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
Since f(x) is a probability mass function,
Σ p = 1
⇒ 2p + p + 4p = 1
⇒ 7 p = 1
The discrete probability distribution of "x" with the values of "p" replaced
Calculations for Mean and Standard Deviations
x 
P (X = x) 
px [x × P (X = x)] 
x^{2} 
px^{2} [x^{2} × P (X = x)] 
1 


1 

2 


4 

3 


9 

Total 
1 





= 2.29


= 6

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

= 
2.29 
Variance of the distribution
⇒ var (x) 
= 
E (x^{2}) − (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) 

= 


= 
