## RS Aggarwal Class 9 Solutions Chapter 14 Statistics Ex 14G

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G.

**Other Exercises**

- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14A
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14B
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14C
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14D
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14E
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14F
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G
- RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14H

**Question 1.**

**Solution:**

Arranging in ascending order, we get:

2,2,3,5,7,9,9,10,11

Here, number of terms is 9 which is odd.

∴ Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 9+1 }{ 2 } \) th term = 5th term = 7 Ans.

(ii) Arranging in ascending order, we get: 6, 8, 9, 15, 16, 18, 21, 22, 25

Here, number of terms is 9 which is odd.

∴ Median = \(\frac { n+1 }{ 2 } \) th term = \(\frac { 9+1 }{ 2 } \) th term = 5th term = 16 Ans.

(iii) Arranging in ascending order, we get: 6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25

Here, number of terms is 11 which is odd.

∴ Median = \(\frac { 11+1 }{ 2 } \) th term = \(\frac { 12 }{ 2 } \) th term = 6th term = 16 Ans.

(iv) Arranging in ascending order, we get:

0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10

Here, number of terms is 13, which is odd.

Median = \(\frac { 13+1 }{ 2 } \) th term = \(\frac { 14 }{ 2 } \) th term = 7th term = 4 Ans.

**Question 2.**

**Solution:**

Arranging in ascending order, we get 9, 10, 17, 19, 21, 22, 32, 35

Here, number of terms is 8 which is even

∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } th\quad term+\left( \frac { 8 }{ 2 } +1 \right) th\quad term \right] \)

= \(\frac { 1 }{ 2 } \) [4th term + 5th term] = \(\frac { 1 }{ 2 } \) (19 + 21) = \(\frac { 1 }{ 2 } \) x 40 = 20

(ii) Arranging in ascending order, we get:

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

Here number of terms is 10 which is even

∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right] \)

= \(\frac { 1 }{ 2 } \) (60 + 63) = \(\frac { 1 }{ 2 } \) x 123 = 61.5 Ans.

(iii) Arranging in ascending order we get

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

Here number of terms is 12 which is even.

∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 12 }{ 2 } th\quad term+\left( \frac { 12 }{ 2 } +1 \right) th\quad term \right] \)

= \(\frac { 1 }{ 2 } \) (6th term + 7th term) = \(\frac { 1 }{ 2 } \) (15 + 17)= \(\frac { 1 }{ 2 } \) x 32

= 16 Ans.

**Question 3.**

**Solution:**

Arranging the given data in ascending order, we get :

17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40

∴ Median = \(\frac { 15+1 }{ 2 } \) th term = \(\frac { 16 }{ 2 } \) th term = 8th term = 23

∴ Median score = 23 Ans.

**Question 4.**

**Solution:**

Arranging in ascending order, we get:

143.7, 144.2, 145, 146.5, 147.3, 148.5, 149.6, 150, 152.1

Here, number of terms is 9 which is odd.

Median = \(\frac { 9+1 }{ 2 } \) th term = \(\frac { 10 }{ 2 } \) th term = 5th term = 147.3 cm

Hence, median height = 147.3 cm Ans.

**Question 5.**

**Solution:**

Arranging in ascending order, we get:

9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2

Here number of terms is 8 which is even

∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 8 }{ 2 } th\quad term+\left( \frac { 8 }{ 2 } +1 \right) th\quad term \right] \)

= \(\frac { 1 }{ 2 } \)[4th term + 5th term]

= \(\frac { 1 }{ 2 } \) (13.4 + 14.3) = \(\frac { 1 }{ 2 } \) (27.7) = 13.85

∴ Median weight = 13.85 kg. Ans.

**Question 6.**

**Solution:**

Arranging in ascending order, we get:

32, 34, 36, 37, 40, 44, 47, 50, 53, 54

Here, number of terms is 10 which is even.

∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right] \)

= \(\frac { 1 }{ 2 } \) [5th term + 6th term ] = \(\frac { 1 }{ 2 } \) (40 + 44) = \(\frac { 1 }{ 2 } \) x 84 = 42 .

∴ Median age = 42 years.

**Question 7.**

**Solution:**

The given ten observations are 10, 13, 15, 18, x + 1, x + 3, 30, 32, 35, 41

These are even

∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 10 }{ 2 } th\quad term+\left( \frac { 10 }{ 2 } +1 \right) th\quad term \right] \)

= \(\frac { 1 }{ 2 } \) [5th term + 6th term ] = \(\frac { 1 }{ 2 } \)(x + 1 + x + 3) = \(\frac { 1 }{ 2 } \)(2x + 4)

= x + 2

But median is given = 24

∴ x + 2 = 24 => x = 24 – 2 = 22

Hence x = 22.

**Question 8.**

**Solution:**

Preparing the cumulative frequency table, we have:

Here, number of terms (n) = 41, which is odd,

Median = \(\frac { 41+1 }{ 2 } \) th term = \(\frac { 42 }{ 2 } \) th term = 21st term = 50 (∵ 20th to 28th term = 50)

Hence median weight = 50 kg Ans.

**Question 9.**

**Solution:**

Arranging first in ascending order, we get:

Now preparing its cumulative frequency table

Here, number of terms is 37 which is odd.

Median = \(\frac { 37+1 }{ 2 } \) th term = \(\frac { 38 }{ 2 } \) th term = 19 th term = 22 (∵18th to 21st = 22)

Hence median – 22 Ans.

**Question 10.**

**Solution:**

first arranging in ascending order we get

Now preparing its cumulative frequency table,we find:

Here, number of terms is 43, which if odd.

Median = \(\frac { 43+1 }{ 2 } \) th term = \(\frac { 44 }{ 2 } \) th term = 22nd term = 25 25 (∵ 11th to 26th = 25)

**Question 11.**

**Solution:**

Arranging in ascending order,we get

Now preparing its cumulative frequency table, we find :

Here, number of terms = 50 which is even

∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 50 }{ 2 } th\quad term+\left( \frac { 50 }{ 2 } +1 \right) th\quad term \right] \)

= \(\frac { 1 }{ 2 } \) (154 + 155) = \(\frac { 1 }{ 2 } \) (309) = 154.5 (∵ 22nd to 25th = 154, 26th to 34th= 155)

**Question 12.**

**Solution:**

Arranging in ascending order, we get:

Now, preparing its cumulative frequency table.

Here, number of terms is 60 which is even

∴Median = \(\frac { 1 }{ 2 } \left[ \frac { 60 }{ 2 } th\quad term+\left( \frac { 60 }{ 2 } +1 \right) th\quad term \right] \)

= \(\frac { 1 }{ 2 } \) (30th term + 31st term)

= \(\frac { 1 }{ 2 } \) (20 + 23) = \(\frac { 1 }{ 2 } \) x 43 = 21.5 (∵ 18th to 30th term = 20, 31st term to 34th = 23)

Hence median = 21.5 Ans.

Hope given RS Aggarwal Solutions Class 9 Chapter 14 Statistics Ex 14G are helpful to complete your math homework.

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