 # RS Aggarwal Class 9 Solutions Chapter 14 -Statistics Ex 14F (14.6)

## RS Aggarwal Class 9 Chapter 14 -Statistics Ex 14F (14.6) Solutions Free PDF

Q.1: Find the median of the following

(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, 9, 2

(iv) 7, 4, 2, 5, 1, 4, 0, 10, 3, 8, 5, 9, 2

Sol:

(i) Arranging the data in ascending order, we have: 2, 2, 3, 5, 7, 9, 9, 10, 11. Here n = 9, which is odd.

Median = 12(n+1)$\frac{1}{2}(n+1)$th term

= 12(9+1)$\frac{1}{2}(9+1)$th term = 5th term (value) = 7

Therefore, the median = 7

(ii) Arranging the data in ascending order, we have: 6, 8, 9, 15, 16, 18, 21, 22, 25. Here n = 9, which is odd.

Median = 12(n+1)$\frac{1}{2}(n+1)$th term

= 12(9+1)$\frac{1}{2}(9+1)$th term = 5th term (value) = 16

Therefore, the median = 16

(iii) Arranging the data in ascending order, we have: 6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25. Here n = 11 (odd).

Median = 12(n+1)$\frac{1}{2}(n+1)$th term

= 12(11+1)$\frac{1}{2}(11+1)$th term = 6th term (value) = 16

Therefore, the median = 16

(iv) Arranging the data in ascending order, we have: 0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10. Here n = 13 (odd).

Median = =12(n+1)$=\frac{1}{2}(n+1)$th term

=12(13+1)$\frac{1}{2}(13+1)$th term = 7th term (value) = 4

Therefore, the median = 4

Q.2: Find the median of the following:

(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

Sol:

(i) Arranging the data in ascending order, we have:

9, 10, 17, 19, 21, 22, 32, 35

Here n = 8, which is even

Median = 12[n2]thterm÷[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; \div \left [ \frac{n}{2} +1\right ]th\; term$

= 12[4thterm+5thterm]$\frac{1}{2}[4th \; term + 5th \; term]$  (since n=8)

= 12(19+21)$\frac{1}{2}(19+21)$

= 12(40)=20$\frac{1}{2}(40)=20$

Therefore, the Median = 20

(ii) Arranging the data in ascending order, we have

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

Here n=10, which is even

Median = 12[n2]thterm+[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; + \left [ \frac{n}{2} +1\right ]th\; term$

=12[5thterm+6thterm]$\frac{1}{2}[5th \; term + 6th \; term]$  (since n=10)

= 12(60+63)$\frac{1}{2}(60+63)$

=12(123)=20$\frac{1}{2}(123)=20$

Therefore, the Median = 61.5

(iii) Arranging the data in ascending order, we have:

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

Here n=12, which is even

Median = 12[n2]thterm+[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; + \left [ \frac{n}{2} +1\right ]th\; term$

= 12[6thterm+7thterm]$\frac{1}{2}[ 6th \; term + 7th \; term]$  (since, n=12)

= 12(15+17)$\frac{1}{2}(15+17)$

= 12(32)=16$\frac{1}{2}(32)=16$

Therefore, the Median = 16

Q.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.

Sol:

Arranging the data in ascending order, we have: 17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40

Here n = 15, which is odd.

Median = 12(n+1)$\frac{1}{2}(n+1)$th term

=12(15+1)$\frac{1}{2}(15+1)$th term = value of 8th term = 23

Therefore, the Median score = 23

Q.4: The heights (in cm) of 9 girls are: 144.2, 148.5, 143.7, 149.6, 150, 146.5, 145, 147.3, 152.1. Find the mean height.

Sol:

Arranging the heights of 9 girls in ascending order, we have: 143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1

Here n = 9, which is odd

Median = 12(n+1)$\frac{1}{2}(n+1)$th term

=12(9+1)$\frac{1}{2}(9+1)$th term = value of 5th term = 147.3

Therefore, the Median score = 147.3 cm

Q5: 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.

Sol:

Arranging the weights of 8 children in ascending order, we have: 9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2

Here, n = 8, which is even.

Median = 12[n2]thterm+[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; + \left [ \frac{n}{2} +1\right ]th\; term$

= 12[4thterm+5thterm]$\frac{1}{2}[ 4th \; term + 5th \; term]$  (since n=8)

= 12(13.4+14.3)$\frac{1}{2}(13.4+14.3)$

= 12(27.7)=13.85$\frac{1}{2}(27.7)= 13.85$

Therefore, the Median = 13.85 kg

Q.6: The ages (in years) of 10 teachers in a school are: 32, 44, 53, 47, 97, 54, 34, 54, 34, 36, 40, 50.

Find the median age.

Sol:

Arranging the ages of teachers in ascending order, we have: 32, 34, 36, 37, 40, 44, 47, 50, 53, 54

Here, n = 10 which is even.

Median = 12[n2]thterm+[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; + \left [ \frac{n}{2} +1\right ]th\; term$

= 12[5thterm+6thterm]$\frac{1}{2}[ 5th \; term + 6th \; term]$  (since n=10)

= 12(40+44)$\frac{1}{2}(40+44)$

= 12(84)=42$\frac{1}{2}(84)= 42$

Therefore, the Median age = 42 years

Q.7: If 10, 13, 15, 18, x+1, x+3, 30, 32, 35, 41 are ten observations in an ascending order with median 24, find the value of x.

Sol:

The ten observations in ascending order: 10, 13, 15, 18, x+1, x+3, 30, 32, 35, 41

Here, n = 10 which is even.

Median = 12[n2]thterm+[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; + \left [ \frac{n}{2} +1\right ]th\; term$

= 12[5thterm+6thterm]$\frac{1}{2}[ 5th \; term + 6th \; term]$  (since n=10)

= 12(x+1+x+3)$\frac{1}{2}(x+1+x+3)$

= 12(2x+4)=x+2$\frac{1}{2}(2x+4)= x+2$

Therefore, Median = x+2

But the given median = 24

Therefore, x+2 =24

Thus, x = 22

Q.8: Find the median weight for the following data.

 Weight (in kg) 14 16 48 50 52 54 55 No. of students 8 5 6 9 7 4 2

Sol:

Let us now prepare the cumulative frequency table:

 Weight (in kg) No. of students Cumulative Frequency 45 8 8 46 5 13 48 6 19 50 9 28 52 7 35 54 4 39 55 2 41

Total n = 41, which is odd.

Median Weight = =12(n+1)$=\frac{1}{2}(n+1)$th term

==12(41+1)$=\frac{1}{2}(41+1)$th term

= value of 21st term

Therefore, Median weight = weight of 21st student

But the above table shows that each one of the student from 20th to 28th has 50kg as his weight.

Therefore, the weight of the 21st student will be 50kg

Hence, the median weight = 50 kg

Q.9: Find the median for the following distribution:

 Variants 17 20 22 15 30 25 Frequency 5 9 4 3 10 6

Sol:

Arranging the terms in ascending order, we have:

 Variants 15 17 20 22 25 30 Frequency 3 5 9 4 6 10

Now preparing the cumulative frequency, we have:

 Variants Frequency Cumulative frequency 15 3 3 17 5 8 20 9 17 22 4 21 25 6 27 30 10 37

Here n = 37, which is odd.

Median = =12(n+1)$=\frac{1}{2}(n+1)$th term

= 12(37+1)$\frac{1}{2}(37+1)$th term = value of 19th term = frequency of 19th variant

But the above table shows that the frequency of 18th to 21st term is 22

Therefore, the frequency of 19th term will be 22

Hence, the median weight = 22

Q.10: Calculate the median for the following data.

 Marks 20 9 25 50 40 80 No. of students 6 4 16 7 8 2

Sol:

 Marks 9 20 25 40 50 80 No. of students 4 6 16 8 7 2

Now preparing the cumulative frequency, we have:

 Marks No. of students (Frequency) Cumulative frequency 9 4 4 20 6 10 25 16 26 40 8 34 50 7 41 80 2 43

Here no. of students = 43, which is odd.

Median = =12(n+1)$=\frac{1}{2}(n+1)$th term

==12(43+1)$=\frac{1}{2}(43+1)$th term

= value of 22th term = Marks of 22nd student

But the above table shows that each one of the students from 11th to 26th gets 25 marks.

Therefore, the 22nd student gets 25 marks

Hence, the median of marks = 25

Q.11: The heights (in cm) of 50 students of a class are given below:

 Height (in cm) 156 154 155 151 157 152 153 No. of students 8 4 10 6 7 3 12

Find the median height.

Sol:

Arranging the terms in ascending order, we have:

 Height (in cm) 151 152 153 154 155 156 157 No. of students 6 3 15 4 10 8 7

Now preparing the cumulative frequency, we have:

 Height (in cm) No. of students frequency Cumulative Frequency 151 6 6 152 3 9 153 12 21 154 4 25 155 10 35 156 8 43 157 7 50

Here, n= 50, which is even.

Median = 12[n2]thterm+[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; + \left [ \frac{n}{2} +1\right ]th\; term$

= 12[25thterm+26thterm]$\frac{1}{2}[ 25th \; term + 26th \; term]$  (since n=50)

= 12(154+155)$\frac{1}{2}(154+155)$

= 12(309)=154.5$\frac{1}{2}(309)= 154.5$ cm

Therefore, the Median height = 154.5 cm

Q12: Find the median for the following data:

 Variants 16 18 20 23 25 26 28 30 Frequency 9 8 13 4 4 6 11 5

Sol:

Arrange the terms in ascending order, we have:

 Variants 16 18 20 23 25 26 28 30 Frequency 9 8 13 4 4 6 11 5

Now preparing the cumulative frequency, we have:

 Variants Frequency Cumulative Frequency 16 9 9 18 8 17 20 13 30 23 4 34 25 4 38 26 6 44 28 11 55 30 5 60

Here, n= 60, which is even

Median = 12[n2]thterm+[n2+1]thterm$\frac{1}{2}\left [ \frac{n}{2} \right ]th \; term \; + \left [ \frac{n}{2} +1\right ]th\; term$

= 12[30thterm+31thterm]$\frac{1}{2}[ 30th \; term + 31th \; term]$  (since n=60)

= 12(20+23)$\frac{1}{2}(20+23)$

= 12(43)=21.5$\frac{1}{2}(43)= 21.5$

Therefore, the Median= 21.5