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Statistics, 2ADV S3 EQ-Bank 24

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

\begin{align*}
f(x)=\left\{\begin{array}{cl}
k x(1-x)^5, & \text { for } 0 \leq x \leq 1 \\
0, & \text { for all other values of } x
\end{array}\ \ \ , \text { where } k\right. \text { is a constant. }
\end{align*}

It is given that

\(\displaystyle \int_0^a x(1-x)^5\, d x=\frac{1}{42}+\frac{(1-a)^7}{7}-\frac{(1-a)^6}{6}\)

and \(\displaystyle\int_0^1 x^m(1-x)^5\, d x=\dfrac{120}{(m+1)(m+2)(m+3)(m+4)(m+5)(m+6)}\)

where  \(a>0\)  and  \(m>0\).

  1. Show that  \(k=42\).   (1 mark)

  2. Show that  \(E (X)=0.25\).   (2 marks)

  3. Show that the median of \(X\) is less than the expected value of \(X\).   (3 marks)

Show Answers Only

a.    \(k \displaystyle \int_0^1 x(1-x)^5 d x=1\)

\(k\left[\dfrac{1}{42}+\dfrac{(1-1)^7}{7}-\dfrac{(1-1)^6}{6}\right]=1\)

\(\dfrac{k}{42}\) \(=1\)  
\(k\) \(=42\)  

 

b.     \(E (X)\) \(=\displaystyle \int_0^1 x \times f(x)\, d x\)
    \(=\displaystyle \int_0^1 42 x^2(1-x)^5\, d x\)
    \(=42 \times \dfrac{120}{3 \times 4 \times 5 \times 6 \times 7 \times 8}\)
    \(=0.25\)

 

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

\(P(X\leqslant m) = 0.5\ \ \Rightarrow\ \ \displaystyle \int_0^m 42 x(1-x)^5\, d x=0.5 \)

\(E(X)=0.25\)

\(\text{Calculate }\ P(X\leqslant 0.25):\)

\(\displaystyle \int_0^{0.25} 42 x(1-x)^5\, d x\) \(=42\left[\frac{1}{42}+\dfrac{(1-0.25)^7}{7}-\dfrac{(1-0.25)^6}{6}\right]\)  
  \(=0.555 \ldots\ \text{(3 dp)}\)  

 
\(\therefore \displaystyle \int_0^m 42 x(1-x)^5 d x=0.5 \ \ \text{requires the median to be less than 0.25.}\)

Show Worked Solution

a.    \(k \displaystyle \int_0^1 x(1-x)^5 d x=1\)

\(k\left[\dfrac{1}{42}+\dfrac{(1-1)^7}{7}-\dfrac{(1-1)^6}{6}\right]=1\)

\(\dfrac{k}{42}\) \(=1\)  
\(k\) \(=42\)  

 

b.     \(E (X)\) \(=\displaystyle \int_0^1 x \times f(x)\, d x\)
    \(=\displaystyle \int_0^1 42 x^2(1-x)^5\, d x\)
    \(=42 \times \dfrac{120}{3 \times 4 \times 5 \times 6 \times 7 \times 8}\)
    \(=0.25\)

 

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

\(P(X\leqslant m) = 0.5\ \ \Rightarrow\ \ \displaystyle \int_0^m 42 x(1-x)^5\, d x=0.5 \)

\(E(X)=0.25\)

\(\text{Calculate }\ P(X\leqslant 0.25):\)

\(\displaystyle \int_0^{0.25} 42 x(1-x)^5\, d x\) \(=42\left[\frac{1}{42}+\dfrac{(1-0.25)^7}{7}-\dfrac{(1-0.25)^6}{6}\right]\)  
  \(=0.555 \ldots\ \text{(3 dp)}\)  

 
\(\therefore \displaystyle \int_0^m 42 x(1-x)^5 d x=0.5 \ \ \text{requires the median to be less than 0.25.}\)

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)

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 EQ-Bank 16

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)

Show Answers Only

`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 EQ-Bank 17

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)

Show Answers Only

\(-5 \log _e\left(\dfrac{1}{2}\right)\) 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 EQ-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. \(m^4-16 m^2+24=0\)
Show Answers Only

\(D\)

Show Worked Solution

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

\(\Rightarrow D\)