smc-732-30-Var(X) / Std Dev

Probability, MET2 2017 VCAA 14 MC

The random variable `X` has the following probability distribution, where `0 < p < 1/3`.

The variance of `X` is

`2p(1 - 3p)`
`1 - 4p`
`(1 - 3p)^2`
`6p - 16p^2`
`p(5 - 9p)`

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`D`

Show Worked Solution

`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`

Probability, MET1 2009 VCAA 7

The random variable `X` has this probability distribution

Find

`text(Pr) (X > 1 | X <= 3)` (2 marks)

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`text(Var)(X),` the variance of `X.` (3 marks)
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`2/3`
`1.2`

Show Worked Solution

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`

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`