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

4 


16 

5 


25 

6 


36 

Total 
1 





= 3.5


= 15.167

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

= 
3.5 
Variance of the distribution
⇒ var (x) 
= 
E (x^{2}) − (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 