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

 Problem Back to Problems Page
 The p.m.f. of a random variable (x) is given by f(x) = 2p for x = 1, f(x) = p for x = 2, f(x) = 4p for x = 3, f(x) = 0 otherwise
Find the value of P, E(x) and Var(x). Also calculate P(X > 1).

 [Expectation: 2.29 ; Variance: 0.7559 ; Standard Deviation : +0.869 ; P(X>1):0.714]

 Solution

where x = 1,   f(x) = 2p   ⇒ f(1) = 2p
where x = 2,   f(x) = p    ⇒ f(2) = p
where x = 3,   f(x) = 4p   ⇒ f(3) = 4p

Since f(x) represents the probability mass function, the discrete probability distribution of "x" would be

The probability distribution of "x" which is a discrete probability distribution would be
 x P(X = x) 1 2 3 2p p 4p

Since f(x) is a probability mass function,
Σ p = 1
⇒ 2p + p + 4p = 1
⇒ 7 p = 1
⇒ p
 1 7

The discrete probability distribution of "x" with the values of "p" replaced
x 1 2 3
P(X = x)
 2 7
 1 7
 4 7

Calculations for Mean and Standard Deviations

x P (X = x) px
[x × P (X = x)]
x2 px2
[x2 × P (X = x)]
1
 2 7
 2 7
1
 2 7
2
 1 7
 2 7
4
 4 7
3
 4 7
 12 7
9
 36 7
Total 1
 16 7
 42 7
= 2.29 = 6

Expectation/Mean of the distribution
 ⇒ E (x) (Or) x = Σ px = 2.29

Variance of the distribution
 ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = 6 − (2.29)2 = 6 − 5.2441 = 0.7559
Standard deviation of the distribution
 ⇒ SD (x) = + √ Var (x) = + √ 0.7559 = + 0.869
P (x > 1) = P (x = 2) + P (x = 3)
=
 1 7
+
 4 7
=
 5 7
(Or) 0.714

 Credit : Vijayalakshmi Desu