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

 Problem Back to Problems Page
For what value of "a" will the function f(x) = ax; x = 1, 2, ..., n be the probability mass function of a discrete random variable x? find the mean and the variance of x.

Net Answers :
 [ Variance: 3.5304 ; Standard Deviation : +1.879]

 Solution

 f(x) = ax ⇒ where x = 1,   f(1) = a × 1 ⇒   f(1) = a where x = 2,   f(2) = a × 2 ⇒   f(2) = 2a where x = 3,   f(3) = a × 3 ⇒   f(3) = 3a ... ... ... where x = n,   f(n) = a × n ⇒   f(n) = na

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

Since f(x) is a probability mass function,
Σ p = 1
⇒ a + 2a + 3a + .... + na = 1
⇒ a (1 + 2 + 3 + .... + n) = 1
⇒ a
 n(n+1) 2
= 1
⇒ a n(n + 1) = 2
⇒ a =
 2 n(n+1)
Therefore, for a =
 2 n(n+1)
, f(x) represents a probability mass function

Calculations for Mean and Standard Deviations

x P (X = x) px
[x × P (X = x)]
x2 px2
[x2 × P (X = x)]
1 a a 1 a
2 2a 22a 22 23a
3 3a 32a 32 33a
... ... ... ... ...
... ... ... ... ...
n na n2a n2 n3a
Total 1.00 4.36 22.54

Expectation/Mean of the distribution
⇒ E (x) (Or) x = Σ px
= a + 22a + 32a + ... + n2a
= a (1 + 22 + 32 + ... + n2)
= a (12 + 22 + 32 + ... + n2)
=
a
 n(n+1)(2n+1) 6
=
a
 n(n+1)(2n+1) 6

Variance of the distribution
 ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = Σ px2 − (Σ px)2 = 22.54 − (4.36)2 = 22.54 − 19.0096 = 3.5304
Standard Deviation of the distribution
 ⇒ SD (x) = + √ Var (x) = + √ 3.5304 = + 1.879

 Credit : Vijayalakshmi Desu

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