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The random variable `X` has the following probability distribution, where `0 < p < 1/3`.
The variance of `X` is
`D`
`text(Var)(X)` | `= text(E)(X^2) – [text(E)(X)]^2` |
`= [(−1)^2p + 0^2 xx 2p + 1^2(1 – 3p)] – [−p + 0 + 1 – 3p]^2` | |
`= 6p – 16p^2` |
`=> D`
Show Answers Only
a. `text(Pr) (X > 1 | X <= 3)` `= (text{Pr} (X = 2) + text{Pr} (X = 3))/(1 – text{Pr} (X = 4))` `= (0.4 + 0.2)/(1 – 0.1)` `= 0.6/0.9` `= 2/3` Show Worked Solution
b. `text(E) (X)`
`= 0.1 (0) + 1 (0.2) + 2 (0.4) + 3 (0.2) + 4 (0.1)`
`= 0 + 0.2 + 0.8 + 0.6 + 0.4`
`= 2`
`text(E) (X^2)`
`= 0^2 (0.1) + 1^2 (0.2) + 2^2 (0.4) + 3^2 (0.2) + 4^2 (0.1)`
`= 0 + 0.2 + 1.6 + 1.8 + 1.6`
`= 5.2`
`:.\ text(Var)\ (X)`
`= text(E) (X^2) – [text(E) (X)]^2`
`= 5.2 – (2)^2`
`= 1.2`