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