Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X

#### Solution

First five positive integers are 1, 2, 3, 4 and 5

Two numbers from these can be selected in 5 × 4 = 20 ways

Let X denote the larger of the two numbers, so X can take values 2, 3, 4, 5

For X=2, the possible observations are (1,2) and (2,1)

`P(X = 2) = 2/20`

For X=3, the possible observations are (1,3),(3,1),(2,3) and (3,2)

`P(X = 3) = 4/20`

For X=4, the possible observations are (1,4),(4,1),(2,4),(4,2),(3,4) and (4,3)

`P(X = 4) = 6/20`

For X=5, the possible observations are (1,5),(5,1),(2,5),(5,2),(3,5),(5,3),(4,5) and (5,4)

`P(X = 5) = 8/20`

Therefore, the required probability distribution is as follows :

X = | 2 | 3 | 4 | 5 |

P(X) = | 2/20 | 4/20 | 6/20 | 8/20 |

Then Mean = `sumX_i P(X_i)`

`= 2xx2/20 + 3xx 4/20 + 4xx 6/20 + 5 xx8/20`

= 4

So the mean is 4

Now variance is `Var(X) = E(X^2) - {E(X)}^2`

`E(X^2) = 2^2 xx 2/20 + 3^3 xx 4/20 + 4^2 xx 6/20 + 5^2 xx 8/20`

`= 340/20 = 17`

Var(X) = `17 - (4)^2`

= 17 - 16

= 1

Hence, the variance is 1 and the mean is 4.