"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) = CK2 ⇒ |
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)] |
x2 |
px2 [x2 × 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 (x2) − (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 |