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)] |
x2 |
px2 [x2 × 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 (x2) − (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) |
|
= |
|
|
= |
|