smc-732-20-E(X) / Mean

Probability, MET2 2023 VCAA 12 MC

The probability mass function for the discrete random variable \(X\) is shown below.

\begin{array} {|c|c|c|c|c|}
\hline X &\ \ \ \ \ \ -1\ \ \ \ \ \ &\ \ \ \ \ \ 0\ \ \ \ \ \ &\ \ \ \ \ \ 1\ \ \ \ \ \ & 2 \\
\hline {Pr}(X=x) & k^2 & 3k & k & -k^2-4k+1 \\
\hline \end{array}

The maximum possible value for the mean of \(X\) is:

\(0\)
\(\dfrac{1}{3}\)
\(\dfrac{2}{3}\)
\(1\)
\(2\)

Show Answers Only

\(E\)

Show Worked Solution

\(E(X)\)	\( =(-1)\times k^2+1\times k+2\times(-k^2-4k+1)\)
	\(=-3k^2-7k+2\)

\(\text{From graph or solving using CAS for }k\geq 0\)

\(\text{Max possible value for mean of X is when}\ k=0\)

\(\therefore\ E(X)=2\)

\(\Rightarrow E\)

♦♦ Mean mark 29%.
MARKER’S COMMENT: 26% of students incorrectly chose C →\(\dfrac{2}{3}\).

Probability, MET2-NHT 2019 VCAA 5 MC

Consider the probability distribution for the discrete random variable `X` shown in the table below.

The value of `text(E)(X)` is

`(76)/(65)`
`1`
`0`
`(2)/(13)`
`(86)/(65)`

Show Answers Only

`A`

Show Worked Solution

`1`	`= b + b + b + (3)/(5) – b + (3b)/(5)`
`(2)/(5)`	`= (13b)/(5)`
`b`	`= (2)/(13)`

`text(E)(X)`	`= -(2)/(13) + 0 + (2)/(13) + 2((3)/(5) – (2)/(13)) +3 ((6)/(65))`
	`= (76)/(65)`

Probability, MET2 2019 VCAA 7 MC

The discrete random variable `X` has the following probability distribution.

	`qquad x`	`qquad 0 qquad`	`qquad 1 qquad`	`qquad 2 qquad`	`qquad 3 qquad`
	`qquad Pr(X = x) qquad`	`a`	`3a`	`5a`	`7a`

The mean of `X` is

`1/16`
`1`
`35/16`
`17/8`
`2`

Show Answers Only

`D`

Show Worked Solution

`16a = 1 qquad => qquad a = 1/16`

`text(E)(X)`	`= 3/16 xx 1 + 5/16 xx 2 + 7/16 xx 3`
	`=34/16`
	`= 17/8`

`=> D`

Probability, MET2 2018 VCAA 12 MC

The discrete random variable `X` has the following probability distribution.

Let `mu` be the mean of `X`.

`text(Pr) (X < mu)` is

`1/2`
`1/4`
`17/20`
`4/5`
`7/10`

Show Answers Only

`E`

Show Worked Solution

`mu`	`= (1 xx 9/20) + (2 xx 1/10) + (3 xx 1/20) + (6 xx 3/20)`
	`= 17/10`

`text(Pr) (X < 17/10)`	`= text(Pr) (X = 0) + text(Pr) (X = 1)`
	`= 7/10`

`=> E`

On any given day, the number `X` of telephone calls that Daniel receives is a random variable with probability distribution given by

Find the mean of `X`.(2 marks)
What is the probability that Daniel receives only one telephone call on each of three consecutive days? (1 mark)
Daniel receives telephone calls on both Monday and Tuesday.

What is the probability that Daniel receives a total of four calls over these two days? (3 marks)

Show Answers Only

`1.5`
`0.008`
`29/64`

Show Worked Solution

a.	`text(E) (X)`	`= 0 xx 0.2 + 1 xx 0.2 + 2 xx 0.5 + 3 xx 0.1`
		`= 0 + .2 + 1 + 0.3`
		`= 1.5`

b. `text(Pr) (1, 1, 1)`

MARKER’S COMMENT: Many students understood that the calculation of 0.2³ was required but were not able evaluate correctly.

`= 0.2 xx 0.2 xx 0.2`

`= 0.008`

c. `text(Conditional Probability:)`

♦ Mean mark 36%.

`text(Pr) (x = 4 | x >= 1\ text{both days})`

`= (text{Pr} (1, 3) + text{Pr} (2, 2) + text{Pr} (3, 1))/(text{Pr}(x>=1\ text{both days}))`

`= (0.2 xx 0.1 + 0.5 xx 0.5 + 0.1 xx 0.2)/(0.8 xx 0.8)`

`= (0.02 + 0.25 + 0.02)/0.64`

`= 0.29/0.64`

`= 29/64`

`5`	`=(0 xx a) + (2 xx 0.2) + (4 xx 0.2) + (6 xx 0.3) + 8b`
`8b`	`=2`
`:. b`	`=0.25`

`1 xx b + 2 xx 0.4`	`=1`
`b`	`=0.2`

`a + 0.2 + 0.4`	`= 1`
`a`	`=0.4`

`0.2 + 0.6p^2 + 0.1 + 1 – p + 0.1`	`= 1`
`0.6p^2 – p + 0.4`	`= 0`
`6p^2 – 10p + 4`	`= 0`
`3p^2 – 5p + 2`	`= 0`
`(p – 1) (3p – 2)`	`= 0`

b.i.	`text(E) (X)`	`= sum x text(Pr) (X = x)`
		`= 1 xx (3/5 xx 2^2/3^2) + 2 (1/10) + 3 (1-2/3) + 4 (1/10)`
		`= 4/15 + 1/5 + 1 + 2/5`
		`= 28/15`

`:.\ text(Smallest)\ text(E)(X)`	`=0.8 xx -1 + 0.2 xx 4`
	`=0`

ii.	`text(Pr) (X >= 28/15)`	`= text(Pr) (X = 2) + text(Pr) (X = 3) + text(Pr) (X = 4)`
		`= 1/10 + 1/3 + 1/10`
		`= 8/15`

`p + 2p + 3p + 4p + 5p`	`= 1`
`:. p`	`= 1/15`

`text(E)(X)`	`= 1 xx p + 2(2p) + 3(3p) + 4(4p) + 5(5p)`
	`= 55p`
	`= 55 xx (1/15)`
	`= 11/3`