RS AGGARWAL CLASS 9 CHAPTER 18 MEAN, MEDIAN AND MODE OF UNGROUPED DATA EXERCISE 18C

 EXERCISE 18C

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Question 1:

Find the median of
(i) 2, 10, 9, 9, 5, 2, 3, 7, 11
(ii) 15, 6, 16, 8, 22, 21, 9, 18, 25
(iii) 20, 13, 18, 25, 6, 15, 21, 9, 16, 8, 22
(iv) 7, 4, 2, 5, 1, 4, 0, 10, 3, 8, 5, 9, 2

Answer 1:

(i) Arranging the numbers in ascending order, we get:
      2, 2, 3, 5, 7, 9, 9, 10, 11
Here, n is 9, which is an odd number.
If n is an odd number, we have:
$\mathrm{Median}= \mathrm{Value} \mathrm{of} \left(\frac{n+1}{2}\right)\mathrm{th} \mathrm{observation}$
Now,
$$\begin{gathered} \mathrm{Median}=\mathrm{Value} \mathrm{of} \left(\frac{9+1}{2}\right)\mathrm{th} \mathrm{observation} \\ =\mathrm{Value} \mathrm{of} \mathrm{the} 5\mathrm{th} \mathrm{observation} \\ =7 \end{gathered}$$

(ii) Arranging the numbers in ascending order, we get:
6, 8, 9, 15, 16, 18, 21, 22, 25
Here, n is 9, which is an odd number.
If n is an odd number, we have:
$\mathrm{Median}= \mathrm{Value} \mathrm{of} \left(\frac{n+1}{2}\right)\mathrm{th} \mathrm{observation}$
Now,
$$\begin{gathered} \mathrm{Median}=\mathrm{Value} \mathrm{of} \left(\frac{9+1}{2}\right)\mathrm{th} \mathrm{observation} \\ =\mathrm{Value} \mathrm{of} \mathrm{the} 5\mathrm{th} \mathrm{observation} \\ =16 \end{gathered}$$

(iii) Arranging the numbers in ascending order, we get:
    6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25
Here, n is 11, which is an odd number.
If n is an odd number, we have:
$\mathrm{Median}= \mathrm{Value} \mathrm{of} \left(\frac{n+1}{2}\right)\mathrm{th} \mathrm{observation}$
Now,
$$\begin{gathered} \mathrm{Median}=\mathrm{Value} \mathrm{of} \left(\frac{11+1}{2}\right)\mathrm{th} \mathrm{observation} \\ =\mathrm{Value} \mathrm{of} \mathrm{the} 6\mathrm{th} \mathrm{observation} \\ =16 \end{gathered}$$

(iv) Arranging the numbers in ascending order, we get:
   0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10
Here, n is 13, which is an odd number.
If n is an odd number, we have:
$\mathrm{Median}= \mathrm{Value} \mathrm{of} \left(\frac{n+1}{2}\right)\mathrm{th} \mathrm{observation}$
Now,
$$\begin{gathered} \mathrm{Median}=\mathrm{Value} \mathrm{of} \left(\frac{13+1}{2}\right)\mathrm{th} \mathrm{observation} \\ =\mathrm{Value} \mathrm{of} \mathrm{the} 7\mathrm{th} \mathrm{observation} \\ =4 \end{gathered}$$

Question 2:

Find the median of
(i) 17, 19, 32, 10, 22, 21, 9, 35
(ii) 72, 63, 29, 51, 35, 60, 55, 91, 85, 82
(iii) 10, 75, 3, 15, 9, 47, 12, 48, 4, 81, 17, 27

Answer 2:

(i) Arranging the numbers in ascending order, we get:
9, 10, 17, 19, 21, 22, 32, 35
Here, n is 8, which is an even number.
If n is an even number, we have:
$\mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{n}{2}\right)\mathrm{th} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observations}$
Now,
$$\begin{gathered} \mathrm{Median} =\mathrm{Mean} \mathrm{of} \left(\frac{8}{2}\right)\mathrm{th} \& \left(\frac{8}{2}+1\right)\mathrm{th} \mathrm{observations} \\ =\mathrm{Mean} \mathrm{of} \mathrm{the} 4\mathrm{th} \& 5\mathrm{th} \mathrm{observations} \\ =\frac{1}{2}\left(19+21\right) \\ = 20 \end{gathered}$$

(ii) Arranging the numbers in ascending order, we get:
   29, 35, 51, 55, 60, 63, 72, 82, 85, 91
Here, n is 10, which is an even number.
If n is an even number, we have:
$\mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{n}{2}\right)\mathrm{th} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observations}$
Now,
$$\begin{gathered} \mathrm{Median} =\mathrm{Mean} \mathrm{of} \left(\frac{10}{2}\right)\mathrm{th} \& \left(\frac{10}{2}+1\right)\mathrm{th} \mathrm{observations} \\ =\mathrm{Mean} \mathrm{of} \mathrm{the} 5\mathrm{th} \& 6\mathrm{th} \mathrm{observations} \\ =\frac{1}{2}\left(60+63\right) \\ =61.5 \end{gathered}$$

(iii) Arranging the numbers in ascending order, we get:
    3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81
Here, n is 12, which is an even number.
If n is an even number, we have:
$\mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{n}{2}\right)\mathrm{th} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observations}$
Now,
$$\begin{gathered} \mathrm{Median} =\mathrm{Mean} \mathrm{of} \left(\frac{12}{2}\right)\mathrm{th} \& \left(\frac{12}{2}+1\right)\mathrm{th} \mathrm{observations} \\ =\mathrm{Mean} \mathrm{of} \mathrm{the} 6\mathrm{th} \& 7\mathrm{th} \mathrm{observations} \\ =\frac{1}{2}\left(15+17\right) \\ =16 \end{gathered}$$

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Question 3:

The marks of 15 students in an examination are:
25, 19, 17, 24, 23, 29, 31, 40, 19, 20, 22, 26, 17, 35, 21.
Find the median score.

Answer 3:

Arranging the marks of 15 students in ascending order, we get:
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40
Here, n is 15, which is an odd number.
We know:
 $\mathrm{Median}=\mathrm{Value} \mathrm{of} \left(\frac{n+1}{2}\right)\mathrm{th} \mathrm{observation}$
Thus, we have:
$$\begin{gathered} \mathrm{Median} \mathrm{score}=\mathrm{Value} \mathrm{of} \left(\frac{15+1}{2}\right) \mathrm{th} \mathrm{observation} \\ =\mathrm{Value} \mathrm{of} \mathrm{the} 8\mathrm{th} \mathrm{observation} \\ =23 \end{gathered}$$

Question 4:

The heights (in cm) of 9 students of a class are 148, 144, 152, 155, 160, 147, 150, 149, 145.
Find the median height.

Answer 4:

Arranging the given data in ascending order:
144, 145, 147, 148, 149, 150, 152, 155, 160

Number of terms = 9 (odd)
$$\begin{gathered} \therefore \mathrm{Median}={\left(\frac{n+1}{2}\right)}^{\mathrm{th}} \mathrm{term} \\ ={\left(\frac{9+1}{2}\right)}^{\mathrm{th}} \mathrm{term} \\ =5^{\mathrm{th}} \mathrm{term} \\ =149 \end{gathered}$$

Hence, the median height is 149.

Question 5:

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 5:

Arranging the weights (in kg) in ascending order, we have:
9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2
Here, n is 8, which is an even number.
Thus, we have:
$\mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{n}{2}\right)\mathrm{th} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observations}$
$$\begin{gathered} \mathrm{Median} \mathrm{weight}=\mathrm{Mean} \mathrm{of} \left(\frac{8}{2}\right)\mathrm{th} \& \left(\frac{8}{2}+1\right)\mathrm{th} \mathrm{observations} \\ =\mathrm{Mean} \mathrm{of} 4\mathrm{th} \& 5\mathrm{th} \mathrm{observations} \\ =\frac{1}{2}\left(13.4+14.3\right) \\ =13.85 \end{gathered}$$

Hence, the median weight is 13.85 kg.

Question 6:

The ages (in years) of 10 teachers in a school are:
32, 44, 53, 47, 37, 54, 34, 36, 40, 50.
Find the median age.

Answer 6:

Arranging the ages (in years) in ascending order, we have:
32, 34, 36, 37, 40, 44, 47, 50, 53, 54
Here, n is 10, which is an even number.
Thus, we have:
$\mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{n}{2}\right)\mathrm{th} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observations}$
$$\begin{gathered} \mathrm{Median} \mathrm{age} =\mathrm{Mean} \mathrm{of} \left(\frac{10}{2}\right)\mathrm{th} \& \left(\frac{10}{2}+1\right)\mathrm{th} \mathrm{observations} \\ = \mathrm{Mean} \mathrm{of} 5\mathrm{th} \& 6\mathrm{th} \mathrm{observations} \\ =\frac{1}{2}\left(40+44\right) \\ =42 \\ \mathrm{Hence}, \mathrm{the} \mathrm{median} \mathrm{age} \mathrm{is} 42 \mathrm{years}. \end{gathered}$$

Question 7:

If 10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41 are ten observation in an ascending order with median 24, find the value of x.

Answer 7:

10, 13, 15, 18, x+1, x+3, 30, 32, 35 and 41 are arranged in ascending order.
Median = 24
We have to find the value of x.
Here, n is 10, which is an even number.
Thus, we have:
$\mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{n}{2}\right)\mathrm{th} \& \left(\frac{n}{2}+1\right)\mathrm{th} \mathrm{observations}$
$$\begin{gathered} \therefore \mathrm{Median}=\mathrm{Mean} \mathrm{of} \left(\frac{10}{2}\right)\mathrm{th} \& \left(\frac{10}{2}+1\right)\mathrm{th} \mathrm{observations} \\ =\mathrm{Mean} \mathrm{of} 5\mathrm{th} \& 6\mathrm{th} \mathrm{observations} \\ =\frac{1}{2}\left(x+1+x+3\right) \\ =\frac{1}{2}\left(2x+4\right) \\ =(x+2) \\ \mathrm{Given}: \\ \mathrm{Median}=24 \\ \Rightarrow x+2=24 \\ \Rightarrow x=22 \end{gathered}$$

Question 8:

The following observations are arranged in ascending order:
26, 29, 42, 53, x, x + 2, 70, 75, 82, 93.
If the median is 65, find the value of x.

Answer 8:

Arranging the given data in ascending order:
26, 29, 42, 53, x, x + 2, 70, 75, 82, 93

Number of terms = 10 (even)

$$\begin{gathered} \therefore \mathrm{Median}=\mathrm{mean} \mathrm{of} \left[{\left(\frac{n}{2}\right)}^{\mathrm{th}} \mathrm{term} \mathrm{and} {\left(\frac{n}{2}+1\right)}^{\mathrm{th}} \mathrm{term}\right] \\ \Rightarrow 65=\mathrm{mean} \mathrm{of} \left[{\left(\frac{10}{2}\right)}^{\mathrm{th}} \mathrm{term} \mathrm{and} {\left(\frac{10}{2}+1\right)}^{\mathrm{th}} \mathrm{term}\right] \\ \Rightarrow 65=\mathrm{mean} \mathrm{of} \left[{\left(5\right)}^{\mathrm{th}} \mathrm{term} \mathrm{and} {\left(6\right)}^{\mathrm{th}} \mathrm{term}\right] \\ \Rightarrow 65=\mathrm{mean} \mathrm{of} \left[x \mathrm{and} x+2\right] \\ \Rightarrow 65=\frac{x+x+2}{2} \\ \Rightarrow 65\times 2=2x+2 \\ \Rightarrow 130=2x+2 \\ \Rightarrow 2x=130-2 \\ \Rightarrow 2x=128 \\ \Rightarrow x=64 \end{gathered}$$

Hence, the value of x is 64.

Question 9:

The numbers 50, 42, 35, (2x + 10), (2x – 8), 12, 11, 8 have been written in a descending order. If their median is 25, find the value of x.

Answer 9:

Arranging the given data in ascending order:
8, 11, 12, (2x – 8), (2x + 10), 35, 42, 50

Number of terms = 8 (even)

$$\begin{gathered} \therefore \mathrm{Median}=\mathrm{mean} \mathrm{of} \left[{\left(\frac{n}{2}\right)}^{\mathrm{th}} \mathrm{term} \mathrm{and} {\left(\frac{n}{2}+1\right)}^{\mathrm{th}} \mathrm{term}\right] \\ \Rightarrow 25=\mathrm{mean} \mathrm{of} \left[{\left(\frac{8}{2}\right)}^{\mathrm{th}} \mathrm{term} \mathrm{and} {\left(\frac{8}{2}+1\right)}^{\mathrm{th}} \mathrm{term}\right] \\ \Rightarrow 25=\mathrm{mean} \mathrm{of} \left[{\left(4\right)}^{\mathrm{th}} \mathrm{term} \mathrm{and} {\left(5\right)}^{\mathrm{th}} \mathrm{term}\right] \\ \Rightarrow 25=\mathrm{mean} \mathrm{of} \left[2x-8 \mathrm{and} 2x+10\right] \\ \Rightarrow 25=\frac{2x-8+2x+10}{2} \\ \Rightarrow 25\times 2=4x+2 \\ \Rightarrow 50=4x+2 \\ \Rightarrow 4x=50-2 \\ \Rightarrow 4x=48 \\ \Rightarrow x=12 \end{gathered}$$

Hence, the value of x is 12.

Question 10:

Find the median of the data
46, 41, 77, 58, 35, 64, 87, 92, 33, 55, 90.
In the above data, if 41 and 55 are replaced by 61 and 75 respectively, what will be the new median?

Answer 10:

Arranging the given data in ascending order:
33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

Number of terms = 11 (odd)

$$\begin{gathered} \therefore \mathrm{Median}={\left(\frac{n+1}{2}\right)}^{\mathrm{th}} \mathrm{term} \\ ={\left(\frac{11+1}{2}\right)}^{\mathrm{th}} \mathrm{term} \\ =6^{\mathrm{th}} \mathrm{term} \\ =58 \end{gathered}$$

Hence, the median of the data is 58.

Now, In the above data, if 41 and 55 are replaced by 61 and 75 respectively.
Then, new data in ascending order is:
33, 35, 46, 58, 61, 64, 75, 77, 87, 90, 92

Number of terms = 11 (odd)

$$\begin{gathered} \therefore \mathrm{Median}={\left(\frac{n+1}{2}\right)}^{\mathrm{th}} \mathrm{term} \\ ={\left(\frac{11+1}{2}\right)}^{\mathrm{th}} \mathrm{term} \\ =6^{\mathrm{th}} \mathrm{term} \\ =64 \end{gathered}$$

Hence, the new median of the data is 64.