In the p.m.f. of r.v. X
X | 1 | 2 | 3 | 4 | 5 |
P (X) | `1/20` | `3/20` | a | 2a | `1/20` |
Find a and obtain c.d.f. of X.
Solution
For p.m.f. of a r.v. X
`sum_("i" = 1)^5` P(X = 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`
∴ The p.m.f. of the r.v. X is
X | 1 | 2 | 3 | 4 | 5 |
P(X = x) | `1/20` | `3/20` | `5/20` | `10/20` | `1/20` |
Let F(x) be the c.d.f. of X.
Then F(x) = P(X ≤ x)
∴ F(1) = P(X ≤ 1) = P(X = 1) = `1/20`
F(2) = P(X≤ 2) = P(X = 1) + P(X = 2)
= `1/20 + 3/20`
= `4/20`
= `1/5`
F(3) = P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3)
= `1/20 + 3/20 + 5/20`
= `9/20`
F(4) = P(X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= `1/20 + 3/20 + 5/20 + 10/20`
= `19/20`
F(5) = P(X ≤ 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
= `1/20 + 3/20 + 5/20 + 10/20 + 1/20`
= `20/20`
= 1
Hence, the c.d.f. of the random variable X is as follows :
xi | 1 | 2 | 3 | 4 | 5 |
F(xi) | `1/20` | `1/5` | `9/20` | `19/20` | 1 |