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

 Problem Back to Problems Page
Evaluate K if f(x) = K, x = 1, 2, 3, 4, 5, 6; f(x) = 0, elsewhere is a probability mass function. Also find its mean and standard deviation.

 [Expectation: 3.5 ; Variance: 2.917 ; Standard Deviation : +1.708]

 Solution

 f(x) = k ⇒ where x = 1,   f(1) = k where x = 2,   f(2) = k where x = 3,   f(3) = k where x = 4,   f(4) = k where x = 5,   f(5) = k where x = 6,   f(6) = k

The probability distribution of "x" which is a discrete probability distribution would be
 x P(X = x) 1 2 3 4 5 6 k k k k k k

Since f(x) is a probability mass function,
Σ p = 1
⇒ k + k + k + k + k + k = 1
⇒ 6 k = 1
⇒ k
 1 6

The discrete probability distribution of "x" with the values of "k" replaced
x 1 2 3 4 5 6
P(X = x)
 1 6
 1 6
 1 6
 1 6
 1 6
 1 6

Calculations for Mean and Standard Deviations

x P (X = x) px
[x × P (X = x)]
x2 px2
[x2 × P (X = x)]
1
 1 6
 1 6
1
 1 6
2
 1 6
 2 6
4
 4 6
3
 1 6
 3 6
9
 9 6
4
 1 6
 4 6
16
 16 6
5
 1 6
 5 6
25
 25 6
6
 1 6
 6 6
36
 36 6
Total 1
 21 6
 91 6
= 3.5 = 15.167

Expectation/Mean of the distribution
 ⇒ E (x) (Or) x = Σ px = 3.5

Variance of the distribution
 ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = 15.167 − (3.5)2 = 15.167 − 12.25 = 15.167 − 12.25 = 2.917
Standard deviation of the distribution
 ⇒ SD (x) = + √ Var (x) = + √ 2.917 = + 1.708

 Credit : Vijayalakshmi Desu