smc-637-20-Median (old)

Probability, MET2 2018 VCAA 15 MC

A probability density function, `f`, is given by

`f(x) = {(1/12 (8x -x^3)), (\ 0):} qquad {:(0 <= x <= 2), (text(elsewhere)):}`

The median, `m`, of this function satisfies the equation

A. `-m^4 + 16m^2 - 6 = 0`

B. `-m^4 + 4m^2 - 6 = 0`

C. `m^4 - 16m^2 = 0`

D. `m^4 - 16m^2 + 24 = 0.5`

E. `m^4 - 16m^2 + 24 = 0`

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`E`

Show Worked Solution

♦ Mean mark 49%.

`int_0^m 1/12(8x – x^3) dx`	`= 0.5`
`[1/12(4x^2 – x^4/4)]_0^m`	`= 0.5`
`(4m^2 – m^4/4)`	`= 6`
`16m^2 – m^4`	`= 24`
`m^4 – 16m^2 + 24`	`= 0`

`=> E`

Probability, MET1 2014 VCAA 8

A continuous random variable, `X`, has a probability density function given by

`f(x) = {{:(1/5e^(−x/5),x >= 0),(0, x < 0):}`

The median of `X` is `m`.

Determine the value of `m`. (2 marks)
The value of `m` is a number greater than 1.

Find `text(Pr)(X < 1 | X <= m)`. (2 marks)

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`−5log_e(1/2)\ \ text(or)\ \ 5log_e(2)\ \ text(or)\ \ log_e 32`
`2(1 – e^(−1/5))`

Show Worked Solution

a.	`1/5 int_0^m e^(−x/5)dx`	`= 1/2`
	`1/5 xx (−5)[e^(−x/5)]_0^m`	`= 1/2`
	`[-e^(- x/5)]_0^m`	`= 1/2`
	`-e^(−m/5) + 1`	`= 1/2`
	`e^(−m/5)`	`= 1/2`
	`- m/5`	`= log_e(1/2)`

`:. m = −5log_e(1/2)\ \ \ (text(or)\ \ 5log_e(2),\ text(or)\ \ log_e 32)`

b. `text(Using Conditional Probability:)`

♦ Part (b) mean mark 38%.
MARKER’S COMMENT: A common error was assuming `m` obtained in part (a) was equivalent to `text(Pr)(X<=m)`.

`text(Pr)(X < 1 \| X <= m)`	`= (text(Pr)(X < 1))/(text(Pr)(X <= m))`
	`= (1/5 int_0^1 e^(−x/5)dx)/(1/2)`
	`= (1/5(−5)[e^(−x/5)]_0^1)/(1/2)`
	`= −2[(e^(−1/5)) – e^0]`
	`= 2(1 – e^(−1/5))`