# Porbability Distribution Functions – Problems

### Porbability Distribution Functions – Problems

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.

Solution: P (X = 0) + p (X = 1) + p (X = 2) = 1

3c³ + 4c – 10c² + 5c – 1 = 1

3c³ + 9c – 10c² – 2 = 0

Put c = 1 then 3 – 10 + 9 – 2 = 12 – 12 = 0

c = 1 satisfy 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, c = ⅓

c = 2 ⇒ p (X = 0) = 3.2³ = 24 which is not possible

Therefore c = ⅓

2) Is the probability distribution of a random variable X. find the value of K and the variable of X.

 X = x -2 -1 0 1 2 3 P (X = x) 0.1 K 0.2 2k 0.3 K

Solution: Sum of the probability = 1

0.1 + k + 0.2 + 2k + 0.3 + k = 1

4k + 0.6 = 1

4k = 1 – 0.6 = 0.4

K = 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

μ = 0.8

Variance (σ²) = $$\sum\limits_{i=1}^{n}{{{x}_{i}}^{2}P(X={{x}_{i}})-{{\mu }^{2}}}$$

Therefore variance = 4 (0.1) + k (1) + 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

= 1.2 + 1.6 – 0.64

= 2.8 – 0.64

σ² = 2.16