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 :

x_{i} |
1 | 2 | 3 | 4 | 5 |

F(x_{i}) |
`1/20` | `1/5` | `9/20` | `19/20` | 1 |