Median, Quartiles & Mode — Exercise 26(A)

Find the median of each of the following sets of numbers:

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

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

(iii) 40, 31, 25, 36, 27, 38, 28, 35

(iv) 56, 81, 51, 42, 69, 85, 72, 35, 66, 92

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

Answer

(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,

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

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,

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

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,

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

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,

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

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,

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

Hence, median = 16.

The marks of 15 students in an examination are given below :

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

Find the median score.

Answer

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,

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

Hence, median score = 23.

The heights (in cm) of 9 girls in a class are given below:

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

Find the median height.

Answer

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,

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

Hence, median height = 147.3 cm.

The weights (in kg) of 8 children are given below:

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

Find the median weight.

Answer

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,

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

Hence, median weight = 13.85 kg.

(i) The median of the observations 11, 12, 14, 18, (x + 4), 30, 32, 35, 41 arranged in ascending order is 24. Find the value of x.

(ii) If 10, 13, 15, 18, x + 1, x + 3, 31, 36, 38, 42 are the observations arranged in ascending order with median 28, find the value of x.

Answer

(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.

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

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.

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

Hence, the value of x = 26.

Calculate the median of the following frequency distribution:

Answer

Cumulative frequency distribution table :

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

By formula,

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

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

∴ Weight of 21st student = 50.

Hence, median weight = 50 kg.

Find the median of the following frequency distribution:

Answer

The given varieties are arranged in ascending order.

Cumulative frequency distribution table :

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

By formula,

$\Rightarrow \text{Median} = \dfrac{37 + 1}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = \dfrac{38}{2} \text{th} \text{ observation} \\[1em] \Rightarrow \text{Median} = 19 \text{ th} \text{ observation}$

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

∴ 19th observation = 22.

Hence, median = 22.

50 persons were examined through X-ray and observations were noted as under:

Find :

(i) The mean diameter of heart,

(ii) The median diameter of heart.

Answer

(i) We know that,

$\Rightarrow \text{Mean} = \dfrac{\text{Σfx}}{\text{Σf}} \\[1em] \Rightarrow \text{Mean} = \dfrac{6133}{50} \\[1em] \Rightarrow \text{Mean} = 122.66 \text{ mm}.$

Hence, mean diameter of heart = 122.66 mm.

(ii) Here,

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

By formula,

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

From table,

Diameter of heart corresponding to 25th term is 122

Diameter of heart corresponding to 26th term is 123

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

Hence, median diameter of heart = 122.5 mm.

The marks scored by 15 students in a class test are:

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

Find :

(i) Median

(ii) Lower quartile (Q₁)

(iii) Upper quartile (Q₃)

(iv) Interquartile range

(v) Semi-interquartile range

Answer

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,

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

Hence, median = 17.

(ii) By formula,

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

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

= 4th term

= 8

Hence, lower quartile (Q₁) = 8.

(iii) By formula,

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

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

= $\Big(\dfrac{3 \times 16}{4}\Big) = \dfrac{48}{4}$ 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 = $\dfrac{1}{2}$ × Inter quartile range

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

= 8.5

Hence, semi-interquartile range = 8.5.

Find :

(i) Median

(ii) Lower quartile (Q₁)

(iii) Upper quartile (Q₃)

(iv) Interquartile range

(v) Semi-interquartile range for the following series :

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

Answer

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,

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

Hence, Median = 13.5.

(ii) By formula,

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

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

= 3 rd term

= 5.

Hence, lower quartile (Q₁) = 5.

(iii) By formula,

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

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

= $\Big(\dfrac{36}{4}\Big)$ 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 = $\dfrac{1}{2}$ × Inter quartile range

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

= 6.5

Hence, semi-interquartile range = 6.5.

From the following frequency distribution, find:

(i) Median

(ii) Lower quartile

(iii) Upper quartile

(iv) Semi-interquartile range

Answer

The given varieties are arranged in ascending order.

Cumulative frequency distribution table :

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

(i) By formula,

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

From table,

30th term is 20

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

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

Hence, median = 21.

(ii) By formula,

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

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

= 15 th term

= 18.

Hence, lower quartile = 18.

(iii) By formula,

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

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

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

= 45 th term

= 22

Hence, Upper Quartile = 22.

(iv) By formula,

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

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

= 2

Hence, semi-interquartile range = 2.

The heights (in nearest cm) of 63 students of a certain school are given in the following frequency distribution table:

Find :

(i) Median

(ii) Lower quartile (Q₁)

(iii) Upper quartile (Q₃)

(iv) Interquartile range from the above data.

Answer

The given varieties are arranged in ascending order.

Cumulative frequency distribution table :

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

(i) By formula,

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

From table,

32 nd term is 153.

Hence, median = 153.

(ii) By formula,

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

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

= 16 th term

From table,

16 th term is 151.

Hence, lower quartile = 151.

(iii) By formula,

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

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

= $\Big(\dfrac{3 \times 64}{4}\Big) = \dfrac{192}{4}$ 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.

From the following frequency distribution find :

(i) Median

(ii) Lower quartile (Q₁)

(iii) Upper quartile (Q₃)

(iv) Interquartile range

Answer

The given varieties are arranged in ascending order.

Cumulative frequency distribution table :

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

(i) By formula,

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

From table,

30th term = 20

31st term = 23

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

Hence, median = 21.5.

(ii) By formula,

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

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

= 15 th term

= 18.

Hence, lower quartile = 18.

(iii) By formula,

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

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

= $\Big(\dfrac{180}{4}\Big)$ 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.