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Probability, 2ADV S1 EQ-Bank 16

The table shows the probability distribution for the discrete random variable \(X\).

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt}& \quad 1 \quad & \quad 2 \quad & \quad 3 \quad & \quad 4 \quad \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt}& 0.3 & 0.4 & 0.2 & 0.1 \\
\hline
\end{array}

Calculate \(\operatorname{Var}(X)\).   (2 marks)

Show Answers Only

\(\operatorname{Var}(X)=0.89\)

Show Worked Solution
\(\operatorname{Var}(X)\) \(=E\left(X^2\right)-\mu^2\)
  \(=1^2(0.3)+2^2(0.4)+3^2(0.2)+4^2(0.1)-[1(0.3)+2(0.4)+3(0.2)+4(0.1)]^2\)
  \(=5.3-2.1^2\)
  \(=0.89\)

Probability, 2ADV S1 2023 HSC 12

The table shows the probability distribution of a discrete random variable.

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & 0 & 1 & 2 & 3 & 4 \\
\hline
\rule{0pt}{2.5ex} P(X = x) \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ 0.3\ \  & \ \ 0.5\ \  & \ \ 0.1\ \  & \ \ 0.1\ \  \\
\hline
\end{array}

  1. Show that the expected value \(E(X)=2\).   (1 mark)

  2. Calculate the standard deviation, correct to one decimal place.   (2 marks)

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a.    `text{See Worked Solutions}`

b.    `0.9`

Show Worked Solution
a.     \(E(X)\) \(=0+1\times 0.3+2\times 0.5+3\times 0.1+4\times 0.1\)
    \(=0.3+1+0.3+0.4=2\)

 

b.     \(\text{Var}(X)\) \(=E(X^2)-[E(X)]^2\)
    \(=(0+1^2\times 0.3+2^2\times 0.5+3^2\times 0.1+4^2\times 0.1)-2^2\)
    \(=(0.3+2+0.9+1.6)-4\)
    \(=0.8\)

 

\(\therefore\ \sigma\) \(=\sqrt{0.8}\)  
  \(=0.8944…=0.9\ \text{(to 1 d.p.)}\)  

Probability, 2ADV S1 EQ-Bank 18

The discrete random variable `X` has the probability distribution shown in the table below.

\begin{array}{|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} X=x \rule[-1ex]{0pt}{0pt}& \quad 4 \quad & \quad 5 \quad & \quad 6 \quad & \quad 7 \quad & \quad 8 \quad \\
\hline
\rule{0pt}{2.5ex} P(x) \rule[-1ex]{0pt}{0pt}& 0.3 & a & 0.1 & 0.15 & 0.2 \\
\hline
\end{array}

 Find the value of  `a`, and hence calculate the the expected value and variance of  `X`.   (3 marks)

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`2.31`

Show Worked Solution

`0.3 + a + 0.1 + 0.15 + 0.2 = 1`

`=> \ a = 0.25`
 

`E(X) = ∑ x P(x)`

\begin{array}{|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} X=x \rule[-1ex]{0pt}{0pt}& \ \ 4 \ \  & \ \ 5 \ \ & \ \ 6 \ \ & \ \ 7 \ \ & \ \ 8 \ \ \\
\hline
\rule{0pt}{2.5ex} P(x) \rule[-1ex]{0pt}{0pt}& 0.3 & 0.25 & 0.1 & 0.15 & 0.2 \\
\hline
\rule{0pt}{2.5ex} x \times P(x) \rule[-1ex]{0pt}{0pt}& 1.2 & 1.25 & 0.6 & 1.05 & 1.6 \\
\hline
\end{array}

`E(X)` `= 1.2 + 1.25 + 0.6 + 1.05 + 1.6`
  `= 5.7`

 

`text(Var)(X)` `= E(X^2)-[E(X)]^2`
  `= (4^2 xx 0.3) + (5^2 xx 0.25) + (6^2 xx 0.1) + (7^2 xx 0.15) + (8^2 xx 0.2)-5.7^2`
  `= 2.31`

Probability, 2ADV S1 2009 MET1 7

The random variable `X` has this probability distribution.

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ 0\ \ \  & \ \ \ 1\ \ \  & \ \ \ 2\ \ \  & \ \ \ 3\ \ \  &\ \ \ 4\ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & 0.1 & 0.2 & 0.4 & 0.2 & 0.1 \\
\hline
\end{array}

Find

  1.  `P (X > 1 | X <= 3)`   (2 marks)

  2.  `P (X),` the variance of  `X.`   (3 marks)
Show Answers Only

a.    `2/3`

b.    `1.2`

Show Worked Solution

a.    `P(X > 1 | X <= 3)`

`= (P(X = 2) + P(X = 3))/(1-P(X = 4))`

`= (0.4 + 0.2)/(1-0.1)`

`= 0.6/0.9 = 2/3`
  

b.    `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`

 

`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)` `= E(X^2)-[E(X)]^2`
  `= 5.2-(2)^2`
  `= 1.2`

Probability, 2ADV S1 2017 MET2 14 MC

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

\begin{array} {|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & \ \ \ -1\ \ \  & \ \ \ \ 0\ \ \ \  & \ \ \ \ 1\ \ \ \ \\
\hline
\rule{0pt}{2.5ex} P(X=x) \rule[-1ex]{0pt}{0pt} & p & 2p & 1-3p \\
\hline
\end{array}

The variance of `X` is

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

`D`

Show Worked Solution
`text(Var)(X)` `= E(X^2)-[E(X)]^2`
  `= [(-1)^2p + 0^2 xx 2p + 1^2(1-3p)]-[-p + 0 + 1-3p]^2`
  `= 6p-16p^2`

 
`=> D`