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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 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)

Show Answers Only
  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 SM-Bank 18

The Lorenz birdwing is the largest butterfly in a habitat.

The probability density function that describes its life span, \(X\), in weeks, is given by
 

\(f(x)= \begin{cases}
\dfrac{4}{625}\left(5 x^3-x^4\right) & 0 \leq x \leq 5 \\
\\
0 & \text {elsewhere }\end{cases}\)
 

In a sample of 80 Lorenz birdwing butterflies, how many butterflies are expected to live longer than two weeks, correct to the nearest integer?  (2 marks)

Show Answers Only

\(73\)

Show Worked Solution

\(\begin{aligned} \operatorname{Pr}(X>2) & =\dfrac{4}{625} \int_2^5 5 x^3-x^4 \, dx \\
& =\dfrac{4}{625}\left[\dfrac{5}{4} x^4-\dfrac{x^5}{5}\right]_2^5 \\
& =\dfrac{4}{625}\left[\left(\dfrac{5^5}{4}-\dfrac{5^5}{5}\right)-\left(\dfrac{5}{4} \times 2^4-\dfrac{2^5}{5}\right)\right] \\
& =\dfrac{4}{625}\left[\dfrac{625}{4}-\dfrac{68}{5}\right] \\
& =0.9129 \ldots\end{aligned}\)

 

\(\begin{aligned} \therefore \text { Expected number } & =80 \times 0.9129 \ldots \\ & \approx 73.03 \\ & \approx 73\end{aligned}\)

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.    \(f(x)\) \( =\dfrac{3}{4}(x-2)(4-x)\)
  \(\displaystyle \int_2^4 f(x) \) \( =\dfrac{3}{4} \int_2^4-x^2+6 x-8 d x\)
    \( =\dfrac{3}{4}\left[\dfrac{-x^3}{3}+3 x^2-8 x\right]_2^4\)
    \( =\dfrac{3}{4}\left[\left(-\dfrac{64}{3}+48-32\right)-\left(-\dfrac{8}{3}+12-16\right)\right] \)
    \( =\dfrac{3}{4}\left(-\dfrac{16}{3}+\dfrac{20}{3}\right) \)
    \(= 1\)

 

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

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

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

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

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 8

The continuous random variable `X` has a distribution with probability density function given by
 

`f(x) = {(ax(5 - x), \ text(if)\ \ 0 <= x <= 5), (0,\ text (if)\ \ x < 0\ \ text(or if)\ \ x > 5):}`
 

where `a` is a positive constant.

  1. Find the value of  `a`.  (3 marks)

  2. Express  `P(X < 3)`  as a  definite integral. (Do not evaluate the definite integral.)  (1 mark)

Show Answers Only
  1. `6/125`
  2. `int_0^3 ax(5 – x)\ dx`
Show Worked Solution
a.   `text(Total Area under curve)` `= 1`
  `a int_0^5 (5x – x^2)\ dx` `= 1`
  `a [5/2 x^2 – 1/3 x^3]_0^5` `= 1`
  `a [(125/2 – 125/3) – (0)]` `= 1`
  `125/6 a` `= 1`
  `:. a` `= 6/125`

 

b.   `P(X < 3) = int_0^3 ax(5 – x)\ dx`

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. \(m^4-16 m^2+24=0\)
Show Answers Only

\(D\)

Show Worked Solution

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

\(\Rightarrow D\)