"x" represent the value in the range of the random variable "X".
Therefore, f(x) = P(X=x) represents the probability mass function
f(x) = P(X = k) = CK^{2} ⇒ 
where K = 1, f(1) = C (1)^{2} 
⇒ f(1) = C 

where K = 2, f(2) = C (2)^{2} 
⇒ f(2) = 4C 

where x = 3, f(3) = C (3)^{2} 
⇒ f(3) = 9C 

where x = 4, f(4) = C (4)^{2} 
⇒ f(4) = 16C 
Since f(x) represents the probability mass function, the discrete probability distribution of "x" would be
x 
1 
2 
3 
4 
P(X = x) 
C 
4C 
9C 
16C 
Since f(x) is a probability mass function,
Σ p = 1
⇒ C + 4C + 9C + 16C = 1
⇒ 30C = 1
The discrete probability distribution of "x" with the values of "C" replaced
The probability distribution replacing the values of of "k" would be

x 
P (X = x) 

In C terms 
Calculations 
Probability 

1 
C 



2 
4C 



3 
9C 



4 
16C 


Therefore, the distribution would be
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 

Total 
1 





= 3.33


= 11.8

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

= 
3.33 
Variance of the distribution
⇒ var (x) 
= 
E (x^{2}) − (E(x))^{2} 
⇒ var (x) 
= 
11.8 − (3.33)^{2} 

= 
11.8 − 11.09 

= 
0.71 
Standard deviation of the distribution
⇒ SD (x) 
= 
+ √ Var (x) 

= 
+ √ 0.71 

= 
+ 0.843 