Samacheer Kalvi 10th Maths Book Solutions Chapter 8 – Statistics and Probability are available here. The Samacheer Kalvi 10th Maths book answers of Chapter 8, available at BYJU’S, contain step by step explanations designed by our mathematics experts. All these important questions are based on the new pattern prescribed by the Tamil Nadu board. Students can also get the solutions of other chapters on samacheer kalvi 10th maths solutions.
Chapter 8 of the Samacheer Kalvi 10th Maths guide will help the students to solve problems related to the measures of central tendency, measures of dispersion, coefficient of variation, probability of an event, algebra of events.
Samacheer Kalvi 10th Maths Chapter 8: Statistics and Probability Book Exercise 8.1 Questions and Solutions
Question 1: Find the range and coefficient of range of the following data: 63, 89, 98, 125, 79, 108, 117, 68.
Solution:
L – 125 and S = 63
Range = L – S
= 125 – 63
Range = L – S = 62
Coefficient of range = (L – S) / (L + S)
L + S = 125 + 63 = 188
Coefficient of range = 62 / 188
= 31 / 94
= 0.33
Question 2: Find the variance and standard deviation of the wages of 9 workers given below:
₹310, ₹290, ₹320, ₹280, ₹300, ₹290, ₹320, ₹310, ₹280.
Solution:
First, let us write the wages of 9 workers in ascending order.
280, 280, 290, 290, 300, 310, 310, 320, 320
x |
d = x – A d = x – 300 |
d^{2} |
280 |
-20 |
400 |
280 |
-20 |
400 |
290 |
-10 |
100 |
290 |
-10 |
100 |
300 |
0 |
0 |
310 |
10 |
100 |
310 |
10 |
100 |
320 |
20 |
400 |
320 |
20 |
400 |
Σd = 0 |
Σd^{2} = 2000 |
Σd^{2 }/ n = 2000 / 9
(Σd / n)^{2} = (0 / 9)^{2} = 0
σ = √(2000 / 9)
σ = √222.22
σ = 14.91
Question 3: If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation.
Solution:
While multiplying or dividing every element by the constant “k” then the standard deviation will be multiplied or divided by some constant “k”.
Since every element is multiplied by 3, it is required to divide the standard deviation by 3 to obtain the standard deviation of the new data set.
= 3.6 / 3
= 1.2
Variance = 1.44
Samacheer Kalvi 10th Maths Chapter 8: Statistics and Probability Book Exercise 8.2 Questions and Solutions
Question 1: The total marks scored by two students Sathya and Vidhya in 5 subjects are 460 and 480 with a standard deviation of 4.6 and 2.4, respectively. Who is more consistent in performance?
Solution:
Sathya:
Sum of marks in 5 subjects = 460
Mean marks of Sathya (x̄) = Σx / n = 460 / 5 = 92
Standard deviation (σ) = 4.6
Coefficient of variation (C.V) = (σ / x̄) ⋅ 100%
= (4.6 / 92) ⋅ 100%
= 0.05 ⋅ 100%
= 5%
Vidhya:
Sum of marks in 5 subjects = 480
Mean marks of Vidhya (x̄) = Σx / n = 480 / 5 = 96
Standard deviation (σ) = 2.4
Coefficient of variation (C.V) = (σ / x̄) ⋅ 100%
= (2.4 / 96) ⋅ 100%
= 0.25 ⋅ 100%
= 25%
So, Vidhya is more consistent.
Question 2: The standard deviation and coefficient of variation of a data are 1.2 and 25.6, respectively. Find the value of the mean.
Solution:
Standard deviation = 1.2
Coefficient of variation = 25.6
Coefficient of variation (C.V) = (σ / x̄) ⋅ 100%
25.6 = (1.2 / x̄) ⋅ 100%
x̄ = (1.2 / 25.6) / 100% = 4.687
x̄ = 4.69
Question 3: If the mean and coefficient of variation of a data are 15 and 48, respectively, then find the value of standard deviation.
Solution:
Mean (x̄) = 15
Coefficient of variation (C.V) = 48
Coefficient of variation (C.V) = (σ / x̄) ⋅ 100%
48 = (σ / 15) ⋅ 100%
σ = (48 ⋅ 15) / 100
= 720 / 100
x̄ = 7.2
Samacheer Kalvi 10th Maths Chapter 8: Statistics and Probability Book Exercise 8.3 Questions and Solutions
Question 1: If A is an event of a random experiment such that P(A) : P (A bar) = 17:15 and n(S) = 640 then find
(i) P (A bar)
(ii) n(A)
Solution:
P(A) : P (A bar) = 17 : 15
P(A) / P (A bar) = 17 / 15
P(A) / [1 – P (A)] = 17 / 15
15 P(A) = 17 [1 – P(A)]
15 P(A) = 17 – 17 P(A)
15 P(A) + 17 P(A) = 17
32 P(A) = 17
P(A) = 17 / 32
P(A) = n(A) / n(S)
P(A) = (17 / 32) ⋅ (20 / 20)
P(A) = 340 / 640
P(A bar) = 1 – P(A)
= 1 – (340 / 640)
= 300 / 640
P(A bar) = 15 / 32
n(A) = 340
Question 2: A coin is tossed thrice. What is the probability of getting two consecutive tails?
Solution:
The required sample space = {HHH. HHT, HTH, HTT, THH, THT, TTH, TTT}
n(S) = 8
Let A be the event of getting two consecutive tails
A = { HTT, TTH, TTT }
n(A) = 3
P(A) = n(A) / n(S)
P(A) = 3 / 8
Question 3: Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3 respectively. They are rolled and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.
Solution:
Sample space = {(1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,1) (2,1) (3,1) (4,1) (5,1) (6,1) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,2) (2,2) (3,2) (4,2) (5,2) (6,2) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) (1,3) (2,3) (3,3) (4,3) (5,3) (6,3) }
n(S) = 36
Let A, B, C, D, E, F, G, H be the event of getting the sum 2, 3, 4, 5, 6, 7, 8 and 9 respectively.
(i) Sum 2
A = {(1, 1) (1, 1)}
n(A) = 2
P(A) = 2 / 36
(ii) Sum 3
B = { (2, 1) (2, 1) (1, 2) (1,2) }
n(B) = 4
P(B) = 4 / 36
(iii) Sum 4
C = { (3,1) (3,1) (2,2) (2,2) (1,3) (1,3) }
n(C) = 6
P(C) = 6 / 36
(iv) Sum 5
D = {(4,1) (4,1) (3,2)(3,2) (2,3) (2,3)}
n(C) = 6
P(C) = 6 / 36
(v) Sum 6
E = {(5, 1) (5, 1) (4, 2) (4, 2) (3, 3) (3, 3) }
n(E) = 6
P(E) = 6 / 36
(vi) Sum 7
F = {(6, 1) (6, 1) (5, 2) (5, 2) (4, 3) (4, 3) }
n(F) = 6
P(F) = 6 / 36
(vii) Sum 8
G = {(6,2) (6,2) (5,3) (5,3)}
n(G) = 4
P(G) = 4 / 36
(viii) Sum 9
H = { (6,3)(6,3)}
n(H) = 2
P(H) = 2 / 36
Samacheer Kalvi 10th Maths Chapter 8: Statistics and Probability Book Exercise 8.4 Questions and Solutions
Question 1: If A and B are two mutually exclusive events of a random experiment and P(not A) = 0.45, P(A U B) = 0.65, then find P(B).
Solution:
A and B are mutually exclusive events, P(A n B) = 0.
P(not A) = 0.45
P(A) = 1 – P(not A)
P(A) = 1 – 0.45
P(A) = 0.55
P(A n B) = P(A) + P(B) – P(AUB)
0 = 0.55 + P(B) – 0.65
0.1 = P(B)
P(B) = 0.1
Question 2: The probability that at least one of A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then find P(A bar) + P(B bar).
Solution:
P(A or B) = P(A U B) = 0.6
P(A n B) = 0.2
P(A n B) = P(A) + P(B) – P(A U B)
0.2 = P(A) + P(B) – 0.6
P(A) + P(B) = 0.2 + 0.6
P(A) + P(B) = 0.8
1 – P(not A) + 1 – P(not B) = 0.8
2 – 0.8 = P(not A) + P(not B)
P(not A) + P(not B) = 1.2
Question 3: The probability of happening of an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then find the probability that neither A nor B happen
Solution:
P(A) = 0.5, P(B) = 0.3
P(A n B) = 0 (Since A and B are mutually exclusive events)
P(A bar n B bar)
P(A U B) = P(A) + P(B) – P(A n B)
P(A U B) = 0.5 + 0.3 – 0
P(A U B) = 0.8
P[(AUB) whole bar] = 1 – P(A U B)
P[(AUB) whole bar] = 1 – 0.8
P[(AUB) whole bar] = 0.2