Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.2 Pdf

Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.2 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.2: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.2 – Statistics and Probability on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 8 Exercise 8.2 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 8.2

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

Chapter 1 Relations and Functions Chapter 2 Numbers and Sequences
Chapter 3 Algebra Chapter 4 Geometry
Exercise 4.1 Solution

Exercise 4.2 Solution

Exercise 4.3 Solution

Exercise 4.4 Solution

Exercise 4.5 Solution

Exercise 4 Solution

 

Chapter 5 Coordinate Geometry Chapter 6 Trigonometry
Exercise 5.1 Solution

Exercise 5.2 Solution

Exercise 5.3 Solution

Exercise 5.4 Solution

Exercise 5.5 Solution

Unit Exercise 5 Solution

 

Exercise 6.1 Solution

Exercise 6.2 Solution

Exercise 6.3 Solution

Exercise 6.4 Solution

Exercise 6.5 Solution

Unit Exercise 6 Solution

Chapter 7 Mensuration Chapter 8 Statistics and Probability
Exercise 7.1 Solution

Exercise 7.2 Solution

Exercise 7.3 Solution

Exercise 7.4 Solution

Exercise 7.5 Solution

Unit Exercise 7 Solution

Exercise 8.1 Solution

Exercise 8.2 Solution

Exercise 8.3 Solution

Exercise 8.4 Solution

Exercise 8.5 Solution

Unit Exercise 8 Solution

 

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Updated: March 4, 2022 — 2:40 pm

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