Samacheer Kalvi Class 9 Maths Chapter 8 Statistics Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 9th class students by providing Solutions for Samacheer Kalvi Class 9 Maths chapter 8 Statistics. Here students can easily find all the solutions for Statistics Exercise 8.1, 8.2, 8.3 and 8.4. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 8 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.
Statistics Exercise 8.1 Solutions :
(1) In a week, temperature of a certain place is measured during winter are as follows 26°C, 24°C, 28°C, 31°C, 30°C, 26°C, 24°C. find the mean temperature of the week
Solution :
Mean temperature of the week,
= 26 + 24 + 28 + 31 + 30 + 26 + 24/7
= 27°c
Thus, the mean temperature of the week is 27°c
(2) The mean weight of 4 members of a family is 60 kg. three of them have the weight 56 kg, 68 kg and 72 kg respectively. find the weight of the fourth member.
Solution :
Mean = 60 kg
Let, the weight of the fourth member = x
Again, 56 + 68 + 72 + x/4
= 60
= 196 + x = 240
= x = 44 kg
Therefore, the weight of the fourth member is 44 kg.
(3) In a class test in mathematics, 10 students scored 75 marks, 12 students scored 60 marks, 8 students scored 40 marks and 3 students scored 30 marks. Find the mean of their score.
Solution :
Total score of 33 students,
= 10 × 75 + 12 × 60 + 40 × 8 + 3 × 30
= 1880
Total number of students,
= 10 + 12 + 40 + 3
= 33
∴ Mean of their score,
= 1880/33
= 56.96
Thus, the mean of their score is 33.
(4) In a research laboratory scientists treated 6 mice with lung cancer using natural medicine. Ten days later, they measured the volume of the tumor in each mouse and given the results in the table.
| Mouse marking | 1 | 2 | 3 | 4 | 5 | 6 |
| Tumor volume (mm³) | 145 | 148 | 142 | 141 | 139 | 140 |
Find the mean.
Solution :
Mean = 145 + 148 + 142 + 141 + 139 + 140/6
= 855/6
= 142.5
Therefore, the mean will be 142.5.
(5) If the mean of the following data is 20.2, then find the value of p
| Marks | 10 | 15 | 20 | 25 | 30 |
| No of students | 6 | 8 | P | 10 | 6 |
Solution :
Mean = 10 × 6 + 15 × 8 + 20 × D + 10 × 25 + 30 × 6/6 + 8 + P + 10 + 6
= 20.2 + 20.2P = 610 + 20P/30 + P
= 606 + 20.2P = 610 + 20P
= 0.2P = 4
∴ P = 4 × 10/0.2
∴ P = 20
Thus, the value of p is 20.
(6) In the class, weight of students is measured for the class records. Calculate mean weight of the class students using Direct method.
| Weight in kg | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 |
| No. of students | 4 | 11 | 19 | 14 | 0 | 2 |
Solution :
Weight : 15 – 25, 25 – 35, 35 – 45, 45 – 55, 55 – 65, 65 – 75
Student : 4, 11, 19, 14, 0, 2 = 50
Mid value of x : 20, 30, 40, 50, 60, 70
Fx : 80, 330, 760, 700, 0, 140 = 2010
Therefore the Mean,
= 2010/50
= 40.2
Hence, the mean weight of the class students is 40.2
(7) Calculate the mean of the following distribution using Assumed Mean Method
| Class interval | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 5 | 7 | 15 | 28 | 8 |
Solution :
Class : 0-10, 10-20, 20-30, 30-40, 40-50
Frequency : 5, 7, 15, 28, 8 = 63
Mid x : 5, 15, 25, 35, 45
A = 25 :
D = x-A : -20, -10, 0, 10, 20
FD : -100, -70, 0, 280, 160 = 270
Now, the Mean
= A + 270/63
= 25 + 4.29
= 29.29
(8) Find the Arithmetic Mean of the following data using Step Deviation Method
| Age | 15-19 | 20-24 | 25-29 | 30-34 | 35-39 | 40-44 |
| No. of persons | 4 | 20 | 38 | 24 | 10 | 9 |
Solution :
Mean = A + [Σfd/Σf ×c] d = x-A/c
Class interval : 15-19, 20-24, 25-29, 30-34, 35-39, 40-44
X : 14.5 – 19.5, 20.5 – 24.5, 25.5 – 29.5, 30.5 – 34.5, 35.5 – 39.5, 40.5 – 44.5
No. of person : 17, 22, 27, 32, 37, 42 = 105
D = x-A/c : -3, -2, -1, 0, 1, 2
Fd : -12, -40, -38, 0, 10, 18
= -62
Now the Mean,
= 32 + (-62/105 × 5)
= 32 – 2.952
= 29.05
Statistics Exercise 8.2 Solutions :
(1) Find the median of the given values : 47, 53, 62, 71, 83, 21, 43, 47, 41.
Solution :
We write the values in according order
21, 41, 43, 47, 47, 53, 62, 71, 82
Since, value are in odd number
Median = (9 + 1/2)th
= 5th number value
= 47
(2) Find the Median of the given data : 36, 44, 86, 31, 37, 44, 86, 35, 60, 51
Solution :
Writing the values in according order,
31, 35, 36, 37, 44, 44, 51, 60, 86, 86
Median = (10/2)th + (10/2 + 1)th/2
= 5th + 6th/2
= 44 + 44/2
= 44
Thus, the Median of the given data is 44.
(3) The median of observation 11, 12, 14, 18, x+2, x+4, 30, 32, 35, 41 arranged in ascending order is 24. Find the values of x.
Solution :
Number of values = 10.
So it is an even number.
Median = 5th + 6th/2
24 = x + 2 + x + 4/2
= 48 = 2x + 6
= 2x = 42
= x = 21
Therefore, the values of x is 21.
(4) A researcher studying the behavior of mice has recorded the time (in seconds) taken by each mouse to locate its food by considering 13 different mice as 31, 33, 63, 33, 28, 29, 33, 27, 27, 34, 35, 28, 32. Find the median time that mice spent in searching its food.
Solution :
Writing in ascending order,
27, 27, 28, 28, 29, 31, 32, 33, 33, 33, 34, 35, 63
Number of values = 13 which is odd
So the median = (13 + 1/2)th
= 7th
= 32
Hence, the median time that mice spent in searching its food is 32
(6) The mean of five positive integers is twice their median. If four of the integers are 3, 4, 6, 9 and median is 6, then find the fifth integer.
Solution :
Mean of the integers,
= 3 + 4 + 6 + 9 + x/5
= 22 + x/5
Median = 6
We know, mean = twice of the median
= 22 + x/5 = 2×6
= x = 60-22
= x = 30
Therefore, the fifth integer will be 38
Statistics Exercise 8.3 Solutions :
(1) The monthly salary of 10 employees in a factory are given below : ₹5000, ₹7000, ₹5000, ₹7000, ₹8000, ₹7000, ₹7000, ₹8000, ₹7000, ₹5000. find the mean, median and mode.
Solution :
First, writing in ascending order = ₹5000, ₹5000, ₹5000, ₹7000, ₹ 7000, ₹7000, ₹8000, ₹8000, ₹17000
Since no of salary is 10
So, median = 5th + 6th/2
= 14000/2
= 7000
∴ Mean = (5000 + 5000 + 5000 + 7000 + 7000 + 7000 + 7000 + 8000 + 8000 + 17000)/10
= 66000/10
= 6600
(2) Find the mode of the given data : 3.1, 3.2, 3.3, 2.1, 1.3, 3.3, 3.1
Solution :
From the given data,
3.1, 3.2, 3.3, 2.1, 1.3, 3.3, 3.1
3.1 occurs maximum number of times.
Hence, the mode of data is 3.1
(3) For the data 11, 15, 17, x + 1, 19, x – 2, 3 if the mean is 14 , find the value of x. Also find the mode of the data.
Solution :
Mean = 14
14 = (11 + 15 + 17 + x + 1 + 19 + x – 2 + 3)/7
= 64 + 2x = 98
= 2x = 34
x = 17
So, the numbers = 11, 15, 17, 18, 19, 15, 3
So, mode = 15
Therefore, the value of x is 17 and mode of the data is 15.
(4) The demand of track suit of different sizes as obtained by a survey is given below :
| Size | 38 | 39 | 40 | 41 | 42 | 43 | 44 | 45 |
| No. of persons | 36 | 15 | 37 | 13 | 26 | 8 | 6 | 2 |
Which size is in greater demanded?
Solution :
Size : 38, 39, 40, 41, 42, 43, 44, 45
Number of persons : 36, 15, 37, 13, 26, 8, 6, 2
Size 40 …… maximum frequency 37
So, the mode is 40
Hence, size 40 is in greater demand.
Statistics Exercise 8.4 Solutions :
Multiple choice questions :
(1) Let m be the mid point and b be the upper limit of a class in a continuous frequency distribution. the lower limit of the class is
(1) 2m – b
(2) 2m + b
(3) m – b
(4) m – 2b
Solution :
Correct option – (1)
Hence, the lower limit of the class is 2m-b
(2) The mean of a set of seven numbers is 81. If one of the numbers is discarded, the mean of the remaining numbers is 78. The value of discarded number is
(1) 101
(2) 100
(3) 99
(4) 98
Solution :
Correct option – (3)
The value of discarded number is 99
(3) A particular observation which occurs maximum number of times in a given data is called its
(1) Frequency
(2) range
(3) mode
(4) Median
Solution :
Correct option – (3)
A particular observation which occurs maximum number of times in a given data is called its Mode.
(4) For which set of numbers do the mean, median and mode all have the same values?
(1) 2,2,2,4
(2) 1,3,3,3,5
(3) 1,1,2,5,6
(4) 1,1,2,1,5.
Solution :
Correct option – (2)
1, 3, 3, 3, 5 – This is her correct set of numbers.
(5) The algebraic sum of the deviations of a set of n values from their mean is
(1) 0
(2) n – 1
(3) n
(4) n + 1
Solution :
Correct option – (1)
The algebraic sum of the deviations of a set of n values from their mean is 0
(6) The mean of a,b,c,d and e is 28. If the mean of a, c and e is 24, then mean of b and d is
(1) 24
(2) 36
(3) 26
(4) 34
Solution :
Correct option – (4)
a + b + c + d + e/5 = 28
= a + b + c + d + e
= 5 × 28
= 140
a + c + e/3 = 24
= a + c + e = 72
∴ b + d = 68
= b + d/2 = 34
So, the mean of b and d is 34
(7) If the mean of five observations x, x+2, x+4, x+6, x+8, is 11, then the mean of first three observations is
(1) 9
(2) 11
(3) 13
(4) 15
Solution :
Correct option – (3)
= (x + x + 2 + x + 4 + x + 6 + x + 8)/5 = 11
= x = y
∴ 11 + 13 + 15/3
= 13
So, the mean of first three observations is 13
(8) The mean of 5, 9, x, 17, and 21 is 13, then find the value of x
(1) 9
(2) 13
(3) 17
(4) 21
Solution :
Correct option – (2)
5 + 9 + x + 17 + 21/5 = 13
= x = 165 – 52
= x = 13
So, the value of x is 13
(9) The mean of the square of first 11 natural numbers is
(1) 26
(2) 46
(3) 48
(4) 52
Solution :
Correct option – (2)
The mean of the square of first 11 natural numbers is 46.
(10) The mean of a set of numbers is X . If each number is multiplied by z, the mean is
(1) X + z
(2) X – z
(3) z X
(4) X
Solution :
Correct option – (3)
The mean of a set of numbers is X. if each number is multiplied by z, the mean is z x̄.
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