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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)

Show Answers Only
  1. `text{See Worked Solutions}`
  2. `0.9`
Show Worked Solution
a.     `E(X)` `=0+1xx0.3+2xx0.5+3xx0.1+4xx0.1`
    `=0.3+1+0.3+0.4`
    `=2`

 

b.     `text{Var}(X)` `=E(X^2)-[E(X)]^2`
    `=(0+1^2xx0.3+2^2xx0.5+3^2xx0.1+4^2xx0.1)-2^2`
    `=(0.3+2+0.9+1.6)-4`
    `=0.8`

 

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

Probability, 2ADV S1 EQ-Bank 42

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

     `X = x` `4` `5` `6` `7` `8`
  `P(x)` `0.3` `a` `0.1` `0.15` `0.2`

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

Show Answers Only

`2.31`

Show Worked Solution

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

`=> \ a = 0.25`
 

`E(X) = ∑ x P(x)`
 

  `qquad X = x` `4` `5` `6` `7` `8`
  `qquad P(x) qquad` `0.3` `0.25` `0.1` `0.15` `0.2`
  `qquad x xx P(x) qquad` `1.2` `1.25` `0.6` `1.05` `1.6`

 

`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
  1. `2/3`
  2. `1.2`
Show Worked Solution

i.   `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`

 

ii.   `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`