Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.2 Pdf
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TN Samacheer Kalvi 10th Maths Solutions Chapter 8 – Statistics and Probability
Board  TNSCERT Class 10th Maths 
Class  10 
Subject  Maths 
Chapter  8 (Exercise 8.2) 
Chapter Name  Statistics and Probability 
TNSCERT Class 10th Maths Pdf  all Exercise Solution
Exercise – 8.2
(1) The standard deviation and mean of a data are 6.5 and 12.5 respectively. Find thecoefficient of variation.
Solution:
Given that σ = 6.5 x = 12.5
We know that,
Coefficient of variation (c.v) = σ/x × 100%
C.V = 6.5/12.5 × 100%
C.V = 52%
Therefore, the coefficient of variation is 52%
(2) The standard deviation and coefficient of variation of a data are 1.2 and 25.6 respectively. Find the value of mean.
Solution:
Let, the mean = x
Given standard deviation σ = 1.2 and C.V = 25.6
Now, C.V = σ/x × 100%
25.6 = 1.2/x × 100
x = 1.225.6 × 100
x = 75/16
x = 4.687
x = 4.69
Thus, the mean (x) = 4.69
(3) If the mean and coefficient of variation of a data are 15 and 48 respectively, then find the value of standard deviation.
Solution:
Given that mean (x) = 15 and coefficient (C.V) = 48
Now, C.V = σ/x × 100%
48 = σ/15 × 100
σ = 48×15/100
σ = 24×3/10
σ = 72/10
σ = 7.2
Thus, the standard deviation σ = 7.2
(4) If n = 5, x = 6, Σx^{2} = 765, then cellular the coefficient of variation.
Solution:
Given that n = 5, mean (x) = 6, Σx^{2} = 765
Standard deviation σ = √Σxi^{2}/n – (Σxi/n)^{2}
σ = √765/5 – (6)^{2}
σ = √153 – 36
σ = √117
σ = 10.82
Now, CV = σ/x × 100%
= 10.82/6 × 100%
= 180.33%
Thus, the coefficient of variation = 180.33%
(5) Find the coefficient of variation of 24, 26, 33, 37, 29, 31.
Solution:
Let, the mean (A) = 31
x_{i}  di = x_{i} – A
= x_{i} – 31 
di^{2} 
24  7  49 
26  5  25 
33  2  4 
37  6  36 
29  2  4 
31  0  0 
Σxi = 180  Σdi = 6  Σdi^{2} = 118 
Now, n = 6
Then, we know that x = Σxi/x = 180/6 = 30
∴ = 30
We know that standard deviation
σ = √Σdi^{2}/n – (Σdi/n)^{2}
σ = √118/6 – (6/6)^{2}
σ = √19.66 – 1
σ = √18.66
σ = 4.319
σ = 4.32
Thus, the C.V = σ/x × 100% = 4.32/30 × 100%
= 1.44 × 10%
= 14.4%
Therefore, the coefficient of variation 14.4%
(6) The time taken (in minutes) to complete a homework by 8 students in a day are given by 38, 40, 47, 44, 46, 43, 49, 53. Find the coefficient of variation.
Solution:
Let the mean (A) = 44
x_{i}  di = x_{i} – A
= x_{i} – 44 
di^{2} 
38  6  36 
40  4  16 
47  3  9 
44  0  0 
46  2  4 
43  1  1 
49  5  25 
53  9  81 
Σx_{i} = 360  Σdi = 8  Σdi^{2}= 172 
Now, n = 8, then x = Σxi/n = 360/8 = 45
∴ x = 45
Standard deviation σ = √Σdi^{2}/n – (Σdi/n)^{2}
σ = √172/8 – (8/8)^{2}
σ = √21.65 – 1
σ = √20.5
σ = 4.527 = 4.53
Coefficient of variation (C.V) = σ/x × 100%
C.V = 4.53/45 × 100%
C.V = 1006 × 100%
C.V = 10.06%
Therefore the coefficient of variation 10.06%
(7) The total marks scored by two students Sathya and Vidhya in 5 subjects are 460 and 480 with standard deviation 4.6 and 2.4 respectively. Who is more consistent in performance?
Solution:
Given that,
Sathya 5 subjects total scored = 460
Vidhya 5 subjects total scored = 480
∴ Satya scored mean (x) 460/5 = 92
∴ Vidhya scored mean (x_{2}) = 480/5 = 96
Sathya standard deviation (σ_{1}) = 4.6
Vidhya standard deviation (σ_{2}) = 2.4
Now, Sathya coefficient of variation
(C.V_{1}) = σ_{1}/x_{1}× 100%
= 4.6/92 × 100%
= 460/92 %
= 5%
Vidhya coefficient of variation
(C.V_{2}) = σ_{2}/x_{2}× 100%
= 2.4/96 × 100%
= 240/96 %
= 2.5 %
Now, C.V_{1}> C.V_{2}
Satya coefficient of variation is greater than vidhya.
Thus, The vidhya is more consistent in performance.
(8) The mean and standard deviation of marks obtained by 40 students of a class in three subjects Mathematics, Science and Social Science are given below.
Subject  Mean  SD 
Mathematics  56  12 
Science  65  14 
Social Science  60  10 
Which of the three subjects shows more consistent and which shows less consistent in marks?
Solution:
Mathematics mean (x_{1}) = 56
Mathematics S.D (σ_{1}) = 12
∴ Mathematics (C.V_{1}) = σ_{1}/x_{1}× 100%
= 12/50 × 100%
= 2143 × 100%
= 21.43%
Science mean (x_{2}) = 65 and S.D (σ_{2}) = 14
Science coefficient of variation (C.V_{2}) = σ_{2}/x_{2 }× 100%
C.V_{2} = 14/65 × 100%
C.V_{2} = .2154×100%
C.V_{2} = 21.54%
Social science mean (x_{3}) = 60 and S.D (σ_{3}) = 10
Social science (C.V_{3}) = σ_{3}/x_{3}× 100%
= 10/60 × 100%
= 16.67%
Now, C.V_{2}> C.V_{1}> C.V_{3}
∴Thus science subject more consistent and social science subject less consistent in mark.
Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 8 Statistics and Probability Exercise 8.2
Chapter 1 Relations and Functions  Chapter 2 Numbers and Sequences 
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Exercise 4.1 Solution

Chapter 5 Coordinate Geometry  Chapter 6 Trigonometry 
Exercise 5.1 Solution

Exercise 6.1 Solution 
Chapter 7 Mensuration  Chapter 8 Statistics and Probability 
Exercise 7.1 Solution  Exercise 8.1 Solution 
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