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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)

Show Answers Only

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 SM-Bank 20

A continuous random variable `X` has a probability density function given by
 

`f(x) = {{:(Cx + D),(0):}\ \ \ \ {:(2 <= x <= 5),(text(elsewhere)):}:}`
 

where `C` and `D` are constants.

Find the exact values of `C` and `D`, given the median of  `X`  is 4.  (4 marks)

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`C = 1/6`

`D = −1/4`

Show Worked Solution

`int_2^5 Cx + D\ dx = 1`

`[C/2 x^2 + Dx]_2^5 = 1`

`[(25/2 C + 5D) – (2C + 2D)]` `= 1`
`21/2C + 3D` `= 1`
`21C + 6D` `= 2\ \ …\ (1)`

 
`text(Using median)\ \ X = 4:`

`[C/2 x^2 + Dx]_2^4 = 0.5`

`[(8C + 4D) – (2C + 2D)]` `= 0.5`
`6C + 2D` `= 0.5\ \ …\ (2)`

  
`text(Multiply:)\ (2) xx 3`

`18C + 6D = 1.5\ \ …\ (3)`
 

`text(Subtract:)\ \ (1) – (3)`

`3C` `= 1/2`
`:. C` `= 1/6`

 
`text{Substitute into (1):}`

`21/6 + 6D` `= 2`
`6D` `= −9/6`
`:. D` `= −1/4`

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 7 MC

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

\(f(x)= \begin{cases}
\dfrac{1}{12}\left(8 x-x^3\right) & 0 \leq x \leq 2 \\
\ \\
0 & \text{elsewhere }
\end{cases}\)

The median  \(m\)  of this function satisfies the equation

  1. \(-m^4+16 m^2-6=0\)
  2. \(m^4-16 m^2=0\)
  3. \(m^4-16 m^2+24=0.5\)
  4. \(m^4-16 m^2+24=0\)
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\(D\)

Show Worked Solution

\(\begin{aligned} \int_0^m \dfrac{1}{12}\left(8 x-x^3\right) d x & =0.5 \\
{\left[\dfrac{1}{12}\left(4 x^2-\dfrac{x^4}{4}\right)\right]_0^m } & =0.5 \\
4 m^2-\dfrac{m^4}{4} & =6 \\
16 m^2-m^4 & =24 \\ m^4-16 m^2+24 & =0
\end{aligned}\)

\(\Rightarrow D\)