Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(a) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Exercise 10(a)

I.

Question 1.
A p.d.f of a discrete random variable is zero except at the points x = 0, 1, 2. At these points it has the value P(0) = 3c3, P(1) = 4c – 10c2, P(2) = 5c – 1 for some c > 0. Find the value of c.
Solution:
P(X = 0) + P(X = 1) + P(X = 2) = 1
3c3 + 4c – 10c2 + 5c – 1 = 1
3c3 – 10c2 + 9c – 2 = 0
Put c = 1, then 3 – 10 + 9 – 2 = 12 – 12 = 0
c = 1 satisfies the above equation
c = 1 ⇒ P(X = 0) = 3 which is not possible
Dividing with c – 1, we get
3c2 – 7c + 2 = 0
(c – 2) (3c – 1) = 0
∴ c = 2 or c = \(\frac{1}{3}\)
c = 2
⇒ P(X = 0) = 3 . 23 = 24 which is not possible
∴ c = \(\frac{1}{3}\)

Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Ex 10(a)

Question 2.
Find the constant C, so that F(x) = \(C\left(\frac{2}{3}\right)^x\), x = 1, 2, 3,……… is the p.d.f of a discrete random variable X.
Solution:
Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Ex 10(a) I Q2

Question 3.
Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Ex 10(a) I Q3
is the probability distribution of a random variable X. Find the value of K and the variance of X.
Solution:
Sum of the probabilities = 1
⇒ 0.1 + k + 0.2 + 2k + 0.3 + k = 1
⇒ 4k + 0.6 = 1
⇒ 4k = 1 – 0.6 = 0.4
⇒ k = \(\frac{0.4}{4}\) = 0.1
Mean = (-2) (0.1) + (-1) (k) + 0 (0.2) + 1 (2k) + 2(0.3) + 3k
= – 0.2 – k + 0 + 2k + 0.6 + 3k
= 4k + 0.4
= 4(0.1) + 0.4
= 0.4 + 0.4
= 0.8
μ = 0.8
Variance (σ2) = \(\sum_{i=1}^n x_i^2 P\left(x=x_i\right)-\mu^2\)
∴ Variance = 4(0.1) + 1(k) + 0(0.2) + 1 (2k) + 4 (0.3) + 9k – μ2
= 0.4 + k + 0 + 2k + 4(0.3) + 9k – μ2
= 12k + 0.4 + 1.2 – (0.8)2
= 12(0.1) + 1.6 – 0.64
= 1.2 + 1.6 – 0.64
= 2.8 – 0.64
= 2.16
∴ σ2 = 2.16

Question 4.
Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Ex 10(a) I Q4
is the probability distribution of a random variable X. Find the variance of X.
Solution:
Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Ex 10(a) I Q4.1

Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Ex 10(a)

Question 5.
A random variable X has the following probability distribution.
Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Ex 10(a) I Q5
Find (i) k (ii) the mean and (iii) P(0 < X < 5)
Solution:
Sum of the probabilities = 1
⇒ 0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
⇒ 10k2 + 9k = 1
⇒ 10k2 + 9k – 1 = 0
⇒ 10k2 + 10k – k – 1 = 0
⇒ 10k(k + 1) – 1(k + 1) = 0
⇒ (10k – 1) (k + 1) = 0
⇒ k = \(\frac{1}{10}\), -1 Since k > 0
∴ k = \(\frac{1}{10}\)

(i) k = \(\frac{1}{10}\)

(ii) Mean = \(\sum_{i=1}^n x_i P\left(x=x_i\right)\)
Mean (μ) = 0(0) + 1(k) + 2(2k) + 3(2k) + 4(3k) + 5(k2) + 6(2k2) + 7 (7k2 + k)
= 0 + k + 4k + 6k + 12k + 5k2 + 12k2 + 49k2 + 7k
= 66k2 + 30k
= 66(\(\frac{1}{100}\)) + 30(\(\frac{1}{10}\))
= 0.66 + 3
= 3.66

(iii) P(0 < x < 5)
P(0 < x < 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= k + 2k + 2k + 3k
= 8k
= 8(\(\frac{1}{10}\))
= \(\frac{4}{5}\)

II.

Question 1.
The range of a random variable X is {0, 1, 2}. Given that P(X = 0) = 3c3, P(X = 1) = 4c – 10c2, P(X = 2) = 5c – 1
(i) Find the value of c
(ii) P(X < 1), P(1 ≤ X < 2) and P(0 < X ≤ 3)
Solution:
P(X = 0) + P(X = 1) + P(X = 2) = 1
3c3 + 4c – 10c2 + 5c – 1 = 1
3c3 – 10c2 + 9c – 2 = 0
c = 1 satisfy this equation
c = 1 ⇒ P(X = 0) = 3 which is not possible
Dividing with c – 1, we get
3c2 – 7c + 2 = 0
(c – 2) (3c – 1) = 0
c = 2 or c = \(\frac{1}{3}\)
c = 2 ⇒ P(X = 0) = 3 . 23 = 24 which is not possible
∴ c = \(\frac{1}{3}\)

(i) P(X < 1) = P(X = 0)
= 3 . c3
= 3 . \(\left(\frac{1}{3}\right)^3\)
= 3 . \(\frac{1}{2}\)
= \(\frac{1}{9}\)

(ii) P(1 < X ≤ 2) = P(X = 2)
= 5c – 1
= \(\frac{5}{3}\) – 1
= \(\frac{2}{3}\)

(iii) P(0 < X ≤ 3) = P(X = 1) + P(X = 2)
= 4c – 10c2 + 5c – 1
= 9c – 10c2 – 1
= 9 . \(\frac{1}{3}\) – 10 . \(\frac{1}{9}\) – 1
= 3 – \(\frac{10}{9}\) – 1
= \(\frac{8}{9}\)

Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Ex 10(a)

Question 2.
The range of a random variable X is {1, 2, 3, …..} and P(X = k) = \(\frac{C^K}{K !}\), (k = 1, 2, 3, ……), Find the value of C and P(0 < X < 3)
Solution:
Sum of the probabilities = 1
Inter 2nd Year Maths 2A Random Variables and Probability Distributions Solutions Ex 10(a) II Q2