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)] |
x2 |
px2 [x2 × 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 (x2) − (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 |