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 Aggarwal 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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