f(x) = ax ⇒ |
where x = 1, f(1) = a × 1 ⇒ f(1) = a |
|
where x = 2, f(2) = a × 2 ⇒ f(2) = 2a |
|
where x = 3, f(3) = a × 3 ⇒ f(3) = 3a |
|
... ... ... |
|
where x = n, f(n) = a × n ⇒ f(n) = na |
The probability distribution of "x" which is a discrete probability distribution would be
x |
1 |
2 |
3 |
... |
... |
n |
P(X = x) |
a |
2a |
3a |
... |
... |
na |
Since f(x) is a probability mass function,
Σ p = 1
⇒ a + 2a + 3a + .... + na = 1
⇒ a (1 + 2 + 3 + .... + n) = 1
⇒ a n(n + 1) = 2
Therefore, for a = |
|
, f(x) represents a probability mass function |
Calculations for Mean and Standard Deviations
x |
P (X = x) |
px [x × P (X = x)] |
x2 |
px2 [x2 × P (X = x)] |
1 |
a |
a |
1 |
a |
2 |
2a |
22a |
22 |
23a |
3 |
3a |
32a |
32 |
33a |
... |
... |
... |
... |
... |
... |
... |
... |
... |
... |
n |
na |
n2a |
n2 |
n3a |
Total |
1.00 |
4.36 |
|
22.54 |
Expectation/Mean of the distribution
⇒ E (x) (Or) x |
= |
Σ px |
|
= |
a + 22a + 32a + ... + n2a |
|
= |
a (1 + 22 + 32 + ... + n2) |
|
= |
a (12 + 22 + 32 + ... + n2) |
|
= |
|
|
= |
|
Variance of the distribution
⇒ var (x) |
= |
E (x2) − (E(x))2 |
⇒ var (x) |
= |
Σ px2 − (Σ px)2 |
|
= |
22.54 − (4.36)2 |
|
= |
22.54 − 19.0096 |
|
= |
3.5304 |
Standard Deviation of the distribution
⇒ SD (x) |
= |
+ √ Var (x) |
|
= |
+ √ 3.5304 |
|
= |
+ 1.879 |