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

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 = \(\frac{1}{3}\)

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:

Question 3.

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.

is the probability distribution of a random variable X. Find the variance of X.
Solution:

Question 5.
A random variable X has the following probability distribution.

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
= \(\frac{8}{9}\)

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