ML Aggarwal Statistics Exe-20.1 Class 9 ICSE Maths APC Understanding Solutions. Solutions of Exe-20.1. 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 Exe-20.1 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 Exe-20.1 Questions |
| Edition | 2021-2022 |
Exe-20.1 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 Exercise-20.1
ML Aggarwal Class 9 ICSE Maths Solutions
Page 479
Question 1. Find the mean of 8, 6, 10, 12, 1, 3, 4, 4.
Answer:
8, 6, 10, 12, 1, 3, 4, 4
Here, n = 8
∴ Mean (x̄) = Ʃ xi/n
= (8 + 6 + 10 + 12 + 1+ 3 + 4 + 4)/8
= 48/8 = 6
Hence, mean of the given data is 6.
Question 2. 5 people were asked about the time in a week they spend in doing social work in their community. They replied 10, 7, 13, 20 and 15 hours, respectively. Find the mean time in a week devoted by them for social work.
Answer:
10, 7, 13, 20, 15
Here, n = 5
∴ Mean (x̄) = Ʃ xi/n
= (10 + 7 + 13 + 20 + 15)/5
= 65/5 = 13
Hence, the mean time in a week devoted by them for social work is 13 hours.
Question 3. The enrollment of a school during six consecutive years was as follows:
1620, 2060, 2540, 3250, 3500, 3710,
Find the mean enrollment.
Answer:
1620, 2060, 2540, 3250, 3500, 3710
Here, n = 6
∴ Mean (x̄) = Ʃ xi/n
= (1620 + 2060 + 2540 + 3250 + 3500 + 3710)/5
= 16680/6 = 2780
Hence, the mean enrollment is 2780.
Question 4. Find the mean of the first twelve natural numbers.
Answer:
The first twelve natural numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Here, n = 12
∴ Mean (x̄) = Ʃ xi/n
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)/12
= 78/12 = 6.5
Hence, the mean of the first twelve natural numbers is 6.5
Question 5.
(i) Find the mean of the first six prime numbers.
(ii) Find the mean of the first seven odd prime numbers.
Answer:
(i) First 6 prime numbers are 2, 3, 5, 7, 11, 13
Here, n = 6
∴ Mean (x̄) = Ʃ xi/n
= (2 + 3 + 5 + 7 + 11 + 13)/6
= 41/6
Hence, the mean of the first six prime numbers is 41/6.
(ii) First seven odd prime numbers are 3, 5, 7, 11, 13, 17, 19
Here, n = 7
∴ Mean (x̄) = Ʃ xi/n
= (3 + 5 + 7 + 11 + 13 + 17 + 19)/7
= 75/7
Hence, the mean of the first six prime numbers is 75/7.
Question 6.
(i)The marks (out of 100) obtained by a group of students in a Mathematics test
are 81, 72, 90, 90, 85, 86, 70, 93 and 71. Find the mean marks obtained by the
group of students.
(ii) The mean of the age of three students Vijay, Rahul and Rakhi is 15 years. If their
ages are in the ratio 4 :5:6 respectively, then find their ages.
Answer:
(i) The marks obtained by the group of students
81, 72, 90, 90, 85, 86, 70, 93, 71
Here, n = 9
∴ Mean (x̄) = Ʃ xi/n
= (81 + 72 + 90 + 90 + 85 + 86 + 70 + 93 + 71)/9
= 738/9 = 82
Hence, the mean marks obtained by the group of students is 82.
(ii) the mean of the age of three students Vijay, Rahul and Rakhi is 15 years.
So, n = 3
Now, the sum of ages of the 3 students = 15×3 = 45
Also given, ratio of their ages is 4 : 5 : 6
Sum of ratios = 4 + 5 + 6 = 15
Hence,
Vijay’s age = (45/15)×4 = 12 years
Rahul’s age = (45/15)×5 = 15 years
Rakhi’s age = (45/15)×6 = 18 years
Question 7. The mean of 5 numbers is 20. If one number is excluded, mean of the remaining
numbers becomes 23. Find the excluded number.
Answer:
The mean of 5 numbers = 20
So, the total sum of the numbers = 20×5 = 100
After excluding one number,
The mean of the remaining 4 numbers = 23
So, the total sum of these numbers = 23×4 = 92
Hence, the excluded number is = 100 – 92 = 8.
Question 8. The mean of 25 observations is 27. If one observation is included, the mean still remains 27. Find the included observation.
Answer:
The mean of 25 observations is 27.
So, the total sum of all the 25 observations = 27×25 = 675
After one observation is included,
Now the mean of 26 (25 + 1) numbers = 27
So, the total sum of all the 26 observations = 27×26 = 702
Hence, the included observation = 702 – 675 = 27
Question 9. The mean of 5 observations is 15. If the mean of first three observations is 14 and that
of the last three is 17, find the third observation.
Answer:
The mean of 5 observations = 15
So, total sum of the 5 observations = 15×5 = 75
Also given,
Mean of first 3 observations = 14
So, the sum of the 3 observations = 14×3 = 42
And, the mean of last 3 observations = 17
So, the sum of last 3 observations = 17×3 = 51
Thus, the total of 3 + 3 observations = 42 + 51 = 93
Hence, The third observation = 93 – 75 = 18.
Question 10. The mean of 8 variate is 10-5. If seven of them are 3, 15, 7, 19, 2, 17 and 8, then find the 8th variate.
Answer:
Seven out of eight variates are: 3, 15, 7, 19, 2, 17 and 8
Mean of 8 variates = 10.5
So, the total of 8 variates = 10.5×8 = 84
Sum of seven variates = (3 + 15 + 7 + 19 + 2 + 17 + 8) = 71
Therefore, the 8th variate = 84 – 71 = 13.
Question 11. The mean weight of 8 students is 45-5 kg. Two more students having weights 41-7 kg and 53-3 kg join the group. What is the new mean weight?
Answer:
The mean weight of 8 students = 45.5 kg
So, the total weight of 8 students = 45.5 x 8 = 364 kg
Weight of two more students are 41.7 kg and 53.3 kg
The total weight of 10 (8 + 2) students = 364 + 41.7 + 53.3
= 364 + 95
= 459 kg
Therefore, the new mean weight of all the 10 students = 459/10 = 45.9 kg
Question 12. Mean of 9 observations was found to be 35. Later on, it was detected that an observation 81 was misread as 18. Find the correct mean of the observations.
Answer:
Mean of 9 observations = 35
So, the sum of all 9 observations = 35×9 = 315
Now, the difference due to misread = 81 – 18 = 63
Thus, the actual sum = 315 + 63 = 378
Therefore, the actual mean = 378/ 9 = 42.
Question 13. A student scored the following marks in 11 questions of a question paper :
7, 3, 4, 1, 5, 8, 2, 2, 5, 7, 6.
Find the median marks.
Answer:
Marks scored in 11 questions of a question paper by the student are:
7, 3, 4, 1, 5, 8, 2, 2, 5, 7, 6
Arranging it in descending order, we have
1, 2, 2, 3, 4, 5, 5, 6, 7, 7, 8
Here, n = 11 which is odd
∴ Median = (n + 1)/2th term
= (11 + 1)/2 = 12/2 = 6th term i.e 5
Therefore, the median mark is 5.
Question 14. Calculate the mean and the median of the numbers:
2, 3, 4, 3, 0, 5, 1, 1, 3, 2.
Answer:
First arrange the number in descending order
0, 1, 1, 2, 2, 3, 3, 3, 4, 5
So n = 10 which is even
Mean (x̄) = Ʃ xi/n
= (0 + 1 + 1 + 2 + 2 + 3 + 3 + 3 + 4 + 5)/10
= 24/10
= 2.4
Median = ½ [10/2th + (10/2 + 1)th terms]
It can be written as
= ½ (5th + 6th) term
= ½ (2 + 3)
= 5/2
= 2.5
Question 15. A group of students was given a special test in Mathematics. The test was completed by the various students in the following time in (minutes)
24, 30, 28, 17, 22, 36, 30, 19, 32, 18, 20, 24.
Find the mean time and median time taken by the students to complete the test.
Answer:
First arrange the data in descending order
17, 18, 19, 20, 22, 24, 24, 28, 30, 30, 32, 36
So n = 12 which is even
Mean (x̄) = Ʃ xi/n
= (17 + 18 + 19 + 20 + 22 + 24 + 24 + 28 + 30 + 30 + 32 + 36)/2
= 300/12
= 25
Median = ½ [12/2th + (12/2 + 1)th terms]
= ½ (6th + 7th) terms
= ½ (24 + 24)
= ½ (48)
= 24
Question 16. In a Science test given to a group of students, the marks scored by them (out of 100)are
41, 39, 52, 48, 54, 62, 46, 52, 40, 96, 42, 40, 98, 60, 52.
Find the mean and median of this data.
Answer:
On arranging the marks obtained by the students, we have
39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98
Here, n = 15 which is odd
∴ Mean (x̄)
Ʃ xi/n = (39 + 40 + 40 + 41 + 42 + 46 + 48 + 52 + 52 + 52 + 54 + 60 + 62 + 96 + 98)/15
= 822/15 = 54.8
And,
Median = (15 + 1)/2th term
= 16/2 = 8th term i.e. 52
Hence, for the given data mean = 54.8 and median = 52.
Question 17. The points scored by a Kabaddi team in a series of matches are as follow
17, 2, 5, 27. 15. 8. 14, 10, 48, 10, 7, 24, 8, 28, 18.
Find the mean and the median of the points scored by the Kabaddi team
Answer:
Let’s arrange the given data in descending order:
2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48
Here, n = 16 when is even
∴ Mean (x̄)
Ʃ xi/ n = (2 + 5 + 7 + 7 + 8 + 8 + 10 + 10 + 14 + 15 + 17 + 18 + 24 + 27 + 28 + 48)/15
= 248/16 = 15.5 points
And,
Median = ½ [(16/2)th term + (16/2 + 1)th term]
= ½ (8th term + 9th term)
= ½ (10 + 14)
= ½ x 24 = 12 points
Hence, the mean and the median of the points scored by the Kabaddi team are 15.5 and 12 respectively.
Statistics Exercise-20.1
ML Aggarwal Class 9 ICSE Maths Solutions
Page 480
Question 18. The following observations have been arranged in ascending order. If the median the data is 47-5, find the value of x.
17, 21, 23, 29, 39, 40, x, 50, 51, 54, 59, 67, 91, 93
Answer:
17, 21, 23, 29, 39, 40, x, 50, 51, 54, 59, 67, 91, 93
Here, n = 14 which is even
As the given data is arranged in descending order
Median = ½ [(14/2)th term + (14/2 + 1)th term]
= ½ (7th term + 8th term)
⇒ 47.5 = ½ (x + 50)
95 = x + 50
x = 95 – 50 = 45
Therefore, the value of x is 45.
Question 19. The following observations have been arranged in ascending order. If the median the data is 13, find the value of x: ….. 0S
3, 6, 7, 10, x, x + 4, 19, 20, 25, 28.
Answer:
3, 6, 7, 10, x, x + 4, 19, 20, 25, 28
Here, n = 10 which is even and median = 13
So, Median = ½ [(10/2)th term + (10/2 + 1)th term]
= ½ (5th term + 6th term)
= ½ (x + x + 4)
= (2x + 4)/2
= x + 2
⇒ x + 2
= 13
x = 13 – 2
= 11
Hence, the value of x is 11.
— : End of ML Aggarwal Statistics Exe-20.1 Class 9 ICSE Maths Solutions :–
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