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, Σx2 = 765, then cellular the coefficient of variation.
Solution:
Given that n = 5, mean (x) = 6, Σx2 = 765
Standard deviation σ = √Σxi2/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
| xi | di = xi – A
= xi – 31 |
di2 |
| 24 | -7 | 49 |
| 26 | -5 | 25 |
| 33 | 2 | 4 |
| 37 | 6 | 36 |
| 29 | -2 | 4 |
| 31 | 0 | 0 |
| Σxi = 180 | Σdi = -6 | Σdi2 = 118 |
Now, n = 6
Then, we know that x = Σxi/x = 180/6 = 30
∴ = 30
We know that standard deviation
σ = √Σdi2/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
| xi | di = xi – A
= xi – 44 |
di2 |
| 38 | -6 | 36 |
| 40 | -4 | 16 |
| 47 | 3 | 9 |
| 44 | 0 | 0 |
| 46 | 2 | 4 |
| 43 | -1 | 1 |
| 49 | 5 | 25 |
| 53 | 9 | 81 |
| Σxi = 360 | Σdi = 8 | Σdi2= 172 |
Now, n = 8, then x = Σxi/n = 360/8 = 45
∴ x = 45
Standard deviation σ = √Σdi2/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 (x2) = 480/5 = 96
Sathya standard deviation (σ1) = 4.6
Vidhya standard deviation (σ2) = 2.4
Now, Sathya coefficient of variation
(C.V1) = σ1/x1× 100%
= 4.6/92 × 100%
= 460/92 %
= 5%
Vidhya coefficient of variation
(C.V2) = σ2/x2× 100%
= 2.4/96 × 100%
= 240/96 %
= 2.5 %
Now, C.V1> C.V2
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 (x1) = 56
Mathematics S.D (σ1) = 12
∴ Mathematics (C.V1) = σ1/x1× 100%
= 12/50 × 100%
= 2143 × 100%
= 21.43%
Science mean (x2) = 65 and S.D (σ2) = 14
Science coefficient of variation (C.V2) = σ2/x2 × 100%
C.V2 = 14/65 × 100%
C.V2 = .2154×100%
C.V2 = 21.54%
Social science mean (x3) = 60 and S.D (σ3) = 10
Social science (C.V3) = σ3/x3× 100%
= 10/60 × 100%
= 16.67%
Now, C.V2> C.V1> C.V3
∴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
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| Chapter 3 Algebra | Chapter 4 Geometry |
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Exercise 4.1 Solution
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| Chapter 5 Coordinate Geometry | Chapter 6 Trigonometry |
| Exercise 5.1 Solution
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Exercise 6.1 Solution |
| Chapter 7 Mensuration | Chapter 8 Statistics and Probability |
| Exercise 7.1 Solution | Exercise 8.1 Solution |
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