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Question
Solve the following problem :
In the following probability distribution of a r.v.X.
| x | 1 | 2 | 3 | 4 | 5 |
| P (x) | `(1)/(20)` | `(3)/(20)` | a | 2a | `(1)/(20)` |
Find a and obtain the c.d.f. of X.
Solution
Since the given table represents a p.m.f. of r.v. X,
\[\sum\limits_{x=1}^{5}\text{P}(x) = 1\]
∴ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1
∴ `(1)/(20) + (3)/(20) + "a" + 2"a" + (1)/(20)` = 1
∴ 3a = `1 - (5)/(20)`
∴ 3a = `1 - (1)/(4)`
∴ 3a = `(3)/(4)`
∴ a = `(1)/(4)`
By definition of c.d.f.,
F(x) = P(X ≤ x)
F(1) = P(X ≤ 1) = P(1) = `(1)/(20)`
F(2) = P(X ≤ 2) = F(1) + P(2)
= `(1)/(20) + (3)/(20) = (4)/(20)`
F(3) = P(X ≤ 3)
= F(2) + P(3)
= `(4)/(20) + "a" = (4)/(20) + (1)/(4) = (4)/(20) + (5)/(20) = (9)/(20)`
F(4) = P(X ≤ 4)
= F(3) + P(4)
= `(9)/(20) + 2"a" = (9)/(20) + (1)/(2) = (9)/(20) + (10)/(20) = (19)/(20)`
F(5) = P(X ≤ 5)
= F(4) + P(5)
= `(19)/(20) + (1)/(20)` = 1
∴ c.d.f. of X is as follows:
| xi | 1 | 2 | 3 | 4 | 5 |
| F(xi) | `(1)/(20)` | `(4)/(20)` | `(9)/(20)` | `(19)/(20)` | 1 |
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