**ML Aggarwal Statistics MCQs Class 9 ICSE Maths APC Understanding Solutions.** Solutions of MCQs . This post is the Solutions of **ML Aggarwal** Chapter 20 – Statistics for **ICSE** Maths **Class-9.** APC Understanding **ML Aggarwal** Solutions (APC) Avichal Publication Solutions of Chapter – 20 Statistics for** ICSE **Board** Class-9****. **Visit official website **CISCE** for detail information about ICSE Board Class-9.

## ML Aggarwal Statistics MCQs Class 9 ICSE Maths Solutions

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 9th |

Chapter-20 | Statistics |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of MCQs Questions |

Edition | 2021-2022 |

**MCQs Solutions of ML Aggarwal for ICSE Class-9 Ch-20, Statistics**

Note:- Before viewing Solutions of Chapter – 20 Statistics Class-9 of ML Aggarwa**l **Solutions . Read the Chapter Carefully. Then solve all example given in Exercise-20.1, Exercise-20.2, Exercise-20.3, MCQs, Chapter Test.

**Statistics MCQs **

ML Aggarwal Class 9 ICSE Maths Solutions

Page 497

**Question 1.The marks obtained by 17 students in a mathematics test (out of 100) are given below:**

91, 82, 100, 100, 96, 65, 82, 76, 79,90, 46, 64, 72, 66, 68, 48, 49

The range of the data is

(a) 46

(b) 54

(a) 46

(d) 100

**Answer :**

(b) 54 is correct

**Question 2. The class mark of the class 90-120 is**

(a) 90

(b) 105

(c) 115

(d) 120

**Answer :**

Class mark = (upper limit + lower limit) / 2

Then, Class mark = 120 + 90 / 2

= 210 / 2 = 105.

(b) 105 is correct

**Question 3. In a frequency distribution, the mid-value of a class is 10 and the width of the class is 6. The lower limit of the class is**

(a) 6

(b) 7

(c) 8

(d) 12

**Answer :**

Mid value = 10 cm

Width = 6

Lower Limit = mid value of class – __width
__ 2

__= 10 –__

__6__

2

= 10 – 3

= 7

(b) 7 is correct

**Question 4. The width of each of 5 continuous classes in a frequency distribution is 5 and the lower limit of the lowest class is 10. The upper limit of the highest class is**

(a) 15

(b) 25

(c) 35

(d) 40

**Answer :**

Upper limit of highest class is 35

As 5 width for five consecutive classes

10-15

15-20

20-25

25-30

30-35

(c) 35 is correct

**Question 5. The class marks of a frequency distribution are given as follows:**

15, 20, 25, …… The class corresponding to the class mark 20 is

………………..

**Answer :**

Class size = 20 – 15 = 5

Consider a as the lower limit of the required class

Upper limit can be written as a + 5

(a + (a + 5))/ 2 = 20

2a + 5 = 40

2a = 35

a = 17.5

Upper limit a + 5 = 17.5 + 5 = 22.5

Hence, the required class is 17.5-22.5

**Statistics MCQs **

ML Aggarwal Class 9 ICSE Maths Solutions

Page 498

**Question 6. In the class intervals 10 – 20, 20 – 30, the number 20 is included in**

(a) 10-20

(b) 20 – 30

……………………

**Answer :**

(d) none of these

**Question 7. A grouped frequency distribution table with class intervals of equal size ………… 250 270 (270 not included in this interval) as one of the class intervals is constructed for the following data:**

268, 220, 368, 258, 242, 310, 272, 342, 310, 290, 300, 320, 319, 304, 402, 318, 406 ………..

(a) 4

(b) 5

(c) 6

(d) 7

**Answer :**

(c) 6 is correct

**Question 8. The mean of x- 1, x + 1, x + 3 and x + 5 is…….**

**Answer :**

(b) x + 2

**Question 9. The mean of five numbers is 30. If one number is excluded, their mean ….. The excluded number is……**

**Answer :**

Let a,b,c,d and e are five numbers

mean = (a +b+ c+ d+e)/5 = 30

⇒ (a +b+ c+ d+e) = 150_______(1)

now, Let the number excluded be a

then, new mean = (b+ c+ d+e)/4 = 28

⇒ (b+ c+ d+e)= 112

putting this value in (1),

⇒a + 112 = 150

⇒a = 150 -112 = 38

**∴ excluded number = 38 is correct**

**Question 10. If the mean of x1, x, is 7.5, and the mean of x1, X2, x3 is………….**

**Answer :**

Mean = sum of variables/number of variables.

First equation

X1 + X2 = 7.5

2

X1 + X2 = 15 —— (1)

Second equation

X1 + X2 + X3 = 8

3

X1 + X2 + X3 = 24 ——– (2)

Subtract equation 1 from equation 2

(X1 + X2 + X3) – (X1 + X2) = 24 – 15

X3 = 9

(a) 9 is correct

#### Question 11. If each observation of the data is increased by 5, then their mean…………

**Answer :**

(d) is increased by 5 is correct

**Question 12. The mean of 100 observations is 50. If one of the observation which was 50 is replaced by 150, the resulting mean will be**

(a) 50-5

(b) 51

(c) 51-5

(d) 52

**Answer :**

Mean =( sum of the observations )/( number of observations )

According to the question

(i) Sum of the 100 observations/ 100

= mean

Sum of the 100 observations

= 50× 100

= 5000 ——( 1 )

(ii) if one of the observation 50

replaced by 150

New Sum of the 100 observations

= 5000 – 50 + 150 [ from ( 1 ) ]

= 5100 —( 2 )

New mean = ( 2 )/ 100

= 5100/ 100

= 51 is correct

**Question 13. For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are respectively:**

(a) upper limits of the classes

(b) lower limits of the classes

(c) class marks of the class

(d) upper limits of preceding classes

**Answer :**

#### (c) class marks of the class is correct

**Question 14. Median of the numbers 4, 4, 5, 7, 6, 7, 7, 3, 12 is**

**Answer :**

(c) 6 is correct

**Question 15. The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54, 84 is**

**Answer :**

(c) 54 is correct

**Question 16. In a data, 10 numbers are arranged in ascending order. If the 8th entry is increased by…………..**

**Answer :**

(a) 0 is correct

— : End of ML Aggarwal Statistics MCQs Class 9 ICSE Maths Solutions :–

Return to :- ** ML Aggarawal Maths Solutions for ICSE Class-9**

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