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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

  1. `2p(1 - 3p)`
  2. `1 - 4p`
  3. `(1 - 3p)^2`
  4. `6p - 16p^2`
  5. `p(5 - 9p)`
Show Answers Only

`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.
 

vcaa-2009-meth-7a

Find

  1. `text(Pr) (X > 1 | X <= 3)`  (2 marks)
  2. `text(Var)\ (X),` the variance of `X.`  (3 marks)
Show Answers Only
  1. `2/3`
  2. `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`