SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

Statistics, 2ADV S3 2025 HSC 21

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

\(f(x)= \begin{cases}\ 0 & \quad x<1 \\ \dfrac{1}{x} & \quad 1 \leq x \leq e \\ \ 0 & \quad x>e\end{cases}\)

  1. Find the mode of the given probability density function. Justify your answer.   (2 marks)

  2. Calculate the value of the 25th percentile \(\left(Q_1\right)\) of this distribution. Give your answer correct to 3 decimal places.   (3 marks)

Show Answers Only

a.    
     

\(\text{Graph is monotonically decreasing.}\)

\(\text{Mode:} \ \ x=1\)

b.   \(Q_1=1.284 \)

Show Worked Solution

a.    
         

\(\text{Graph is monotonically decreasing.}\)

\(f(x)_{\text{max}}\ \text{occurs on interval when}\ \ x=1.\)

\(\text{Mode:} \ \ x=1\)
 

b.    \(\text{Find \(k\) such that} \ P(X<k)=0.25:\)

\(\displaystyle \int_1^{Q_1} \frac{1}{x}\) \(=0.25\)
\(\ln Q_1-\ln 1\) \(=0.25\)
\(Q_1\) \(=e^{0.25}\)
  \(=1.284 \ \text{(3 d.p.)}\)

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)

Show Answers Only

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 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)

Show Answers Only
  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 S2 SM-Bank 14

A probability density function  `f(x)`  is given by
 

`f(x) = {(px(3 - x), \ text(if)\ \ 0 <= x <= 3),(0, \ text(if)\ \ x < 0\ \ text(or if)\ \ x > 3):}`
 

where  `p`  is a positive constant.

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

  2. Find the mode of  `f(x)`.  (2 marks)

Show Answers Only
  1. `2/9`
  2. `3/2`
Show Worked Solution

i.   `text(Total Area under curve = 1)`

`p int_0^3 3x-x^2\ dx` `= 1`
`p[3/2 x^2-(x^3)/3]_0^3` `= 1`
`p[27/2-27/3-0]` `= 1`
`(9p)/2` `= 1`
`p` `= 2/9`

 

ii.   `f(x) = 2/9(3x-x^2)`

`f^{′}(x) = 2/9(3-2x)`

`text(S.P. when)\ \ f^{′}(x) = 0:`

`3-2x` `= 0`
`x` `= 3/2`

 
`f^{″}(x) = -4/9 < 0 \ =>\ text(MAX)`

`:.\ text(Mode) = 3/2`

Statistics, 2ADV S2 SM-Bank 15

A function  \(f(x)\)  is given by

 \(f(x)= \begin{cases}
\dfrac{3}{4}(x-2)(4-x) & \text {if } 2 \leq x \leq 4 \\
\ \\
0 & \text{if } x<2 \text { or if } x>4
\end{cases}\)

  1. Show this curve is a probability density function.  (2 marks)

  2. Find the mode.  (2 marks)

Show Answers Only
  1. \(\text{See Worked Solutions}\)
  2. \(3\)
Show Worked Solution
i.    \(f(x)\) \( =\dfrac{3}{4}(x-2)(4-x)\)
  \(\displaystyle \int_2^4 f(x) \) \( =\dfrac{3}{4} \displaystyle\int_2^4-x^2+6 x-8 d x\)
    \( =\dfrac{3}{4}\left[\dfrac{-x^3}{3}+3 x^2-8 x\right]_2^4\)
    \( =\dfrac{3}{4}\left[\left(-\dfrac{64}{3}+48-32\right)-\left(-\dfrac{8}{3}+12-16\right)\right] \)
    \( =\dfrac{3}{4}\left(-\dfrac{16}{3}+\dfrac{20}{3}\right) \)
    \(= 1\)

 

\(f(2)=0, f(4)=0\)

\(f(x)>0 \text { for } 2<x<4\)

\(f(2) \geq 0 \text { for } 2 \leq x \leq 4\)

 
\(\therefore f(x) \ \text{is a probability density function.}\)
 

ii.    \(f(x\) \(=\dfrac{3}{4}\left(-x^2+6 x-8\right)\)
  \(f^{\prime}(x)\) \(=\dfrac{3}{4}(-2 x+6)\)
  \(f^{\prime\prime}(x)\)  \(=-\dfrac{3}{2}\)

 
\(\text{SP when}\ f'(x) = 0\)

\(\begin{array}{r}-2 x+6=0 \\
x=3\end{array}\)

\(f^{\prime \prime}(x)<0 \Rightarrow \text {MAX}\)

\(\therefore \text {Mode}=3\)

Statistics, 2ADV S3 SM-Bank 13

The time Jennifer spends on her homework each day varies, but she does some homework every day.

The continuous random variable \(T\), which models the time, \(t\), in minutes, that Jennifer spends each day on her homework, has a probability density function \(f\), where

\(f(t)= \begin{cases}
\dfrac{1}{625}(t-20) & 20 \leq t<45 \\
\ \\
\dfrac{1}{625}(70-t) & 45 \leq t \leq 70 \\
\ \\
0 & \text {elsewhere }
\end{cases}\)

  1. Sketch the graph of  \(f(t)\) on the axes provided below.  (3 marks)

     

        
     

  2. Find the mode.  (1 mark)

  3. Find  \(P(25 \leq T \leq 55)\).  (2 marks)

Show Answers Only

  1.  

  2. \(45\)
  3. \(\dfrac{4}{5}\)
Show Worked Solution
i.   

MARKER’S COMMENT: Many did not draw graph along \(t\)-axis between 0 and 20 and for  \(t>70\).

 
ii.     \(\text{Mode \(=45\) (value of \(t\) at highest value of \(f(t))\)}\)
 

iii.    \(P(25 \leq T \leq 55)\)

\(\begin{aligned}
& =\int_{25}^{45} \dfrac{1}{625}(t-20) d t+\int_{45}^{55} \dfrac{1}{625}(70-t) d t \\
& =\dfrac{1}{625}\left[\dfrac{t^2}{2}-20 t\right]_{25}^{45}+\dfrac{1}{625}\left[70 t-\dfrac{t^2}{2}\right]_{45}^{55} \\
& =\dfrac{1}{625}[112.5-(-187.5)]+\dfrac{1}{625}(2337.5-2137.5) \\ & =\dfrac{300}{625}+\dfrac{200}{625} \\
& =\dfrac{4}{5}
\end{aligned}\)