Mean & Median — Exercise 18(B)

12 observations in a data are written in ascending order. If the last (12^th) observation is doubled, the median will increase by :

1. 12

2. 24

3. 0

4. 6

Answer

As we know, the median is the value of the middle term for any given set of data.

If the last (12^th) observation is doubled, the middle term remains unaffected.

Hence, option 3 is the correct option.

12 observations in a data are written in descending order. If the first observation is doubled, the median will :

1. remain same

2. increase by 12

3. decrease by 12

4. change by 12

Answer

As we know, the median is the value of the middle term for any given set of data.

If the first observation is doubled, the middle term remains unaffected.

Hence, option 1 is the correct option.

Median of numbers 10, 12, 9, 8, 10, 12, 10, 6 and 4 is :

1. 12

2. 10

3. 6

4. 9

Answer

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

4, 6, 8, 9, 10, 10, 10, 12, 12

Number of observations, n = 9 (odd)

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

= $\Big[\dfrac{9+1}{2}\Big]^{th}$ term

= $\Big[\dfrac{10}{2}\Big]^{th}$ term

= $5^{th}$ term

= 10

Hence, option 2 is the correct option.

The median of 80 observations is 60. If each observation is doubled, the resulting median will be :

1. 60

2. 20

3. 140

4. 120

Answer

As we know, the median is the value of the middle term for any given set of data.

If each observation is doubled, the middle term will also be doubled.

Given, Median = 60

Therefore, new median = 60 x 2 = 120

Hence, option 4 is the correct option.

If each observation in a data is decreased by two, their :

1. mean decreases by 2, but median remains same.

2. mean remains same, but median decreases, by 2.

3. mean and median both decrease by 2.

4. mean and median both remain the same.

Answer

As we know, the median is the value of the middle term for any given set of data.

If each observation in a data is decreased by 2, both the mean and median will also decrease by 2.

Hence, option 3 is the correct option.

Find the median of :

25, 16, 26, 16, 32, 31, 19, 28 and 35

Answer

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

16, 16, 19, 25, 26, 28, 31, 32, 35

Number of observations, n = 9 (odd)

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

= $\Big[\dfrac{9+1}{2}\Big]^{th}$ term

= $\Big[\dfrac{10}{2}\Big]^{th}$ term

= $5^{th}$ term

= 26

Hence, the median = 26.

Find the median of :

241, 243, 347, 350, 327, 299, 261, 292, 271, 258 and 257

Answer

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

241, 243, 257, 258, 261, 271, 292, 299, 327, 347, 350

Number of observations, n = 11 (odd)

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

= $\Big[\dfrac{11+1}{2}\Big]^{th}$ term

= $\Big[\dfrac{12}{2}\Big]^{th}$ term

= $6^{th}$ term

= 271

Hence, the median = 271.

Find the median of :

63, 17, 50, 9, 25, 43, 21, 50, 14 and 34

Answer

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

9, 14, 17, 21, 25, 34, 43, 50, 50, 63

Number of observations, n = 10 (even)

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

= $\dfrac{1}{2}\Big[\text{the value of} \Big(\dfrac{10}{2}\Big)^{th} + \text{the value of} \Big(\dfrac{10}{2} + 1\Big)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[\text{the value of} (5)^{th} + \text{the value of} (5 + 1)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[25 + 34\Big]$

= $\dfrac{1}{2}\Big[59\Big]$

= 29.5

Hence, the median = 29.5.

Find the median of :

233, 173, 189, 208, 194, 204, 194, 185, 200 and 220

Answer

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

173, 185, 189, 194, 194, 200, 204, 208, 220, 233

Number of observations, n = 10 (even)

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

= $\dfrac{1}{2}\Big[\text{the value of} \Big(\dfrac{10}{2}\Big)^{th} + \text{the value of} \Big(\dfrac{10}{2} + 1\Big)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[\text{the value of} (5)^{th} + \text{the value of} (5 + 1)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[194 + 200\Big]$

= $\dfrac{1}{2}\Big[394\Big]$

= 197

Hence, the median = 197.

The following data have been arranged in ascending order. If their median is 63, find the value of x.

34, 37, 53, 55, x, x + 2, 77, 83, 89 and 100.

Answer

Number of observations, n = 10 (even)

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

⇒ 63 = $\dfrac{1}{2}\Big[\text{the value of} \Big(\dfrac{10}{2}\Big)^{th} + \text{the value of} \Big(\dfrac{10}{2} + 1\Big)^{th}\Big]$ term

⇒ 63 = $\dfrac{1}{2}\Big[\text{the value of} (5)^{th} + \text{the value of} (5 + 1)^{th}\Big]$ term

⇒ 63 = $\dfrac{1}{2}\Big[x + x + 2\Big]$

⇒ 63 = $\dfrac{1}{2}\Big[2x + 2\Big]$

⇒ 63 = x + 1

⇒ x = 63 - 1

⇒ x = 62

Hence, the value of x = 62.

In 10 numbers, arranged in increasing order, the 7th number is increased by 8, how much will the median be changed ?

Answer

Number of observations, n = 10 (even)

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

= $\dfrac{1}{2}\Big[\text{the value of} \Big(\dfrac{10}{2}\Big)^{th} + \text{the value of} \Big(\dfrac{10}{2} + 1\Big)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[\text{the value of} (5)^{th} + \text{the value of} (5 + 1)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[\text{the value of} (5)^{th} + \text{the value of} (6)^{th}\Big]$ term

Since the 7th number is increased by 8, it has no effect on the 5th and 6th terms.

Hence, the median will not change.

Out of 10 students, who appeared in a test, three secured less than 30 marks and 3 secured more than 75 marks. The marks secured by the remaining 4 students are 35, 48, 66 and 40. Find the median score of the whole group.

Answer

Let x1, x2, x3 be the marks less than 30.

Let y1, y2, y3 be the marks greater than 75.

The order of the marks will be:

x1, x2, x3, 35, 40, 48, 66, y1, y2, y3

Number of observations, n = 10 (even)

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

= $\dfrac{1}{2}\Big[\text{the value of} \Big(\dfrac{10}{2}\Big)^{th} + \text{the value of} \Big(\dfrac{10}{2} + 1\Big)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[\text{the value of} (5)^{th} + \text{the value of} (5 + 1)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[\text{the value of} (5)^{th} + \text{the value of} (6)^{th}\Big]$ term

= $\dfrac{1}{2}\Big[40 + 48\Big]$

= $\dfrac{1}{2}\Big[88\Big]$

= 44

Hence, the median score of the whole group is 44.

The median of observations 10, 11, 13, 17, x + 5, 20, 22, 24 and 53 (arranged in ascending order) is 18; find the value of x.

Answer

Number of observations, n = 9 (odd)

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

⇒ 18 = $\Big[\dfrac{9+1}{2}\Big]^{th}$ term

⇒ 18 = $\Big[\dfrac{10}{2}\Big]^{th}$ term

⇒ 18 = $\Big[5\Big]^{th}$ term

⇒ 18 = x + 5

⇒ x = 18 - 5

⇒ x = 13

Hence, the value of x is 13.