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Statistics, 2ADV S3 2020 HSC 23

A continuous random variable, `X`, has the following probability density function.
 

`f(x) = {(sin x, text(for)\ \ 0 <= x <= k),(0, text(for all other values of)\ x):}`
 

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

  2. Find  `P(X <= 1)`. Give your answer correct to four decimal places.  (2 marks)

Show Answers Only
  1. `pi/2`
  2. `0.4597\ \ (text(to 2 d.p.))`
Show Worked Solution
a.    `int_0^k sin x` `= 1`
  `[−cos x]_0^k` `= 1`
  `−cos k + cos 0` `= 1`
  `−cos k` `= 0`
  `cos k` `= 0`
  `k` `= pi/2`

 

♦ Mean mark part (b) 44%.
b.    `P(X <= 1)` `= int_0^1 sin x\ dx`
    `= [−cos x]_0^1`
    `= −cos1 + cos0`
    `= 1 – cos1`
    `= 0.45969…`
    `= 0.4597\ \ (text(to 4 d.p.))`

Statistics, 2ADV S3 SM-Bank 12

The function
 

`f(x) = {{:(k),(0):}{:(sin(pix)qquad\ \ 0<=x<=1),(qquadqquadqquadqquadquadtext(otherwise)):}`
 

is a probability density function for the continuous random variable `X`.

Show that  `k = pi/2`.  (2 marks)

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`text(See Worked Solutions)`

Show Worked Solution

`text(Total Area under curve) = 1\ text(u²)`

`int_0^1 k sin(pix)\ dx` `= 1`
`- k/pi [cos(pix)]_0^1` `= 1`
`- k/pi[cos(pi) – cos(0)]` `= 1`
`- k/pi[-1 – 1]` `= 1`
`2k` `= pi`
`:.k` `= pi/2`

Statistics, 2ADV S3 SM-Bank 6 MC

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

\(f(x)= \begin{cases}
\dfrac{1}{4} \cos \left(\dfrac{x}{2}\right) & 3 \pi \leq x \leq 5 \pi \\
\ & \ \\
0 & \text {elsewhere}
\end{cases}\)

The value of \(a\) such that  \(P(X<a)=\dfrac{\sqrt{3}+2}{4}\)  is

  1. \(\dfrac{19 \pi}{6}\)
  2. \(\dfrac{14 \pi}{3}\)
  3. \(\dfrac{10 \pi}{3}\)
  4. \(\dfrac{29 \pi}{6}\)
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\(B\)

Show Worked Solution

\(\begin{aligned}
\int_{3 \pi}^a \dfrac{1}{4}\, \cos \left(\dfrac{x}{2}\right) d x & =\dfrac{\sqrt{3}+2}{4} \\
{\left[\dfrac{1}{2}\, \sin \left(\dfrac{x}{2}\right)\right]_{3 \pi}^a } & =\dfrac{\sqrt{3}+2}{4} \\
\dfrac{1}{2}\left[\sin \left(\dfrac{a}{2}\right)-\sin \left(\dfrac{3 \pi}{2}\right)\right] & =\dfrac{\sqrt{3}+2}{4} \\
\dfrac{1}{2}\, \sin \left(\dfrac{a}{2}\right)+\dfrac{1}{2} & =\dfrac{\sqrt{3}+2}{4} \\
\sin \left(\dfrac{a}{2}\right) & =\dfrac{\sqrt{3}}{2} \\
\dfrac{a}{2} & =\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{7 \pi}{3}, \ldots \\
\therefore a & =\dfrac{14 \pi}{3} \quad(3 \pi \leq a \leq 5 \pi)
\end{aligned}\)

\(\Rightarrow B\)

Statistics, 2ADV S3 SM-Bank 4

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

\(f(x)= \begin{cases}
\cos(2x)& \text {if}\quad \dfrac{3 \pi}{4}<x<\dfrac{5 \pi}{4} \\
\ \\
0 & \text{elsewhere}
\end{cases}\)
 

Find the value of  \(a\) such that  \(P(X < a) = 0.25\). Give your answer correct to 2 decimal places.   (3 marks)

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\(2.8\)

Show Worked Solution

\(\displaystyle \int_{\frac{3 x}{4}}^a \cos (2 x) d x=0.25\)

\begin{aligned}
\frac{1}{2}[\sin (2 x)]_{\frac{3 \pi}{4}}^a & =0.25 \\
{\left[\sin (2 a)-\sin \left(\frac{3 \pi}{2}\right)\right] } & =0.5 \\
\sin (2 a)+1 & =0.5 \\
2 a & =\sin ^{-1}(-0.5) \\
& =-0.5235
\end{aligned}

\(\text {Since sin is negative in 3rd/4th quadrants: }\)

\begin{aligned}
2 a & =\pi+0.5234 \\
a &=1.832 \ldots \quad \text { (not in range) } \\
&\text { or } \\
2a & =2 \pi-0.5234 \\
 a &=2.87989 \ldots \\
&=2.88 \quad\left(\text{in range: } \frac{3 \pi}{4}<x<\frac{5 \pi}{4}\right)
\end{aligned}

Statistics, 2ADV S3 SM-Bank 3

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

`f(x) = {(pi sin (2 pi x), text(if)\ \ 0 <= x <= 1/2), (0, text(elsewhere)):}`
 

Find the value of  `a`  such that  `P(X > a) = 0.2`. Give your answer to 2 decimal places.   (3 marks)

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`0.35`

Show Worked Solution

`int_a^(1/2) pi sin (2 pi x)\ dx = 0.2`

`-1/2 [cos(2 pi x)]_a^(1/2)` `=0.2`  
`-1/2[cos(pi) -cos(2 pia)]`  `=0.2`  
`-1/2(-1-cos(2pia))` `=0.2`  
`-1-cos(2pia)` `=-0.4`  
`cos(2pia)` `=-0.6`  
`2pia` `=cos^(-1)(-0.6)`  
`:.a` `=cos^(-1)(-0.6)/(2pi)`  
  `=0.3524…`  
  `=0.35\ \ \ text{(to 2 d.p.)}`