RD Sharma solution class 7 chapter 23 Data Handling II(Central values) Exercise 23.3

Exercise 23.3

Page-23.16

Question 1:

Find the median of the following data (1-8)
83, 37, 70, 29, 45, 63, 41, 70, 34, 54

Answer 1:

 Arranging the data in ascending order, we have:
 

              29, 34, 37, 41, 45, 54, 63, 70, 70, 83

Here, the number of observations, n = 10 (Even).

  $$\begin{gathered} \Rightarrow \mathrm{Median}=\frac{{\left({\displaystyle \frac{\mathrm{n}}{2}}\right)}^{\mathrm{th}} \mathrm{observation} +{\left({\displaystyle \frac{\mathrm{n}}{2}} +1\right)}^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median}=\frac{\mathrm{Value} \mathrm{of} 5^{\mathrm{th}} \mathrm{observation} +\mathrm{Value} \mathrm{of} 6^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median}=\frac{(45 + 54)}{2} = 49.5 \end{gathered}$$
Hence, the median of the given data is 49.5.

Question 2:

Find the median of the following data (1-8)
133, 73, 89, 108, 94, 104, 94, 85, 100, 120

Answer 2:

 Arranging the data in ascending order, we have:

 73, 85, 89, 94, 94, 100, 104, 108, 120, 133.

Here, the number of observations n = 10 (Even).

  $$\begin{gathered} \Rightarrow \mathrm{Median}=\frac{{\left({\displaystyle \frac{n}{2}}\right)}^{\mathrm{th}} \mathrm{observation} +{\left({\displaystyle \frac{n}{2}} +1\right)}^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median}=\frac{\mathrm{Value} \mathrm{of} 5^{\mathrm{th}} \mathrm{observation} +\mathrm{Value} \mathrm{of} 6^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median}=\frac{(94 + 100)}{2} = \frac{194}{2} \\ \Rightarrow \mathrm{Median}=97 \end{gathered}$$
Hence, the median of the given data is 97.

Question 3:

Find the median of the following data (1-8)
31, 38, 27, 28, 36, 25, 35, 40

Answer 3:

 Arranging the data in ascending order, we have:

              25,27, 28, 31, 35, 36, 38, 40

Here, the number of observations n = 8 (Even).

  $$\begin{gathered} \Rightarrow \mathrm{Median}=\frac{{\left({\displaystyle \frac{n}{2}}\right)}^{\mathrm{th}} \mathrm{observation} +{\left({\displaystyle \frac{n}{2}} +1\right)}^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median}=\frac{\mathrm{Value} \mathrm{of} 4^{\mathrm{th}} \mathrm{observation} +\mathrm{Value} \mathrm{of} 5^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median}=\frac{(31 + 35)}{2} = \frac{66}{2} \\ \Rightarrow \mathrm{Median}=33 \end{gathered}$$
Hence, the median of the given data is 33.

Question 4:

Find the median of the following data (1-8)
15, 6, 16, 8, 22, 21, 9, 18, 25

Answer 4:

 Arranging the data in ascending order, we have:

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

Here, the number of observations n = 9 (Odd).

  $\Rightarrow \mathrm{Median}=\mathrm{Value} \mathrm{of} \left({\displaystyle \frac{n+1}{2}}\right)$^th observation, i.e., value of the 5^th observation =16. 
Hence, the median of the given data is 16.

Question 5:

Find the median of the following data (1-8)
41, 43, 127, 99, 71, 92, 71, 58, 57

Answer 5:

 Arranging the given data in ascending order, we have:

  41, 43, 57, 58, 71,71, 92, 99, 127 
Here, n = 9, which is odd.

∴ Median = Value of ${\left(\frac{9+1}{2}\right)}^{th}$ observation, i.e., the 5^th observation = 71.

Question 6:

Find the median of the following data (1-8)
25, 34, 31, 23, 22, 26, 35, 29, 20, 32

Answer 6:

 Arranging the given data in ascending order, we have:

  20, 22, 23, 25, 26, 29, 31, 32, 34, 35

Here, n = 10, which is even.

 $$\begin{gathered} \mathrm{Median} = \frac{\mathrm{Value} \mathrm{of} {\left({\displaystyle \frac{n}{2}}\right)}^{\mathrm{th}} \mathrm{observation} + \mathrm{Value} \mathrm{of} {\left({\displaystyle \frac{n}{2} +1}\right)}^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median} =\frac{\mathrm{Value} \mathrm{of} {\left({\displaystyle 5}\right)}^{\mathrm{th}} \mathrm{observation} + \mathrm{Value} \mathrm{of} {\left({\displaystyle 6}\right)}^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median} =\frac{26 +29}{2} = \frac{55}{2} \\ \Rightarrow \mathrm{Median} =27.5 \end{gathered}$$
Hence, the median is 27.5 for the given data.

Question 7:

Find the median of the following data (1-8)
12, 17, 3, 14, 5, 8, 7, 15

Answer 7:

 Arranging the given data in ascending order, we have:

 3,5,7,8,12,14,15,17

Here, n = 8, which is even.

 $$\begin{gathered} \mathrm{Median} = \frac{\mathrm{Value} \mathrm{of} {\left({\displaystyle \frac{n}{2}}\right)}^{\mathrm{th}} \mathrm{observation} + \mathrm{Value} \mathrm{of} {\left({\displaystyle \frac{n}{2} +1}\right)}^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median} =\frac{\mathrm{Value} \mathrm{of} {\left({\displaystyle 4}\right)}^{\mathrm{th}} \mathrm{observation} + \mathrm{Value} \mathrm{of} {\left({\displaystyle 5}\right)}^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median} =\frac{8 +12}{2} \\ \Rightarrow \mathrm{Median} =10. \end{gathered}$$
Hence, the median of the given data is 10.

Question 8:

Find the median of the following data (1-8)
92, 35, 67, 85, 72, 81, 56, 51, 42, 69

Answer 8:

 Arranging the given data in ascending order, we have:

 35, 42, 51, 56, 67, 69, 72, 81, 85, 92

Here, n = 10, which is even.

 $$\begin{gathered} \mathrm{Median} = \frac{\mathrm{Value} \mathrm{of} {\left({\displaystyle \frac{n}{2}}\right)}^{\mathrm{th}} \mathrm{observation} + \mathrm{Value} \mathrm{of} {\left({\displaystyle \frac{n}{2} +1}\right)}^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median} =\frac{\mathrm{Value} \mathrm{of} {\left({\displaystyle 5}\right)}^{\mathrm{th}} \mathrm{observation} + \mathrm{Value} \mathrm{of} {\left({\displaystyle 6}\right)}^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow \mathrm{Median} =\frac{67 +69}{2} = \frac{136}{2} \\ \Rightarrow \mathrm{Median} =68. \end{gathered}$$
Hence, the median of the given data is 68.

Question 9:

Numbers 50, 42, 35, 2x + 10, 2x − 8, 12, 11, 8, 6 are written in descending order and their median is 25, find x.

Answer 9:

Here, the number of observations n is 9. Since n is odd , the median is the ${\left(\frac{n+1}{2}\right)}^{th}$ observation, i.e. the 5^th  observation.
As the numbers are arranged in the descending order, we therefore observe from the last.
Median =  ​5^th  observation.
$\Rightarrow$ 25 = 2x -8
$\Rightarrow$ 2x = 25 +8
$\Rightarrow$ 2x = 33
$\Rightarrow$x = $\frac{33}{2}$
$\Rightarrow$x = 16.5
Hence, ​x = 16.5.

Question 10:

Find the median of the following observations : 46, 64, 87, 41, 58, 77, 35, 90, 55, 92, 33. If 92 is replaced by 99 and 41 by 43 in the above data, find the new median?

Answer 10:

 Arranging the given data in ascending order, we have:

      33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

  Here, the number of observations n is 11 (odd).

Since the number of observations is odd, therefore, 
 
Median = Value of ${\left(\frac{n+1}{2}\right)}^{th}$ observation = Value of the 6^th observation = 58.
Hence, median = 58.
           
If 92 is replaced by 99 and 41 by 43, then the new observations arranged in ascending order are:

 33, 35, 43, 46, 55, 58, 64, 77, 87, 90, 99.

∴ New median =  Value of the 6^th observation = 58.

Question 11:

Find the median of the following data : 41, 43, 127, 99, 61, 92, 71, 58, 57, If 58 is replaced by 85, what will be the new median?

Answer 11:

 Arranging the given data in ascending order, we have:

  41, 43, 57, 58, 61, 71, 92, 99,127

  Here, the number of observations, n, is 9(odd).

∴ Median = Value of ${\left(\frac{n+1}{2}\right)}^{th}$ observation = Value of the 5^th observation = 61.
Hence, the median = 61.
If 58 is replaced by 85 , then the new observations arranged in ascending order are:
 41, 43, 57,  61, 71, 85, 92, 99,12 .
∴ New median =  Value of the 5^th observation = 71.

Question 12:

The weights (in kg) of 15 students are : 31, 35, 27, 29, 32, 43, 37, 41, 34, 28, 36, 44, 45, 42, 30. Find the median. If the weight 44 kg is replaced by 46 kg and 27 kg by 25 kg, find the new median.

Answer 12:

Arranging the given data in ascending order, we have:

  27, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 44, 45.

  Here, the number of observations n is 15(odd).

   Since the number of observations is odd, therefore, 
   Median = Value of ${\left(\frac{n+1}{2}\right)}^{th}$ observation = Value of the 8^th observation = 35.
  Hence, median = 35 kg.

If 44 kg is replaced by 46 kg and 27 kg by 25 kg , then the new observations arranged in ascending order are:
25, 28, 29, 30, 31, 32, 34, 35, 36, 37, 41, 42, 43, 45, 46.
 ∴ New median =  Value of the 8^th observation = 35 kg.

Question 13:

The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x :
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95

Answer 13:

Here, the number of observations n is 10. Since n is even,

 $$\begin{gathered} \mathrm{Median} = \frac{{\left(\frac{n}{2}\right)}^{th} \mathrm{observation}+ {\left(\frac{n}{2}+1\right)}^{th}\mathrm{observation}}{2} \\ \mathrm{Median}=\frac{\mathrm{Value} \mathrm{of} 5^{\mathrm{th}} \mathrm{observation}+\mathrm{Value} \mathrm{of} 6^{\mathrm{th}} \mathrm{observation}}{2} \\ \Rightarrow 63 =\frac{x+ (x+2)}{2} \\ \Rightarrow 63 =\frac{2x+2}{2} = \frac{2(x+1)}{2} \\ \Rightarrow 63 = x+1 \\ \Rightarrow x = 63-1 = 62. \end{gathered}$$
Hence, ​x = 62.