SELINA Solution Class 9 Chapter 19 Means and median (For Unground Data Only) Exercise 19B

Question 1.1

Find the median of: 25, 16, 26, 16, 32, 31, 19, 28 and 35

Sol:

 Firstly arrange the numbers in ascending order
16, 16, 19, 25, 26, 28, 31, 32, 35

Now since
n = 9( odd )

Therefore the Median

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

= ${\displaystyle {\left(\frac{9+1}{2}\right)}^{\text{th}}}$  

= 5^th

Thus the median is 26.

Question 1.2

Find the median of:
241, 243, 347, 350, 327, 299, 261, 292, 271, 258 and 257

Sol:

Firstly arrange the numbers in ascending order

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

Now since n = 11 ( Odd )

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

             = 6^th term
             = 271
Thus the median is 271.

Question 1.3

Find the median of:

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

Sol:

 Firstly arrange the numbers in ascending order

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

Now since n = 10( even )

Median =${\displaystyle \frac{1}{2}[\text{value of}{\left(\frac{n}{2}\right)}^{\text{th}}\text{term}+\text{value of} {(\frac{n}{2}+1)}^{\text{th}}\text{term}]}$

=${\displaystyle \frac{1}{2}[\text{value of}{\left(\frac{10}{2}\right)}^{\text{th}}\text{term}+\text{value of} {(\frac{10}{2} +1)}^{\text{th}}\text{term}]}$

= ${\displaystyle \frac{1}{2}[25+34]}$

= ${\displaystyle \frac{1}{2}\left[59\right]}$

= 29.5

Thus the median is 29.5.

Question 1.4

Find the median of:

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

Sol:

 Firstly arrange the numbers in ascending order

173, 185, 189, 194, 194, 200, 204, 208, 220, 223

Median =${\displaystyle \frac{1}{2}[\text{value of}{\left(\frac{n}{2}\right)}^{\text{th}}\text{term}+\text{value of} {(\frac{n}{2}+1)}^{\text{th}}\text{term}]}$

=${\displaystyle \frac{1}{2}[\text{value of}{\left(\frac{10}{2}\right)}^{\text{th}}\text{term}+\text{value of} {(\frac{10}{2} +1)}^{\text{th}}\text{term}]}$

= ${\displaystyle \frac{1}{2}[200+194]}$

= ${\displaystyle }$${\displaystyle \frac{1}{2}\left[394\right]}$

= 197

Thus the mention is 197.

Question 2

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.

Sol:

Given numbers are 34, 37, 53, 55, x, x+2, 77, 83, 89, 100

Here n = 10(even)

Median = ${\displaystyle \frac{1}{2}[\text{value of}{\left(\frac{n}{2}\right)}^{\text{th}}\text{term}+\text{value of} {(\frac{n}{2}+1)}^{\text{th}}\text{term}]}$

= ${\displaystyle \frac{1}{2}[\text{value of}{\left(\frac{10}{2}\right)}^{\text{th}}\text{term}+\text{value of} {(\frac{10}{2}+1)}^{\text{th}}\text{term}]}$

= ${\displaystyle \frac{1}{2}}$ [ value of ( 5 )^th term + value of ( 5 + 1)^th term ]

= ${\displaystyle \frac{1}{2}}$ [ value of ( 5 )^th term + value of (6)^th term ]

63 = ${\displaystyle \frac{1}{2}}$ [  x + x + 2 ]

⇒ ${\displaystyle \frac{2+2x}{2}}$ = 63

⇒  x + 1 = 63

⇒ x = 62

Question 3

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

Sol:

For any given set of data, the median is the value of its middle term.

Here, total observations = n = 10 (even)

If n is even, we have

Median =${\displaystyle \frac{1}{2}[\text{value of} {\left(\frac{n}{2}\right)}^{\text{th}}\text{term}+\text{value of} {(\frac{n}{2}+1)}^{\text{th}}\text{term}]}$

Thus, for n = 10, we have

Median =${\displaystyle \frac{1}{2}[\text{value of}{\left(\frac{10}{2}\right)}^{\text{th}}\text{term}+\text{value of} {(\frac{10}{2}+1)}^{\text{th}}\text{term}]}$

=${\displaystyle \frac{1}{2}[\text{value of}{5}^{\text{th}}\text{term}+\text{value of}{6}^{\text{th}}\text{term}]}$

Hence, if 7^th number is diminished by 8, there is no change in the median value.

Question 4

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.

Sol:

Here, total observations = n = 10 (even)
Thus, we have

Median =${\displaystyle \frac{1}{2}[\text{value of}{\left(\frac{10}{2}\right)}^{\text{th}}\text{term}+\text{value of}{(\frac{10}{2}+1)}^{\text{th}}\text{term}]}$

=${\displaystyle \frac{1}{2}[\text{valueof}{5}^{\text{th}}\text{term}+\text{value of}{6}^{\text{th}}\text{term}]}$

According to the given information, data in ascending order is as follows:

	1 st Term	2nd Term	3rd Term	4th Term	5th term	6th Term	7th Term	8th Term	9th term	10th term
Marks	Less than 30			35	40	48	66	More than 75

∴ Median =${\displaystyle \frac{1}{2}(40+48)=\frac{88}{2}=44}$

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

Question 5

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.

Sol:

Total number of observations = 9(odd)
Now, if n = odd
Median = ${\displaystyle {\left(\frac{n+1}{2}\right)}^{\text{th}} \text{term}}$

⇒ Median = ${\displaystyle {\left(\frac{9+1}{2}\right)}^{\text{th}}\text{term}}$ = 5^th term = x + 5

Now, Median = 18    ...(given)
∴ x + 5 = 18
⇒ x = 13.