Data Handling - Exercise 24(B)

Find the median of :

(i) 72, 0, 46, 34, 8, 31, 65, 25, 39, 53, 18

(ii) 25, 18, 13, 20, 16, 9, 22, 8, 6, 15, 21, 11, 17

Answer

(i) On arranging the given set of data in ascending order, we get :

0, 8, 18, 25, 31, 34, 39, 46, 53, 65, 72

Number of observations, n = 11 (odd)

Median = $\Big[\dfrac{n+1}{2}\Big]^{\text{th}}$ term

$= \Big[\dfrac{11+1}{2}\Big]^{th} \text{ term} \\[1em] = \Big[\dfrac{12}{2}\Big]^{th} \text{ term} \\[1em] = 6^{th} \text{ term} \\[1em] = 34$

Hence, the median is 34.

(ii) On arranging the given set of data in ascending order, we get :

6, 8, 9, 11, 13, 15, 16, 17, 18, 20, 21, 22, 25

Number of observations, n = 13 (odd)

Median = $\Big[\dfrac{n+1}{2}\Big]^{th}$ term

$= \Big[\dfrac{13+1}{2}\Big]^{th} \text{ term} \\[1em] = \Big[\dfrac{14}{2}\Big]^{th} \text{ term} \\[1em] = 7^{th} \text{ term} \\[1em] = 16$

Hence, the median is 16.

Find the median of :

(i) 22, 9, 32, 17, 35, 10, 19, 21

(ii) 85, 91, 51, 35, 82, 55, 60, 29, 63, 72

Answer

(i) On arranging the given set of data in ascending order, we get :

9, 10, 17, 19, 21, 22, 32, 35

Number of observations, n = 8 (even)

Median = $\dfrac{1}{2}\Big[\Big(\dfrac{n}{2}\Big)^{th} \text{ term} + \Big(\dfrac{n}{2} + 1\Big)^{th} \text{ term}\Big]$

$= \dfrac{1}{2}\Big[\Big(\dfrac{8}{2}\Big)^{th} \text{ term} + \Big(\dfrac{8}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[(4)^{th} \text{ term} + (5)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[19 + 21\Big] \\[1em] = \dfrac{1}{2}\Big[40\Big] \\[1em] = 20$

Hence, the median is 20.

(ii) On arranging the given set of data in ascending order, we get :

29, 35, 51, 55, 60, 63, 72, 82, 85, 91

Number of observations, n = 10 (even)

Median = $\dfrac{1}{2}\Big[\Big(\dfrac{n}{2}\Big)^{th} \text{ term} + \Big(\dfrac{n}{2} + 1\Big)^{th} \text{ term} \Big]$

$= \dfrac{1}{2}\Big[\Big(\dfrac{10}{2}\Big)^{th} \text{ term} + \Big(\dfrac{10}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[(5)^{th} \text{ term} + (6)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[60 + 63\Big] \\[1em] = \dfrac{1}{2}\Big[123\Big] \\[1em] = 61.5$

Hence, the median is 61.5.

The weights of 13 students (in kg) are :

48, 41, 52, 65, 40, 53, 34, 61, 47, 42, 44, 31, 35

Find the median weight.

Answer

On arranging the given set of data in ascending order, we get :

31, 34, 35, 40, 41, 42, 44, 47, 48, 52, 53, 61, 65

Number of observations, n = 13 (odd)

Median = $\Big[\dfrac{n+1}{2}\Big]^{th}$ term

$= \Big[\dfrac{13+1}{2}\Big]^{th} \text{ term} \\[1em] = \Big[\dfrac{14}{2}\Big]^{th} \text{ term} \\[1em] = 7^{th} \text{ term} \\[1em] = 44$

Hence, the median weight is 44 kg.

The marks (out of 50) of 10 students in a class are :

40, 34, 37, 50, 47, 42, 31, 46, 36, 43

Find the median marks.

Answer

On arranging the given set of data in ascending order, we get :

31, 34, 36, 37, 40, 42, 43, 46, 47, 50

Number of observations, n = 10 (even)

Median = $\dfrac{1}{2}\Big[\Big(\dfrac{n}{2}\Big)^{th} \text{ term} + \Big(\dfrac{n}{2} + 1\Big)^{th} \text{ term}\Big]$

$= \dfrac{1}{2}\Big[\Big(\dfrac{10}{2}\Big)^{th} \text{ term} + \Big(\dfrac{10}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[(5)^{th} \text{ term} + (6)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[40 + 42\Big] \\[1em] = \dfrac{1}{2}\Big[82\Big] \\[1em] = 41$

Hence, the median marks is 41.

Find the median of first 15 odd numbers.

Answer

The first 15 odd numbers are :

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29

Number of observations, n = 15 (odd)

Median = $\Big[\dfrac{n+1}{2}\Big]^{th}$ term

$= \Big[\dfrac{15+1}{2}\Big]^{th} \text{ term} \\[1em] = \Big[\dfrac{16}{2}\Big]^{th} \text{ term} \\[1em] = 8^{th} \text{ term} \\[1em] = 15$

Hence, the median of first 15 odd numbers is 15.

Find the median of first 50 whole numbers.

Answer

The first 50 whole numbers are :

0, 1, 2, 3, ..., 49

Number of observations, n = 50 (even)

$Median = \dfrac{1}{2}\Big[\Big(\dfrac{n}{2}\Big)^{th} \text{ term} + \Big(\dfrac{n}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[\Big(\dfrac{50}{2}\Big)^{th} \text{ term} + \Big(\dfrac{50}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[(25)^{th} \text{ term} + (26)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[24 + 25\Big] \\[1em] = \dfrac{1}{2}\Big[49\Big] \\[1em] = 24.5$

Hence, the median of first 50 whole numbers is 24.5.

The daily wages (in ₹) of 100 labourers in a factory are given below :

Find the median wages.

Answer

Arranging the terms in ascending order, we get :

Now, we prepare the cumulative frequency table :

Total number of terms = ∑f_i = N = 100, which is even.

Median = $\dfrac{1}{2}\Big[\Big(\dfrac{N}{2}\Big)^{th} + \Big(\dfrac{N}{2} + 1\Big)^{th}\Big]$

$= \dfrac{1}{2}\Big[\Big(\dfrac{100}{2}\Big)^{th} \text{ term} + \Big(\dfrac{100}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[(50)^{th} \text{ term} + (51)^{st} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[360 + 360\Big] \\[1em] = \dfrac{1}{2}\Big[720\Big] \\[1em] = 360$

[Note that each one of 46th, 47th, ......, 62nd labourer earns ₹ 360.]

Hence, the median wages is ₹ 360.

The heights (in cm) of 35 students of a class are given below :

Find the median height.

Answer

Arranging the terms in ascending order, we get :

Now, we prepare the cumulative frequency table :

Total number of terms = ∑f_i = N = 35, which is odd.

Median = $\Big[\dfrac{N+1}{2}\Big]^{th}$

$= \Big[\dfrac{35+1}{2}\Big]^{th} \text{ term} \\[1em] = \Big[\dfrac{36}{2}\Big]^{th} \text{ term} \\[1em] = 18^{th} \text{ term} \\[1em] = 155$

[Note that each one of 13th, 14th, ......, 21st student has height 155 cm.]

Hence, the median height is 155 cm.