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Statistics, 2ADV S3 EQ-Bank 2

A probability density function for a random variable, \(X\), is defined by the following function
 

\(f(x)=\left\{\begin{array}{rl}
k x(2-x), & 0 \leqslant x \leqslant 2 \\
0, & \text{for all other } x
\end{array}\right.\)
 

Determine the value of \(k\) and hence write the equation of the Cumulative Density Function.   (3 marks)

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\(F(x)=\dfrac{3}{4} x^2-\dfrac{1}{4} x^3\)

Show Worked Solution
  \(\displaystyle \int_0^2 k x(2-x)\) \(=1\)
  \(\displaystyle \int_0^2 2 k x-k x^2\) \(=1\)
  \(\left[k x^2-\dfrac{k}{3} x^3\right]_0^2\) \(=1\)
  \(\left(4 k-\dfrac{8 k}{3}\right)-0\) \(=1\)
  \(4 k\) \(=3\)
  \(k\) \(=\dfrac{3}{4}\)

 

\(F(x)\) \(=\displaystyle \int 2 \times \frac{3}{4} x-\frac{3}{4} x^2\, d x\)
  \(=\displaystyle \int \frac{3}{2} x-\frac{3}{4} x^2\, d x\)
  \(=\dfrac{3}{4} x^2-\dfrac{1}{4} x^3+c\)

 
\(\text{When} \ \ x=0, F(x)=0 \ \Rightarrow \ c=0\)

\(\therefore\ F(x)=\dfrac{3}{4} x^2-\dfrac{1}{4} x^3\)

Statistics, 2ADV S3 2024 HSC 25

A function \(f(x)\) is defined as

\(f(x)=\left\{\begin{array}{ll} 0, & \text { for}\ \ x \lt 0 \\
1-\dfrac{x}{h}, & \text { for}\ \ 0 \leq x \leq h, \\
0, & \text { for}\ \  x \gt h \end{array}\right.\)

where \(h\) is a constant.

  1. Find the value of \(h\) such that \(f(x)\) is a probability density function.   (2 marks)

  2. By first finding a formula for the cumulative distribution function, sketch its graph.   (2 marks)

  3. Find the value of the median of the probability density function \(f(x)\) . Give your answer correct to 3 decimal places.   (2 marks)

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a.   \(h=2\)

b.   

c.   \(\text{Median}\ =0.586\)

Show Worked Solution

a.   \(f(x)\ \text{is a PDF if:}\)

\(\displaystyle \int_{0}^{h} 1-\dfrac{x}{h}\,dx\) \(=1\)  
\(\Big[x-\dfrac{x^2}{2h} \Big]_0^{h}\) \(=1\)  
\(h-\dfrac{h}{2}\) \(=1\)  
\(h\) \(=2\)  

  
b.   \( \displaystyle \int 1-\dfrac{x}{2}\,dx = x-\dfrac{x^2}{4} + c\)

\(\text{At}\ \ x=0, \ \ F(0)=0,\ \ c=0 \)

\(f(x)=\left\{\begin{array}{ll} 0, & \text { for}\ \ x \lt 0 \\
x-\dfrac{x^2}{4}, & \text { for}\ \ 0 \leq x \leq 2, \\
1, & \text { for}\ \  x \gt 2 \end{array}\right.\)
 

♦♦ Mean mark (b) 27%.

c.   \(\text{Let}\ \ m=\ \text{median of}\ f(x) \)

\(\displaystyle \int_0^{m} 1-\dfrac{x}{2}\,dx\)  \(=0.5\)  
\(\Big[ x-\dfrac{x^2}{4} \Big]_0^m\) \(=0.5\)  
\(m-\dfrac{m^2}{4}\) \(=0.5\)  
\(4m-m^2\) \(=2\)  
\(m^2-4m+2\) \(=0\)  
\((m-2)^2\) \(=2\)  
\(m\) \(=2 \pm \sqrt{2}\)  

 

\(\therefore \ \text{Median}\) \(=2-\sqrt{2}\ \ (x \in [0,2]) \)   
  \(=0.586\ \text{(3 d.p.)}\)  
♦♦ Mean mark (c) 38%.

Statistics, 2ADV S3 2023 HSC 29

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

\(f(x)=\left\{\begin{array}{cl} 12 x^2(1-x), & \text { for } 0 \leq x \leq 1 \\ 0, & \text { for all other values of } x \end{array}\right.\)

 

  1. Find the mode of \(X\).  (2 marks)

  2. Find the cumulative distribution function for the given probability density function.  (2 marks)

  3. Without calculating the median, show that the mode is greater than the median.  (2 marks)

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a.   \(x=\dfrac{2}{3} \)

b.   \(F(x)=\left\{\begin{array}{cl} 0, & \text { for } x \lt 0 & \\ 4x^3-3x^4, & \text { for } 0 \leq x \leq 1 \\ 1, & \text { for } x > 1 \end{array}\right.\)

c.    \(\text{See Worked Solutions}\)

Show Worked Solution

a.    \(\text{Mode}\ \rightarrow \ f(x)_\text{max} \)

\(f(x)=12 x^2(1-x)=12x^2-12x^3 \)

\(f^{′}(x)=24x-36x^2=12x(2-3x) \)

\(f^{″}(x)=24-72x \)

♦ Mean mark (a) 45%.

\(\text{Max/min when}\ f^{′}(x)=0 \)

\(2-3x=0\ \ ⇒\ \ x=\dfrac{2}{3} \ \ (x \neq 0) \)

\(\text{At}\ x=\dfrac{2}{3}, \ f^{″}(x)=24-72(\dfrac{2}{3})=-24<0 \)

\(\therefore \ \text{Mode (max) at}\ x=\dfrac{2}{3} \)
  

b.     \(F(x)\) \(= \int 12x^2-12x^3\ dx\)
    \(=4x^3-3x^4+c \)

 
\(\text{At}\ x=0, F(x)=0\ \ ⇒\ \ c=0 \)

\(F(x)=4x^3-3x^4 \)
 

\(F(x)=\left\{\begin{array}{cl} 0, & \text { for } x \lt 0 & \\ 4x^3-3x^4, & \text { for } 0 \leq x \leq 1 \\ 1, & \text { for } x > 1 \end{array}\right.\)

 
c.    \(\text{Find}\ F\Big{(}\dfrac{2}{3}\Big{)}: \)

\(F\Big{(}\dfrac{2}{3}\Big{)} \) \(=4 \times \Big{(}\dfrac{2}{3}\Big{)}^3-3 \times \Big{(}\dfrac{2}{3}\Big{)}^4 \)  
  \(=\dfrac{16}{27}>0.5 \)  

 
\(\therefore\ \text{Mode > median}\)

♦♦♦ Mean mark (c) 17%.

Statistics, 2ADV S3 2022 HSC 30

A continuous random variable \(X\) has cumulative distribution function given by

\(F(x)= \begin{cases}
1 & x>e^3 \\
\ \\
\dfrac{1}{k}\, \ln x & 1 \leq x \leq e^3 . \\
\ \\
0 & x<1\end{cases}\)

  1. Show that  \(k = 3\).  (1 mark)

  2. Given that  \(P(X < c)=2P(X > c)\), find the exact value of \(c\).  (2 marks)

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  1. \(\text{Proof (See Worked Solutions)}\)
  2. \(e^2\)
Show Worked Solution

a.  \(\text{Show} \ \ k=3\)

\begin{aligned}
{\left[\dfrac{1}{k} \, \ln x\right]_1^{e^3} } & =1 \\
\dfrac{1}{k}\, \ln \left(e^3\right)-\dfrac{1}{k}\, \ln 1 & =1 \\
\dfrac{1}{k}(3)-\frac{1}{k}(0) & =1 \\
k & =3 \ldots \text{as required}
\end{aligned}


♦♦ Mean mark (a) 38%.
b.    \(P(X<c)\) \(=\left[\dfrac{1}{3}\, \ln x\right]_0^c\)
    \(=\dfrac{1}{3}\, \ln c\)

\begin{aligned}
2 P(X>c) & =2 P(1-P(X<c)) \\
& =2\left(1-\frac{1}{3} \ln c\right)
\end{aligned}

\(\text { Given } P(X<c)=2 P(X>c)\)

\begin{aligned}
\dfrac{1}{3}\, \ln c & =2-\dfrac{2}{3}\, \ln c \\
\ln c & =2 \\
\therefore c & =e^2
\end{aligned}


♦♦♦ Mean mark (b) 19%.

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 2021 HSC 30

The number of hours for which light bulbs will work before failing can be modelled by the random variable `X` with cumulative distribution function

`F(x) = { {:(1 - e^(-0.01x)),(0):} {: (\ \ x >= 0), (\ \ x < 0):} :}`

Jane sells light bulbs and promises that they will work for longer than exactly 99% of all light bulbs.

Find how long, according to Jane’s promise, a light bulb bought from her should work. Give your answer in hours, rounded to two decimal places.  (2 marks)

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`text(460.52 hours.)`

Show Worked Solution

`text(Find)\ x\ text(when)\ \ F(x) > 0.99:`

♦♦ Mean mark 22%.
`1 – e^(-0.01x)` `= 0.99`
`e^(-0.01x)` `= 0.01`
`-0.01x` `= ln(0.01)`
`x` `= (ln(0.01))/(-0.01)`
  `= 460.517 …`
  `= 460.52\ text(hours)`

 
`:.\ text(Light bulbs should work at least 460.52 hours.)`

Statistics, 2ADV S3 EQ-Bank 1

A probability density function can be used to model the lifespan of a termite, `X`, in weeks, is given by
 

`f(x) = {(k(36 - x^2)),(0):}\ \ \ {:(3 <= x <= 6),(text(otherwise)):}`
 

  1. Show that the value of  `k`  is  `1/45`.  (2 marks)

  2. Find the cumulative distribution function.  (2 marks)

  3. Find the probability that a termite's lifespan is greater than 5 weeks.  (1 mark)

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  1. `text(See Worked Solutions)`
  2. `F(x) = {(0),(1/135 (108t – t^3  – 297)),(1):}\ \ \ {:(x < 3),(3 <= x <= 6),(x > 6):}`
  3. `17/135`
Show Worked Solution
i.    `k int_3^6 36 – x^2\ dx` `= 1`
  `k[36x – (x^3)/3]_3^6` `= 1`
  `k[(216 – 72)-(108 – 9)]` `= 1`
  `45k` `= 1`
  `k` `= 1/45`

 

ii.    `F(t)` `= int_(-∞)^t f(x)\ dx`
    `= int_3^t f(x)\ dx`
    `= 1/45 int_3^t 36 – x^2\ dx`
    `= 1/45 [36x – (x^3)/3]_3^t`
    `= 1/135[108x – x^3]_3^t`
    `= 1/35[(108t – t^3) – (324 – 27)]`
    `= 1/135(108t – t^3 – 297)`

 
`:. F(x) = {(0),(1/135 (108t – t^3  – 297)),(1):}\ \ \ {:(x < 3),(3 <= x <= 6),(x > 6):}`

 

iii.    `P(X > 5)` `= 1 – F(5)`
    `= 1 – 1/135(108 xx 5 – 5^3 – 297)`
    `= 1 – 118/135`
    `= 17/135`

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)

Show Answers Only
  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):}`