = 8.5, mode = 8.

(iii) Arranging given observations in ascending order:

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

Sum of observations = 0 + 1 + 1 + 2 + 2 + 3 + 3 + 3 + 4 + 5 = 24

By formula,

Mean = $\frac{\text{Sum of observation}}{\text{No. of observation}} = \frac{24}{10}$ = 2.4

Here n = 10, which is even.

By formula,

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

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

From set of observations, we see that:

3 has the maximum frequency.

Hence, mean = 2.4, median = 2.5, mode = 3.

(iv) Arranging given observations in ascending order:

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

Sum of observations = 6 + 6 + 7 + 8 + 10 + 10 + 10 + 11 + 13 = 81

By formula,

Mean = $\frac{\text{Sum of observation}}{\text{No. of observation}} = \frac{81}{9}$ = 9

Here n = 9, which is odd.

By formula,

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

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

From set of observations, we see that:

10 has the maximum frequency.

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

Here, n = 10, which is even.

By formula,

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

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

Median = $\frac{x + (x + 4)}{2}$

$$\begin{aligned}\Rightarrow 48 = \frac{\text{x + x + 4}}{2} \\[1em] \Rightarrow 48 = \frac{\text{2x + 4}}{2} \\[1em] \Rightarrow 48 \times 2 = \text{2x + 4}\end{aligned}$$

⇒ 96 = 2x + 4

⇒ 2x = 96 - 4

⇒ 2x = 92

⇒ x = $\frac{92}{2}$

⇒ x = 46

Set of observations : 13, 35, 43, 46, 46, 50, 55, 61, 71, 80

Here, 46 has the maximum frequency.

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

In the given data : 8, 9, 5, 6, 4, 9, 1, 9, 3, 6, 3, 9, 7, 1, 2, 9, 5

9 is repeated more number of times than any other number.

Hence, modal size of the shoes sold = 9.

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

Weight(kg) (x)	No. of students (f)	Cumulative frequency	fx
47	4	4	188
50	3	7 (4 + 3)	150
53	2	9 (7 + 2)	106
56	2	11 (9 + 2)	112
60	4	15 (11 + 4)	240
Total	Σf = 15		Σfx = 796

Total number of observations = 15, which is odd.

(i) By formula,

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{\sum\text{fx}}{\sum\text{f}} \\[1em] \Rightarrow \text{Mean} = \frac{796}{15} \\[1em] \Rightarrow \text{Mean} = 53.06\end{aligned}$$

Hence, mean = 53.06.

(ii) By formula,

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

$$\begin{aligned}= \frac{15 + 1}{2} \text{ th observation} \\[1em] = \frac{16}{2} \text{ th observation} \\[1em] = 8 \text{ th observation} \\[1em]\end{aligned}$$

Cumulative frequencies 8th and 9th corresponds to 53 kg.

Hence, median = 53 kg.

(iii) The highest frequency is 4.

4 corresponds to two weight = 47 kg and 60 kg.

Hence, mode = 47 kg and 60 kg.

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

Number (x)	Frequency (f)	Cumulative frequency	fx
5	1	1	5
10	2	3 (1 + 2)	20
15	5	8 (3 + 5)	75
20	6	14 (8 + 6)	120
25	3	17 (14 + 3)	75
30	2	19 (17 + 2)	60
35	1	20 (19 + 1)	35
Total	Σf = 20		Σfx = 390

Total number of observations = 20, which is even.

By formula,

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{\sum\text{fx}}{\sum\text{f}} \\[1em] \Rightarrow \text{Mean} = \frac{390}{20} \\[1em] \Rightarrow \text{Mean} = 19.5\end{aligned}$$

By formula,

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

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

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

Then,

Median = $\frac{20 + 20}{2} = \frac{40}{2}$ = 20.

As the variate 20 has maximum frequency 6, so mode = 20.

Hence, mean = 19.5, median = 20, mode = 20.

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

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

Total number of observations = 40, which is even.

(i) By formula,

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

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

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

Then,

Median = $\frac{6 + 6}{2} = \frac{12}{2}$ = 6.

Hence, median = 6.

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

Hence, mode = 6.

(i) We construct the following table :

Daily wages (xi)	Number of workers (fi)	Class mark (ui)	Cumulative frequency	fiui
200 - 220	5	210	5	1050
220 - 240	20	230	25 (20 + 5)	4600
240 - 260	10	250	35 (25 + 10)	2500
260 - 280	10	270	45 (35 + 10)	2700
280 - 300	9	290	54 (45 + 9)	2610
300 - 320	6	310	60 (54 + 6)	1860
320 - 340	12	330	72 (60 + 12)	3960
340 - 360	8	350	80 (72 + 8)	2800
Total	Σfi = 80			Σfiui = 22080

By formula,

$$\begin{aligned}\Rightarrow \text{Mean} = \frac{Σ\text{f}_{i}\text{u}_{i}}{Σ\text{f}_{i}}\\[1em] \Rightarrow \text{Mean} = \frac{22080}{80}\\[1em] \Rightarrow \text{Mean} = 276\end{aligned}$$

Hence, the mean is ₹ 276.

(ii) The class 220 - 240 has maximum frequency 20.

Hence, modal class = 220 - 240.

(iii) From table,

Hence, the number of workers getting daily wages below ₹ 300 = 54.

(iv) From table,

Hence, the number of workers getting ₹ 260 or more but less than ₹ 340 as daily wages = 72 - 35 = 37.

Steps :

1. Take 1 cm along x-axis = 5 marks and 1 cm along y-axis = 5 units (frequency).

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. 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 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, modal runs = 57.

Steps :

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

Adjustment factor = $\frac{48 - 47}{2} = \frac{1}{2}$ = 0.5

We construct the continuous frequency table for the given data :

Classes before adjustment	Classes after adjustment	No. of students
44 - 47	43.5 - 47.5	23
48 - 51	47.5 - 51.5	25
52 - 55	51.5 - 55.5	37
56 - 59	55.5 - 59.5	18
60 - 63	59.5 - 63.5	7
64 - 67	63.5 - 67.5	2

2. Take 2 cm along x-axis = 4 kg and 1 cm along y-axis = 4 (frequency).

3. Since, the scale on x-axis starts at 43.5, a break (zig-zag curve) is shown near the origin along x-axis to indicate that the graph is drawn to scale beginning at 43.5 and not at origin itself.

4. Construct rectangles corresponding to the given data.

5. In highest rectangle, draw two straight 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 53 kg.

Hence, the required mode = 53.

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 runs.

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 straight 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 2 cm along x-axis = 10 marks and 2 cm along y-axis = 10 students (frequency).

2. Construct rectangles corresponding to the given data.

3. In highest rectangle, draw two straight 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 35.

Hence, the required mode = 35.

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 KL perpendicular to the horizontal axis.

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

From graph,

L = ₹ 525

Hence, required mode = ₹ 525.

Steps of Construction:

1. Draw a histogram of the given distribution.

2. Inside the highest rectangle, which represents the maximum frequency (the modal class 60-65), draw two lines AC and BD diagonally from the upper corners C and D of the adjacent rectangles (50-55 and 65-70) to the top corners of the modal rectangle.

3. Through the 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 the mode.

From graph,

L = 61.

Hence, modal weight = 61 kg.

(i) 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

(ii) From graph,

The modal class is 60-70.

(iii) From graph,

The mode = 64.

(iv) From graph,

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

(i) From histogram,

The number of batsmen who scored between 500 and 600 runs is 3.

The number of batsmen who scored between 600 and 700 runs is 2.

Thus, the total number of batsmen who scored 500 to 700 runs is 3 + 2 = 5.

Hence, number of batsmen who scored 500 to 700 runs = 5.

(ii) The modal class is the class with the highest frequency.

Hence, the modal class interval is 400 - 500.

(iii) From graph,

Mode = 430.

Hence, mode = 430.

We know that,

The value of the middle-most observation obtained after arranging the data in an ascending order, is called median of the data.

Here no. of observations, n = 35

By formula,

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

$$\begin{aligned}= \frac{35 + 1}{2} \text{ th observation} \\[1em] = \frac{36}{2} \text{ th observation} \\[1em] = 18 \text{ th observation}\end{aligned}$$

Since, given data is arranged in descending order,

Median = 18 th observation.

Hence, Option 3 is the correct option.

Prime numbers : 2, 3, 5, 7, 11, 13, 17, 19

No. of terms (n) = 8, which is even.

By formula,

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

Substituting values we get:

$$\begin{aligned}= \frac{\frac{8}{2} \text{ th observation} + \Big(\frac{8}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \frac{4 \text{ th observation} + \Big(4 + 1\Big) \text{ rth observation}}{2} \\[1em] = \frac{4 \text{ th observation} + 5 \text{ th observation}}{2} \\[1em] = \frac{7 + 11}{2} \\[1em] = \frac{18}{2} \\[1em] = 9.\end{aligned}$$

Hence, Option 2 is the correct option.

We know that,

The value of the middle-most observation obtained after arranging the data in an ascending order, is called median of the data.

Hence, Option 3 is the correct option.

No. of observations = 10, which is even.

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

$\left(\frac{n}{2}\right)^\text{th}\text{term} = \frac{10}{2}$ = 5^th term = x

$\left(\frac{n}{2} + 1\right)^\text{th}\text{term} = \frac{10}{2} + 1$ = 5 + 1 = 6^th term = x + 4

Given,

Median = 64

$$\begin{aligned}\therefore \frac{x + x + 4}{2} = 64 \\[1em] \Rightarrow \frac{2x + 4}{2} = 64 \\[1em] \Rightarrow \frac{2(x + 2)}{2} = 64 \\[1em] \Rightarrow x + 2 = 64 \\[1em] \Rightarrow x = 64 - 2 \\[1em] \Rightarrow x = 62\end{aligned}$$

Hence, Option 3 is the correct option.

Set of observations = 20, 24, 25, 26, 27, 28, 29, 30

Here, n = 8, which is even.

By formula,

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

$$\begin{aligned}= \frac{\frac{8}{2} \text{ th observation} + \Big(\frac{8}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \frac{4 \text{ th observation} + \Big(4 + 1\Big) \text{ th observation}}{2} \\[1em] = \frac{4 \text{ th observation} + 5 \text{ th observation}}{2} \\[1em] = \frac{26 + 27}{2} \\[1em] = \frac{53}{2} \\[1em] = 26.5\end{aligned}$$

If 25 is removed from given set, then we get:

Set of observations = 20, 24, 26, 27, 28, 29, 30

Here, n = 7, which is odd.

By formula,

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

$$\begin{aligned}= \frac{7 + 1}{2} \text{ th observation} \\[1em] = \frac{8}{2} \text{ th observation} \\[1em] = 4 \text{ th observation} \\[1em] = 27\end{aligned}$$

Difference between median = 27 - 26.5 = 0.5

Hence, Option 1 is the correct option.

Given,

Mean of 4, 5, 1, 3, 7, 4 is x.

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

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

Given,

Numbers 2, 4, 3, 2, 3, y, 3 have mean x - 1 = 4 - 1 = 3.

Sum of observations = 2 + 4 + 3 + 2 + 3 + y + 3 = 17 + y

Mean (x) = $\frac{\text{Sum of observations}}{\text{No. of observations}} = \frac{17 + \text{y}}{7}$

$\Rightarrow 3 = \frac{17 + \text{y}}{7}$

⇒ 7 × 3 = 17 + y

⇒ 21 = 17 + y

⇒ y = 21 - 17

⇒ y = 4

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

Here, n = 7, which is odd.

By formula,

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

$$\begin{aligned}= \frac{7 + 1}{2} \text{ th observation} \\[1em] = \frac{8}{2} \text{ th observation} \\[1em] = 4 \text{ th observation} \\[1em] = 3\end{aligned}$$

∴ z = 3.

y + z = 4 + 3 = 7.

Hence, Option 4 is the correct option.

Mode represents the value that occurs most frequently in a dataset.

The mode will determine the most frequently sold shoe size in the shop.

Hence, Option 3 is the correct option.

We know that,

To determine mean we do not need any graphical method, since, it relies on the exact numerical value of every single observation in a dataset and graphs like histograms or frequency polygons represent data in ranges, specific individual values to determine mean is not possible.

Hence, Option 1 is the correct option.

The median of a grouped frequency distribution is typically found graphically using a cumulative frequency curve, also known as an ogive.

Hence, Option 1 is the correct option.

Given,

Set of observations = 0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5, 6

Arranging the numbers in ascending order :

-3, -3, -1, 0, 2, 2, 2, 5, 5, 5, 5, 6, 6, 6.

Here n = 14, which is even.

By formula,

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

$$\begin{aligned}= \frac{\frac{14}{2} \text{ th observation} + \Big(\frac{14}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \frac{7 \text{ th observation} + \Big(7 + 1\Big) \text{ th observation}}{2} \\[1em] = \frac{7 \text{ th observation} + 8 \text{ th observation}}{2} \\[1em] = \frac{2 + 5}{2} \\[1em] = \frac{7}{2} \\[1em] = 3.5\end{aligned}$$

Hence, Option 4 is the correct option.

Cumulative frequency distribution table is:

x	f	Cumulative frequency
10	2	2
20	3	5 (2 + 3)
30	2	7 ( 5 + 2)
40	3	10 (7 + 3)
50	1	11 (10 + 1)

Here, n = 11, which is odd.

By formula,

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

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

From table,

6th observation corresponds to 30.

∴ Median = 30

Hence, Option 1 is the correct option.

Cumulative frequency distribution table is as follows :

Diameter of heart (in mm)	Number of persons	Cumulative frequency
120	5	5
121	9	14 (5 + 9)
122	14	28 (14 + 14)
123	8	36 (28 + 8)
124	5	41 (36 + 5)
125	9	50 (41 + 9)

Here n = 50, which is even.

By formula,

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

$$\begin{aligned}= \frac{\frac{50}{2} \text{ th observation} + \Big(\frac{50}{2} + 1\Big) \text{ th observation}}{2} \\[1em] = \frac{25 \text{ th observation} + \Big(25 + 1\Big) \text{ th observation}}{2} \\[1em] = \frac{25 \text{ th observation} + 26 \text{ th observation}}{2} \\[1em] = \frac{122 + 122}{2} \\[1em] = \frac{244}{2} \\[1em] = 122 \text{ mm}.\end{aligned}$$

Since, all observations from 25th to 26th corresponds to 122 mm.

Hence, Option 1 is the correct option.

We construct the cumulative frequency distribution table as under :

Class	Frequency	Cumulative frequency
0 - 5	8	8
5 - 10	10	18 (10 + 8)
10 - 15	19	37 (18 + 19)
15 - 20	25	62 (37 + 25)
20 - 25	8	70 (62 + 8)

Here n = 70, which is even.

By formula,

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

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

As observation from 19th to 37th lies in the class 10 - 15

∴ Median class = 10 - 15, with upper limit = 15

Hence, Option 2 is the correct option.

From above set of numbers : 8 occurs most of the time.

Mode = 8.

Hence, Option 3 is the correct option.

Arranging set of observations in ascending order : 1, 2, 3, 4, 6, 7, 8

Here, n = 7, which is odd.

By formula,

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

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

Given,

2, 9, 3, 4, 9, 6, 9

From above set of numbers : 9 occurs most of the time.

Mode = 9.

Average of median and mode = $\frac{9 + 4}{2} = \frac{13}{2}$ = 6.5

Hence, Option 2 is the correct option.

The mode of a frequency distribution can be determined graphically from Histogram.

Hence, Option 1 is the correct option.

Given, mode = 7

From above set of numbers : 6 and 7 occurs most of the time (3 times each). For 7 to be the mode, its frequency must be greater than that of 6.

Therefore, the unknown expression 2k + 1 must be equal to 7.

⇒ 2k + 1 = 7

⇒ 2k = 7 - 1

⇒ 2k = 6

⇒ k = $\frac{6}{2}$

⇒ k = 3.

Hence, Option 2 is the correct option.

Since the class 20 - 30 has highest frequency i.e. 13.

∴ Modal class = 20 - 30.

The lower limit of the modal class = 20.

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 (15 + 10)
10 - 15	12	37 (25 + 12)
15 - 20	20	57 (37 + 20)
20 - 25	9	66 (57 + 9)

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

As n is even,

By formula,

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

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

As observation from 26th to 37th 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.

By formula,

Mode = 3 Median - 2 Mean

Hence, Option 2 is the correct option.

By formula,

Mode = 3 Median - 2 Mean

⇒ 12.3 = 3 × Median - 2 × 10.5

⇒ 12.3 = 3 × Median - 21

⇒ 3 × Median = 12.3 + 21

⇒ 3 × Median = 33.3

⇒ Median = $\frac{33.3}{3}$

⇒ Median = 11.1

Hence, Option 3 is the correct option.

By formula,

Mode = 3 Median - 2 Mean

⇒ 50.5 = 3 × 45.5 - 2 Mean

⇒ 50.5 = 136.5 - 2 Mean

⇒ 2 Mean = 136.5 - 50.5

⇒ 2 Mean = 86

⇒ Mean = $\frac{86}{2}$

⇒ Mean = 43.

Hence, Option 1 is the correct option.

By formula,

Mode = 3 Median - 2 Mean

= 3 × 41.25 - 2 × 33.75

= 123.75 - 67.5

= 56.25.

Hence, Option 3 is the correct option.

Given,

Mode - Median = 24

Mode = 24 + Median

By formula,

Mode = 3 Median - 2 Mean

⇒ 24 + Median = 3 Median - 2 Mean

⇒ 3 Median - Median - 2 Mean = 24

⇒ 2 Median - 2 Mean = 24

⇒ 2 (Median - Mean) = 24

⇒ Median - Mean = $\frac{24}{2}$

⇒ Median - Mean = 12.

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	Frequency	Cumulative frequency
0 - 10	2	2
10 - 20	4	6 (2 + 4)
20 - 30	3	9 (6 + 3)
30 - 40	5	14 (9 + 5)

Here, Cumulative frequency = 14, which is even.

By formula,

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

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

All observations from 7th to 9th are equal, each lies in the interval 20 - 30.

So, median class = 20 - 30

Hence, Option 3 is the correct option.

For 11 arrayed data.

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

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

= 6th term, which will be the middle term.

∴ For a collection of 11 arrayed data, the median is the middle number.

∴ Assertion (A) is true.

Numbers = 5, 9, 7, 13, 10, 11, 10

Arranging in ascending order, we get :

5, 7, 9, 10, 10, 11, 13.

n = 7, which is odd.

Median = $\frac{n + 1}{2} = \frac{7 + 1}{2} = \frac{8}{2}$ = 4th term = 10.

∴ Reason (R) is false.

Hence, option 3 is the correct option.

Arranging the observations 15, 14, 21, 11, 19, 10, 18 in ascending order :

10, 11, 14, 15, 18, 19, 21

Here, n = 7, which is odd.

By formula,

Lower quartile (Q₁) = $\Big(\frac{\text{n} + 1}{4}\Big) \text{ th observation} = \frac{7 + 1}{4}$ th observation = 2 nd observation = 11.

∴ Assertion (A) is true.

Considering Reason (R), for an ungrouped data containing n observations, the median is given by $\Big(\frac{\text{n} + 1}{2}\Big)$ th observation, when n is odd.

∴ Reason (R) is true.

But Reason (R) is a statement about the median, whereas Assertion (A) is about the first quartile. So, R is not the correct explanation of A.

Hence, option 2 is the correct option.

Arranging the given data in ascending order :

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

Here, n = 16, which is even.

By formula,

Upper quartile (Q₃) = $\Big(\frac{3\text{n}}{4}\Big) \text{ th observation} = \frac{3 \times 16}{4}$ th observation = 12 th observation = 18.

∴ Assertion (A) is true.

Considering Reason (R), for an ungrouped data containing n observations, the upper quartile, when n is even, is given by Q₃ = $\Big(\frac{3\text{n}}{4}\Big)$ th observation and not $\Big(\frac{3\text{n}}{4} + 1\Big)$ th observation.

∴ Reason (R) is false.

Hence, option 3 is the correct option.

Cumulative frequency distribution table :

Class interval	Class mark	Frequency	Cumulative frequency
20-30	25	1	1
30-40	35	3	4
40-50	45	2	6
50-60	55	6	12
60-70	65	4	16

Median = $\frac{16}{2}$ = 8th term.

The 8th term lies in the class 50-60.

∴ Median class = 50-60

Also, frequency of class 50-60 is highest.

∴ Modal class = 50-60.

∴ Assertion (A) is true.

Modal class and median class are not always the same for a given frequency distribution.

∴ Reason (R) is false.

Hence, Option 3 is the correct option.

Arranging the given data in ascending order :

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

Given, Median = 24.

Here, n = 9, which is odd.

By formula,

$$\begin{aligned}\Rightarrow \text{Median} = \frac{\text{n} + 1}{2} \text{ th observation} \\[1em] \Rightarrow 24 = \frac{9 + 1}{2} \text{ th observation} \\[1em] \Rightarrow 24 = \frac{10}{2} \text{ th observation} \\[1em] \Rightarrow 24 = 5 \text{ th observation}\end{aligned}$$

⇒ 24 = x + 4

⇒ x = 24 − 4

⇒ x = 20.

∴ Assertion (A) is true.

Considering Reason (R), for an ungrouped data containing n observations, the median, when n is odd, is given by Median = $\frac{\text{n} + 1}{2}$ th observation and not $\frac{\text{n}}{2}$ th observation.

∴ Reason (R) is false.

Hence, option 3 is the correct option.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 6 = \frac{\frac{12}{2}\text{ th term} + \Big(\frac{12}{2} + 1\Big)\text{ th term}}{2} \\[1em] \Rightarrow 6 \times 2 = \text{6th term + 7th term} \\[1em] \Rightarrow 12 = x + x + 2 \\[1em] \Rightarrow 12 = 2x + 2 \\[1em] \Rightarrow 12 - 2 = 2x \\[1em] \Rightarrow 2x = 10 \\[1em] \Rightarrow x = \frac{10}{2} = 5.\end{aligned}$$

Numbers :

2, 3, 3, 3, 4, 5, 7, 8, p, q, 8, 9.

Since, mode = 8, it means 8 occurs for the most times in the series.

Since, 3 occurs 3 times in the series,

∴ 8 must occur for atleast 4 times in order to be the mode.

∴ p = q = 8.

Hence, x = 5, p = 8 and q = 8.

Cumulative frequency distribution table :

Daily wages (in ₹)	No. of employees	Cumulative frequency
30-40	8	8
40-50	14	22
50-60	12	34
60-70	17	51
70-80	20	71
80-90	26	97
90-100	13	110
100-110	10	120

Here, n = 120, which is even.

Median = $\frac{n}{2} = \frac{120}{2}$ = 60th term.

Steps of construction :

1. Plot daily wages on x-axis.

2. Plot cumulative frequency on y-axis.

3. Mark points (40, 8), (50, 22), (60, 34), (70, 51), (80, 71), (90, 97), (100, 110) and (110, 120).

4. Draw a free hand curve passing through the points marked, strating from the lower limit of first class and terminating at upper limit of the last class.

5. Mark A = 60 on y-axis, draw a horizontal line which meets curve at B.

6. Through point B, draw a vertical line which meets x-axis at point C. The value of point C on x-axis is the median. ∴ Median wage is ₹74.

7. Mark D = 84 on x-axis, draw a vertical line which meets curve at E.

8. Through point E, draw a horizontal line which meets y-axis at point F. The value of point F on y-axis represents no. of employees earning less than or equal to ₹ 84 per day.

From graph,

F = 81.

No. of employees earning more than ₹ 84 per day = 120 - 81 = 39.

Percentage of employees earning more than ₹ 84 = $\frac{\text{No. of employees earning more than ₹ 84}}{\text{Total employees}} \times 100$

$= \frac{39}{120} \times 100 = \frac{3900}{120}$ = 32.5 %.

9. Mark G = 56 on x-axis, draw a vertical line which meets curve at H.

10. Through point H, draw a horizontal line which meets y-axis at point I. The value of point I on y-axis represents no. of employees earning less than or equal to ₹ 56 per day.

From graph,

I = 30.

No. of employees earning less than or equal to ₹ 56 per day = 30.

(a) Given,

Total no. of policy holders = 100

∴ 7 + 12 + 15 + 22 + f + 14 + 8 + 4 = 100

⇒ 82 + f = 100

⇒ f = 100 - 82 = 18.

Hence, f = 18.

(b) Cumulative frequency distribution table :

Age in years	No. of policy holders	Cumulative frequency
15-20	7	7
20-25	12	19
25-30	15	34
30-35	22	56
35-40	18	74
40-45	14	88
45-50	8	96
50-55	4	100

Since, n = 100, which is even.

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

Steps of construction :

1. Plot class interval on x-axis and cumulative frequency on y-axis.

2. Mark points (20, 7), (25, 19), (30, 34), (35, 56), (40, 74), (45, 88), (50, 96) and (55, 100).

3. Draw a free hand curve passing through the points marked and starting from the lower limit of first class and terminating at upper limit of the last class.

4. From point A = 50 draw a line parallel to x-axis touching the graph at point B. From point B draw a line parallel to y-axis touching x-axis at C.

From graph,

C = 33.75, which lies between 30-35.

Hence, median class = 30-35.

Difference between two consecutive class marks = 475 - 425 = 50.

Adjustment factor = $\frac{\text{Class width}}{2} = \frac{50}{2}$ = 25.

Lower class limit = Class mark - Adjustment factor

Upper class limit = Class mark + Adjustment factor

Class	Frequency
400-450	6
450-500	12
500-550	15
550-600	17
600-650	7
650-700	3

Steps of construction :

1. Take 2 cm along x-axis = ₹50 and 2 cm along y-axis = 5 workers.

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.

4. Let K be the point of intersection of AD and BC. Through K, draw a vertical line to meet the x-axis at L. The abscissa of the point L represents 557.50.

Hence, mode = ₹ 557.50

(a)

Class	Frequency (f)	Cumulative frequency (c.f.)
120-130	6	6
130-140	29	35
140-150	34	69
150-160	22	91
160-170	12	103

(b) From the above table,

The number of students whose height is less than 150 cm = 69.

(c) From the above table,

The total number of students = 103.

(d) From graph,

The modal height = 143 cm.

(e) Mean height = Average height = 145.5

Difference in the modal height and the mean height :

145.5 - 143 = 2.5 cm

Hence, difference between modal height and the mean height = 2.5 cm