Mean & Median of Ungrouped Data & Frequency Polygon — Exercise 16(B)

Find the median of :

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

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

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

Answer

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

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

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

By formula,

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

⇒ $\dfrac{9 + 1}{2}$ th observation

⇒ $\dfrac{10}{2}$th observation

⇒ 5th observation

∴ Median = 16.

Hence, median is 16.

(ii) 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,

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

∴ Median = 16.

Hence, median = 16.

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

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

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

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{10}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{10}{2} + 1\right)^{\text{th}} \text{term}}{2}\\[1em] \Rightarrow \text{Median} = \dfrac{\text{5th term} + \text{6th term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{60} + \text{63}}{2}\\[1em] \Rightarrow \text{Median} = \dfrac{\text{123}}{2}$

∴ Median = 61.5.

Hence, median is 61.5.

The runs scored by 11 members of a cricket team are :

26, 38, 53, 18, 66, 72, 0, 47, 32, 7, 35

Find the median score.

Answer

By arranging data in ascending order, we get :

0, 7, 18, 26, 32, 35, 38, 47, 53, 66, 72

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

By formula,

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

⇒ $\dfrac{11 + 1}{2}$ th observation

⇒ $\dfrac{12}{2}$th observation

⇒ 6th observation

∴ Median = 35.

Hence, median is 35 runs.

The heights (in cm) of 9 girls are :

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

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,

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

⇒ $\dfrac{9 + 1}{2}$ th observation

⇒ $\dfrac{10}{2}$th observation

⇒ 5th observation

∴ Median = 147.3.

Hence, median is 147.3 cm.

The age(in years) of 10 teachers in a school are :

34, 37, 53, 46, 52, 43, 31, 36, 40, 50

Find the median age.

Answer

By arranging data in ascending order, we get :

31, 34, 36, 37, 40, 43, 46, 50, 52, 53

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

By formula,

$\text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow\text{Median} = \dfrac{\left(\dfrac{10}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{10}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{5th term} + \text{6th term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{40} + \text{43}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{83}}{2} \\[1em]$

∴ Median = 41.5.

Hence, median is 41.5 years.

The weights (in kg) of 8 children are :

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

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,

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

∴ Median = 13.85.

Hence, median is 13.85 kg.

Find the median weight for the following data:

Answer

Cumulative frequency table :

Total number of observations = 41, which is odd.

By formula,

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

⇒ $\dfrac{41 + 1}{2}$ th observation

⇒ $\dfrac{42}{2}$th observation

⇒ 21st observation

From the table,

From 20th to 28th position, the weight = 50 kg

∴ Median = 50 kg.

Hence, median weight is 50 kg.

Calculate the median for the following frequency distribution :

Answer

By rearranging the variates in the ascending order along with their frequencies, we construct the cumulative frequency as under

Total number of observations = 40, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{40}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{40}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{20th term} + \text{21st term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{7} + \text{10}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{17}}{2} \\[1em]$

∴ Median = 8.5.

Hence, median is 8.5.

The heart of 60 patients were examined through X-ray and the observations obtained are given below :

Find the median.

Answer

Cumulative frequency table :

Total number of observations = 60, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{60}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{60}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{30th term} + \text{31st term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{122} + \text{122}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{244}}{2} \\[1em]$

∴ Median = 122.

Hence, median = 122 mm.

Find the median for the following data :

Answer

By rearranging the variates in the ascending order along with their frequencies, we construct the cumulative frequency as under

Total number of observations = 60, which is even.

By formula,

$\Rightarrow \text{Median} = \dfrac{\left(\dfrac{n}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{n}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\left(\dfrac{60}{2}\right)^{\text{th}} \text{term} + \left(\dfrac{60}{2} + 1\right)^{\text{th}} \text{term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{30th term} + \text{31st term}}{2} \\[1em] \Rightarrow \text{Median} = \dfrac{\text{20} + \text{23}}{2} \\[1em] \Rightarrow\text{Median} = \dfrac{\text{43}}{2} \\[1em]$

∴ Median = 21.5.

Hence, median is 21.5.