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Question
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.
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