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Statistics, 2ADV S3 SM-Bank 20

A random variable \(X\) has the probability density function \(f(x)\) given by

\(f(x)= \begin{cases}
\dfrac{k}{x^2} & 1 \leq x \leq 2 \\
\ \\
0 & \text {elsewhere }
\end{cases}\)

where \(k\) is a positive real number.

Show that  \(k = 2\).  (2 marks)

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\(\text{See Worked Solution}\)

Show Worked Solution

♦ Mean mark 48%.

\(\begin{aligned} \int_1^2 \dfrac{k}{x^2} d x & =1 \\
k\left[-\dfrac{1}{x}\right]_1^2 & =1 \\
k\left(-\dfrac{1}{2}+1\right) & =1 \\
\dfrac{k}{2} & =1 \\
\therefore k & =2 \quad \ldots \text{ as required }
\end{aligned}\)

 

Statistics, 2ADV S3 2021 HSC 33

People are given a maximum of six hours to complete a puzzle. The time spent on the puzzle, in hours, can be modelled using the continuous random variable \(X\) which has probability density function

\(f(x)= \begin{cases}
\dfrac{A x}{x^2+4} & \text{for } 0 \leq x \leq 6,(\text { where } A>0) \\
\ \\
0 & \text {for all other values of } x
\end{cases}\)

The graph of the probability density function is shown below. The graph has a local maximum.
 

  1. Show that  \(A=\dfrac{2}{\ln 10}\).  (2 marks)

  2. Show that the mode of \(X\) is two hours.  (2 marks)

  3. Show that  \(P(X<2)=\log _{10} 2\).  (2 marks)

  4. The Intelligence Quotient (IQ) scores of people are normally distributed with a mean of 100 and standard deviation of 15.
  5. It has been observed that the puzzle is generally completed more quickly by people with a high IQ.
  6. It is known that 80% of people with an IQ greater than 130 can complete the puzzle in less than two hours.
  7. A person chosen at random can complete the puzzle in less than two hours.
  8. What is the probability that this person has an IQ greater than 130? Give your answer correct to three decimal places.  (2 marks)

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  1. \(\text{See Worked Solution}\)
  2. \(\text{See Worked Solution}\)
  3. \(\text{See Worked Solution}\)
  4. \(0.066\)
Show Worked Solution

a.  \(\displaystyle\int_0^6 \dfrac{A x}{x^2+4} \, d x=1\)

\(\begin{aligned} \dfrac{A}{2} \int_0^6 \dfrac{2 x}{x^2+4} d x & =1 \\
\dfrac{A}{2}\left[\ln \left(x^2+4\right)\right]_0^6 & =1 \\
\dfrac{A}{2}(\ln 40-\ln 4) & =1 \\
\dfrac{A}{2} \ln \left(\dfrac{40}{4}\right) & =1 \\
\dfrac{A}{2} \ln 10 & =1 \\
A & =\dfrac{2}{\ln 10}\end{aligned}\)

b.  \(\text{Mode \(\rightarrow f(x)\) is a MAX}\)

♦♦♦ Mean mark part (b) 24%.

\(\begin{aligned}
f(x) & =\dfrac{A x}{x^2+4} \\
f^{\prime}(x) & =\dfrac{A\left(x^2+4\right)-A x(2 x)}{\left(x^2+4\right)^2} \\
& =\dfrac{A x^2+4 A-2 A x^2}{\left(x^2+4\right)^2} \\
& =\dfrac{A\left(4-x^2\right)}{\left(x^2+4\right)^2}
\end{aligned}\)

\(\text{\(f(x)\) max occurs when \(f^{\prime}(x)=0\) :}\)

\(\begin{aligned}
4-x^2 & =0 \\
x & =2 \quad(x>0)
\end{aligned}\)

♦♦ Mean mark part (c) 30%.
c.    \(P(X<2)\) \(=\displaystyle \int_0^2 \dfrac{A x}{x^2+4} d x\)
    \(=\dfrac{A}{2}\left[\ln \left(x^2+4\right)\right]_0^2\)
    \(=\dfrac{1}{\ln 10}(\ln 8-\ln 4)\)
    \(=\dfrac{1}{\ln 10}\left(\ln \dfrac{8}{4}\right)\)
    \(=\dfrac{1}{\ln 10} \cdot \ln 2\)
    \(=\log _{10} 2\)

 

d.   \(z \text{-score}(130)=\dfrac{x-\mu}{\sigma}=\dfrac{130-100}{15}=2\)

♦♦ Mean mark part (d) 25%.

\(P(z>2)=2.5 \%\)

\begin{aligned}
P(\text { IQ }>130 \mid x<2) & =\dfrac{P(\text { IQ }>130 \cap X<2)}{P(X<2)} \\
& =\dfrac{0.8 \times 0.025}{\log _{10} 2} \\
& =0.0664 \ldots \\
& =0.066
\end{aligned}

Statistics, 2ADV S3 EQ-Bank 16

A probability density function is defined by
 

`f(x) = {(a, \ text(for)\ \ 0 <= x <= 4),(3a, \ text(for)\  4 < x <= 8):}`
 

  1. Find the value of  `a`.  (2 marks)

  2. Sketch the probability density function.  (1 mark)

  3. Find an expression for the cumulative distribution function.  (2 marks)

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  1. `1/16`
  2.  
  3. `F(x) = {(x/16, text(for)\ 0 <= x <= 4),((3x)/16 – 1/2, text(for)\ 4 < x <= 8):}`
Show Worked Solution

i.   `int_0^4 a\ dx = [ax]_0^4 = 4a`

`int_4^8 3a\ dx = [3ax]_4^8 = 24a – 12a = 12a`

`4a + 12a` `= 1`
`a` `= 1/16`

 

ii.   

 

iii.   `text(When)\ \ 0 <= x <= 4:`

`F(x) = int 1/16\ dx = x/16 + C`

`F(0) = 0 \ => \ C = 0`

`F(x) = x/16\ \ …\ (1)`
 

`text(When)\ \ 4 < x <= 8:`

`F(x) = int 3/16\ dx = (3x)/16 + C`

`F(4) = 1/4\ \ (text{see (1) above})`

COMMENT: Understand why you can check your equation here by confirming  `F(8)=1`

`(3 xx 4)/16 + C` `= 1/4`
`C` `= −1/2`

 
`F(x) = (3x)/16 – 1/2`
 

`:. F(x) = {(x/16, \ text(for)\ \ 0 <= x <= 4),((3x)/16 – 1/2, \ text(for)\ \ 4 < x <= 8):}`

Statistics, 2ADV S3 SM-Bank 10

A continuous random variable, \(X\), has a probability density function given by

\(f(x)= \begin{cases}\dfrac{1}{5}\,e^{-\frac{x}{5}} & x \geq 0 \\
\ \\
0 & x<0
\end{cases}\) 

The median of \(X\) is  \(m\).

Determine the value of  \(m\).  (3 marks)

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\(-5 \log _e\left(\dfrac{1}{2}\right)\) or \(5 \log _e(2)\) or \(\log _e 32\)

Show Worked Solution

\(\begin{aligned} \dfrac{1}{5} \int_0^m e^{-\frac{x}{5}} d x & =\dfrac{1}{2} \\
\dfrac{1}{5} \times(-5)\left[e^{-\frac{x}{5}}\right]_0^m & =\dfrac{1}{2} \\
{\left[-e^{-\frac{x}{5}}\right]_0^m } & =\dfrac{1}{2} \\
-e^{-\frac{m}{5}}+1 & =\frac{1}{2} \\ e^{-\frac{m}{5}} & =\dfrac{1}{2} \\
-\frac{m}{5} & =\log _e\left(\dfrac{1}{2}\right)
\end{aligned}\)

\(\therefore m=-5 \log _e\left(\dfrac{1}{2}\right)\) (or \(5 \log _e(2)\), or \(\log _e 32\) )

Statistics, 2ADV S3 SM-Bank 5 MC

The function  `f(x)`  is a probability density function of a continuous random variable with the rule
 

`f(x) = {(ae^x, 0 <= x <= 1), (ae, 1 < x <= 2), (\ 0, text(otherwise)):}`
 

The value of `a` is

A.   `1`

B.   `1/e`

C.   `1/(2e)`

D.   `1/(2e - 1)`

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`D`

Show Worked Solution

`text(Total area) = 1`

`int_0^1 ae^x\ dx + int_1^2 ae\ dx` `= 1`
`[ae^x]_0^1 + [ae*x]_1^2` `=1`
`[ae-a] + [2ae-ae]` `=1`
`2ae-a` `=1`
`a(2e-1)` `=1`
`:. a` `= 1/(2e – 1)`

 
`=>   D`