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

Probability, MET1 2025 VCAA 4

The probability distribution for the discrete random variable \(X\) is given in the table below, where \(k\) is a positive real number.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}x \rule[-1ex]{0pt}{0pt}& \quad \quad 0 \quad \quad & \quad \quad 1 \quad \quad & \quad \quad 2 \quad \quad & \quad \quad 3 \quad \quad \\
\hline
\rule{0pt}{2.5ex}\operatorname{Pr}(X=x) \rule[-1ex]{0pt}{0pt}& \dfrac{4}{k} &\dfrac{2 k}{75} &\dfrac{k}{75} & \dfrac{2}{k} \\
\hline
\end{array}

Show that \(k=10\) or \(k=15\). (2 marks)

--- 8 WORK AREA LINES (style=lined) ---
Let \(k=15\).
1. Find \(\operatorname{Pr}(X>1)\). (1 mark)
  
  --- 3 WORK AREA LINES (style=lined) ---
2. Find \(E (X)\). (1 mark)
  
  --- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a. \(\text{Sum of probabilities\(=1\):}\)

\(\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}\)	\(=1\)
\(\dfrac{6}{k}+\dfrac{3 k}{75}\)	\(=1\)
\(3 k^2+450\)	\(=75k\)
\(3 k^2-75 k+450\)	\(=0\)
\(k^2-25 k+150\)	\(=0\)
\((k-10)(k-15)\)	\(=0\)

\(\therefore k=10 \ \ \text{or 15}\)

b.i. \(\operatorname{Pr}(X>1)=\dfrac{1}{3}\)

b.ii. \(E(X)=\dfrac{90}{75}\)

Show Worked Solution

a. \(\text{Sum of probabilities\(=1\):}\)

\(\dfrac{4}{k}+\dfrac{2 k}{75}+\dfrac{k}{75}+\dfrac{2}{k}\)	\(=1\)
\(\dfrac{6}{k}+\dfrac{3 k}{75}\)	\(=1\)
\(3 k^2+450\)	\(=75k\)
\(3 k^2-75 k+450\)	\(=0\)
\(k^2-25 k+150\)	\(=0\)
\((k-10)(k-15)\)	\(=0\)

\(\therefore k=10 \ \ \text{or 15}\)

b.i.	\(\operatorname{Pr}(X>1)\)	\(=\operatorname{Pr}(X=2)+\operatorname{Pr}(X=3)\)
		\(=\dfrac{15}{75}+\dfrac{2}{15}\)
		\(=\dfrac{1}{3}\)

b.ii.	\(E(X)\)	\(=0 \times \dfrac{4}{15}+1 \times \dfrac{30}{75}+2 \times \dfrac{15}{75}+3 \times \dfrac{2}{15}\)
		\(=\dfrac{30}{75}+\dfrac{30}{75}+\dfrac{30}{75}\)
		\(=\dfrac{90}{75}\)

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 \text{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

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

\(\text{Probabilities in table must be}\ \geqslant 0\ \ \Rightarrow \ k \geqslant 0.\)

\(E(X)_{\max}\ \text{occurs when}\ k=0\)

\(E(X)_{\max}\ =2\)

\(\Rightarrow E\)

♦♦ Mean mark 29%.
MARKER’S COMMENT: 26% of students incorrectly chose C.

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)

--- 4 WORK AREA LINES (style=lined) ---
What is the probability that Daniel receives only one telephone call on each of three consecutive days? (1 mark)

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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)

--- 6 WORK AREA LINES (style=lined) ---

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`