# Probability Distribution Mean (Expectation), Variance :: Problems

 Problem Back to Problems Page
A random variable x has the following probability distribution
 Values of x P(X = x) 0 1 2 3 4 5 6 7 0 2k 2k k 3k k2 2K2 7k2 + k
Find the value of K and then evaluate P(x < 6), P(x ≥ 6), and P(0 < x < 5). Find also the mean and variance of the distribution

 [Expectation: 3.46; Variance: 4.0284 ; Standard Deviation : +2.007]

 Solution

Since X is a random variable.

Assuming the random variable to be discrete..

The given probability distribution would be a discrete probability distribution of "X"

For a discrete probability distribution of a random variable "X", Σ p = 1

From the above distribution

Σ p = 1
⇒ 0 + 2K + 2K + K + 3K + K2 + 2K2 + (7K2 + K) = 1
⇒ 9k + 10 k2 = 1
⇒ 10 K2 + 9K − 1 = 0
⇒ 10 K2 + 10K − k − 1 = 0
⇒ 10 K (K + 1) − 1 (K + 1) = 0
⇒ (10K − 1) (K + 1) = 0
⇒ 10K − 1 = 0 (Or) K + 1 = 0
⇒ 10K = 1 (Or) K = − 1
⇒ K =
 1 10
(Or) 0.1
Since probability cannot be negative, k = −1 is ignored.

The probability distribution replacing the values of of "k" would be

x P (X = x) In K terms Calculations 0 0 0 1 2k 2 × (0.1) 0.2 2 2k 2 × (0.1) 0.2 3 k 1 × (0.1) 0.1 4 3k 3 × (0.1) 0.3 5 k2 (0.1)2 0.01 6 2k2 2 × (0.1) 0.02 7 7k2 + k 7 × (0.1) + (0.1) 0.17

Therefore,
 P(x < 6) = P(x=0) + P (x = 1) + P (x = 2) + P (x = 3) + P (x = 4) + P (x = 5) (Or) = 1 − P(x ≥ 6) = 1 − [P(x=6) + P(x=7)] = 1 − [0.02 + 0.17] = 1 − 0.19 = 0.81
 P(x ≥ 6) = P(x=6) + P(x=7) = 0.02 + 0.17 = 0.19
 P(0 < x < 5) = P(x=1) + P(x=2) + P(x=3) + P(x=4) = 0.2 + 0.2 + 0.1 + 0.3 = 0.8

Calculations for finding the mean and variance of the distribution

x P (X = x) px [x × P (X = x)] x2 px2 [x2 × P (X = x)] 0 0.02 0 0 0 1 0.20 0.20 1 0.20 2 0.20 0.40 4 0.80 3 0.10 0.30 9 0.90 4 0.30 1.20 16 4.80 5 0.01 0.05 25 0.25 6 0.02 0.12 36 0.72 7 0.17 1.19 49 8.33 1.00 3.46 16.00

Mean/Expectation of the distribution
 ⇒ E (x) (Or) x = Σ px = 3.46

Variance of the distribution
 ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = Σ px2 − (Σ px)2 = 16 − (3.46)2 = 16 − 11.9716 = 4.0284
Standard Deviation of the distribution
 ⇒ SD (x) = + √ Var (x) ⇒ SD (x) = + √ 4.0284 ⇒ SD (x) = + 2.007

 Credit : Vijayalakshmi Desu