ML Aggarwal Measures of Central Tendency Exe-21.2 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-21.2 Questions for Measures of Central Tendency as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Measures of Central Tendency Exe-21.2 Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-21 | Measures of Central Tendency |

Writer / Book | Understanding |

Topics | Solutions of Exe-21.2 |

Academic Session | 2024-2025 |

**Measures of Central Tendency Exe-21.2**

ML Aggarwal Class 10 ICSE Maths Solutions

Page 499

** ****Question 1. **A student scored the following marks in 11 questions of a question paper : 3, 4, 7, 2, 5, 6, 1, 8, 2, 5, 7 Find the median marks.

**Answer :**

Arranging in the ascending order, 1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8

Here, n = 11 i.e. odd,

The middle term = (n+1)\2 = (11+1)\2 = 12\6 = 6th term

Median = 5

**Question 2. **For the following set of the number, find the median :

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

**Answer :**

On arranging in ascending order 3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81

Here, n = 12 which is even

Therefore, median = {(n/2)^{th} term + (n/2 + 1)^{th} term}/2

= {(12/2)^{th} term + (12/2 + 1)^{th} term}/2

= (6^{th} term + 7^{th} term)/2

= (15 + 17)/2 = 32/2 = 16

Median = 16.

**Question 3. **Calculate the mean and the median of the numbers : 2, 1, 0, 3, 1, 2, 3, 4, 3, 5

**Answer :**

Writing in ascending order 0, 1, 1, 2, 2, 3, 3, 3, 4, 5

Here, n = 10 which is even

Median = ½ ( n/2 ^{th} term + ((n/2) + 1)^{th} term)

= ½ (10/2 ^{th} term + ((10/2) + 1)^{th} term)

= ½ (5 ^{th} term + (5 + 1)^{th} term)

= ½ (5 ^{th} term + 6^{th} term)

= ½ (2 + 3)

= ½ × 5

= 2.5

Hence, the median is 2.5.

Mean = sum of the observations/number of observations

= Ʃx_{i}/n

= (0 + 1 + 1 + 2 + 2 + 3 + 3 + 3 + 4 + 5)/10

= 24/10

= 2.4

Hence, the mean is 2.4.

**Question 4. **The median of the observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.

**Answer :**

Observation are :

11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47

n = 9

Median = (9+1)\2 th term

i.e, 5th term = x + 4

median = 24

x + 4 = 24

⇒ x = 24 – 4 = 20

Sum of observations = 11 + 12 + 14 + (x – 2) + (x + 4) + (x + 9) + 32 + 38 + 47

= 165 + 3x

Substitute x = 20

Sum of observations = 165 + 3 × 20

= 165 + 60

= 225

Mean = Sum of observations /number of observations

= 225/9 = 25

**Question 5. **The mean of the numbers 1, 7, 5, 3, 4, 4, is m. The numbers 3, 2, 4, 2, 3, 3, p have mean m – 1 and median q. Find

(i) p

(ii) q

(iii) the mean of p and q.

**Answer :**

**(i) Mean of 1, 7, 5, 3, 4, 4 is m.**

Here n = 6

Mean, m = (1 + 7 + 5 + 3 + 4 + 4)/6

⇒ m = 24/6

⇒ m = 4

Given the numbers 3, 2, 4, 2, 3, 3, p have mean m-1.

So,

m-1 = (3 + 2 + 4 + 2 + 3 + 3 + p)/7

⇒ 4 – 1 = (17 + p)/7

⇒ 3 = (17 + p)/7

⇒ 3 × 7 = 17 + p

⇒ 21 = 17 + p

⇒ p = 21 – 17

p = 4

Hence, the value of p is 4.

**(ii)** **Given the numbers have median q.**

Arranging them in ascending order

2, 2, 3, 3, 3, 4, 4

Here n = 7 which is odd

So, median = ((n + 1)/2)^{th} term

⇒ q = ((7 + 1)/2)^{th} term

⇒ q = (8/2)^{th} term

⇒ q = 4^{th }term

⇒ q = 3

So, value of q is 3.

**(iii)** **mean of p and q = (p + q)/2**

= (4 + 3)/2

= 7/2

= 3.5

**Question 6. **Find the median for the following distribution:

**Answer :**

Writing the distribution in cumulative frequency table:

Here total number of observations, n = 47 which is odd.

So median = ((n + 1)/2)^{th }term

= ((47 + 1)/2)^{th} term

= (48/2)^{th} term

= 24^{th} term

= 48 **[Since, 23 ^{rd} to 29^{th} observation is 48]**

**Measures of Central Tendency Exe-21.2**

### ML Aggarwal Class 10 ICSE Maths Solutions

Page 500

**Question 7. **Marks obtained by 70 students are given below :

Calculate the median marks.

**Answer :**

Arranging the variates in ascending order and in c.f. table.

Median = ½ ( n/2^{th} term + ((n/2)+1)^{th} term)

Here, total number of observations, n = 70 which is even.

= ½ (70/2^{th} term + ((70/2)+1)^{th} term)

= ½ (35^{th} term + (35+1)^{th} term)

= ½ (35^{th} term + 36^{th} term)

= ½ (50 + 50) **[Since all observations from 21 ^{st} to 38^{th} are 50]**

= ½ ×100

= 50

**Question 8. **Calculate the mean and the median for the following distribution :

**Answer :**

Writing the distribution in c.f. table :

Mean = Ʃfx/Ʃf

= 390/20

= 19.5

Hence the mean is 19.5.

Here number of observations, n = 20 which is even.

So median = = ½ (n/2 ^{th} term + ((n/2) + 1)^{th} term)

= ½ (20/2 ^{th} term + ((20/2) + 1)^{th} term)

= ½ (10 ^{th} term + (10 + 1)^{th} term)

= ½ (10 ^{th} term + 11^{th} term)

= ½ (20+20) **[Since all observations from 9 ^{th} to 14^{th} are 20]**

= ½ ×140

= 20

**Question 9.**

The daily wages (in rupees) of 19 workers are

41, 21, 38, 27, 31, 45, 23, 26, 29, 30, 28, 25, 35, 42, 47, 53, 29, 31, 35.

Find

(i) the median

(ii) lower quartile

(iii) upper quartile range,

(iv) interquartile range.

**Answer :**

Arranging the observations in ascending order

21, 23, 25, 26, 27, 28, 29, 29, 30, 31, 31, 35, 35, 38, 41, 42, 45, 47, 53

Here n = 19 which is odd.

**(i) **Median = ((n + 1)/2)^{th} term

= (19 + 1)/2

= 20/2

= 10^{th }term

= 31

**(ii)** Lower quartile, Q_{1} = ((n + 1)/4)^{ th} term

= (19 + 1)/4

= 20/4

= 5^{th }term

= 27

**(iii)** Upper quartile, Q_{3} = (3(n + 1)/4)^{th} term

**= (**3 × (19 + 1)/4)^{th} term

= (3 × (20/4))^{th} term

= (3 × 5)^{th }term

**= **15 ^{th }term

= 41

**(iv)** Interquartile range = Q_{3 }– Q_{1}

= 41 – 27

= 14

**Question 10. **From the following frequency distribution, find :

(i) the median

(ii) lower quartile

(iii) upper quartile

(iv) inter quartile range

**Answer :**

Writing frequency distribution in c.f. table :

So median = ½ (n/2^{th} term + ((n/2) + 1)^{th} term)

**(i)** Here number of observations, n = 48 which is even.

= ½ (48/2^{th} term + ((48/2) + 1)^{th} term)

= ½ (24^{th} term + (24 + 1)^{th} term)

= ½ (24^{th} term + 25^{th} term)

= ½ (22 + 22)** [Since all observations from 19 ^{th} to 27^{th} are 22]**

= ½ × 44

= 22

Hence, the median is 22.

**(ii)** Lower quartile, Q_{1} = (n/4)^{th} term

= (48)/4

= 12^{th }term

= 20

Hence the lower quartile is 20.

**(iii)** Upper quartile, Q_{3} = (3n/4)^{th} term

**= (**3 × 48/4)^{th} term

= (3 × 12)^{th} term

**= **36^{th }term

= 27

Hence the upper quartile is 27.

**(iv)** Interquartile range = Q_{3 }– Q_{1}

= 27 – 20

= 7

Hence, the Interquartile range is 7.

**Question 11. **For the following frequency distribution, find :

(i) the median

(ii) lower quartile

(iii) upper quartile

**Answer :**

Writing the distribution in cumulative frequency (c.f.) table :

**(i) **Here number of observations, n = 63 which is odd.

Median = ((n + 1)/2)^{th} term

= (63 + 1)/2

= 64/2

= 32^{th }term

= 40

**(ii)** Lower quartile, Q_{1} = ((n + 1)/4)^{ th} term

= (63 + 1)/4

= 64/4

= 16^{th }term

= 34

**(iii) **Upper quartile, Q_{3} = (3(n + 1)/4)^{th} term

**= (**3 × (63 + 1)/4)^{th} term

= (3 × (64/4))^{th} term

= (3 × 16)^{th }term

**= **48^{th }term

= 48

-: End of ML Aggarwal Measures of Central Tendency Exe-21.2 Class 10 ICSE Maths Solutions :–

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