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Probability, MET2 2022 VCAA 3

Mika is flipping a coin. The unbiased coin has a probability of \(\dfrac{1}{2}\) of landing on heads and \(\dfrac{1}{2}\) of landing on tails.

Let \(X\) be the binomial random variable representing the number of times that the coin lands on heads.

Mika flips the coin five times.

    1. Find \(\text{Pr}(X=5)\).   (1 mark)
    2. Find \(\text{Pr}(X \geq 2).\) (1 mark)
    3. Find \(\text{Pr}(X \geq 2 | X<5)\), correct to three decimal places.   (2 marks)
    4. Find the expected value and the standard deviation for \(X\).   (2 marks)

The height reached by each of Mika's coin flips is given by a continuous random variable, \(H\), with the probability density function
  

\(f(h)=\begin{cases} ah^2+bh+c         &\ \ 1.5\leq h\leq 3 \\ \\ 0       &\ \ \text{elsewhere} \\ \end{cases}\)
  

where \(h\) is the vertical height reached by the coin flip, in metres, between the coin and the floor, and \(a, b\) and \(c\) are real constants.

    1. State the value of the definite integral \(\displaystyle\int_{1.5}^3 f(h)\,dh\).   (1 mark)
    2. Given that  \(\text{Pr}(H \leq 2)=0.35\)  and  \(\text{Pr}(H \geq 2.5)=0.25\), find the values of \(a, b\) and \(c\).   (3 marks)

    3. The ceiling of Mika's room is 3 m above the floor. The minimum distance between the coin and the ceiling is a continuous random variable, \(D\), with probability density function \(g\).
    4. The function \(g\) is a transformation of the function \(f\) given by \(g(d)=f(rd+s)\), where \(d\) is the minimum distance between the coin and the ceiling, and \(r\) and \(s\) are real constants.
    5. Find the values of \(r\) and \(s\).   (1 mark)

  1. Mika's sister Bella also has a coin. On each flip, Bella's coin has a probability of \(p\) of landing on heads and \((1-p)\) of landing on tails, where \(p\) is a constant value between 0 and 1 .
  2. Bella flips her coin 25 times in order to estimate \(p\).
  3. Let \(\hat{P}\) be the random variable representing the proportion of times that Bella's coin lands on heads in her sample.
    1. Is the random variable \(\hat{P}\) discrete or continuous? Justify your answer.   (1 mark)
    2. If \(\hat{p}=0.4\), find an approximate 95% confidence interval for \(p\), correct to three decimal places.   (1 mark)
    3. Bella knows that she can decrease the width of a 95% confidence interval by using a larger sample of coin flips.
    4. If \(\hat{p}=0.4\), how many coin flips would be required to halve the width of the confidence interval found in part c.ii.?   (1 mark)

Show Answers Only

a. i.    `frac{1}{32}`    ii.    `frac{13}{16}`    iii.    `0.806` (3 d.p.)

a. iv    `text{E}(X)=5/2,  text{sd}(X)=\frac{\sqrt{5}}{2}`

b. i.   `1`   

b. ii.    `a=-frac{4}{5},  b=frac{17}{5},  c=-frac{167}{60}`

b. iii. `r=-1,  s=3`

c. i.   `text{Discrete}`   ii.   `(0.208,  0.592)`   iii.   `n=100`

Show Worked Solution

a.i  `X ~ text{Bi}(5 , frac{1}{2})`

`text{Pr}(X = 5) = (frac{1}{2})^5 = frac{1}{32}`
 

a.ii  By CAS:    `text{binomCdf}(5,0.5,2,5)`     `0.8125`

`text{Pr}(X>= 2) = 0.8125 = frac{13}{16}`

  
a.iii  `\text{Pr}(X \geq 2 | X<5)`

`=frac{\text{Pr}(2 <= X < 5)}{\text{Pr}(X < 5)} = frac{\text{Pr}(2 <= X <= 4)}{text{Pr}(X <= 4)}`

  
By CAS: `frac{text{binomCdf}(5,0.5,2,4)}{text{binomCdf}(5,0.5,0,4)}`
 

`= 0.806452 ~~ 0.806` (3 decimal places)
  

a.iv `X ~ text{Bi}(5 , frac{1}{2})`

`text{E}(X) = n xx p= 5 xx 1/2 = 5/2`

`text{sd}(X) =\sqrt{n p(1-p)}=\sqrt((5/2)(1 – 0.5)) = \sqrt{5/4} =\frac{\sqrt{5}}{2}`

  

b.i   `\int_{1.5}^3 f(h) d h = 1`


♦♦ Mean mark (b.i) 40%.
MARKER’S COMMENT: Many students did not evaluate the integral or evaluated incorrectly.

b.ii  By CAS:

`f(h):= a\·\h^2 + b\·\h +c`
 

`text{Solve}( {(\int_{1.5}^2 f(h) d h = 0.35), (\int_{2.5}^3 f(h) d h = 0.25), (\int_{1.5}^3 f(h) d h = 1):})`

`a = -0.8 = frac{-4}{5}, \ b = 3.4 = frac{17}{5},\  c = =-2.78 \dot{3} = frac{-167}{60}`


♦♦ Mean mark (b.ii) 47%.
MARKER’S COMMENT: Many students did not give exact answers.

b.iii  `h + d = 3`

`:.\  f(h) = f(3  –  d) = f(- d + 3)`

`:.\  r = – 1 ` and ` s = 3`


♦♦♦♦ Mean mark (b.iii) 10%.
MARKER’S COMMENT: Many students did not attempt this question.

c.i  `\hat{p}`  is discrete.

The number of coin flips must be zero or a positive integer so  `\hat{p}`  is countable and therefore discrete.
 

c.ii  `\left(\hat{p}-z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p}+z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right)`

`\left(0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\ , 0.4-1.96 \sqrt{\frac{0.4 \times 0.6}{25}}\right)`

`\approx(0.208\ ,0.592)`

 

c.iii To halve the width of the confidence interval, the standard deviation needs to be halved.

`:.\  \frac{1}{2} \sqrt{\frac{0.4 \times 0.6}{25}} = \sqrt{\frac{0.4 \times 0.6}{4 xx 25}} = \sqrt{\frac{0.4 \times 0.6}{100}}`

`:.\  n = 100`

She would need to flip the coin 100 times


♦♦♦ Mean mark (c.iii) 30%.
MARKER’S COMMENT: Common incorrect answers were 0, 10, 11, 50 and 101.

Probability, MET1 2023 VCAA 8

Suppose that the queuing time, \(T\) (in minutes), at a customer service desk has a probability density function given by
 

\(f(t) = \begin {cases}
kt(16-t^2)         &\ \ 0 \leq t \leq 4 \\
\\
0 &\ \ \text{elsewhere}
\end{cases}\)

 
for some  \(K \in R\).

  1. Show that  \(k=\dfrac{1}{64}\).   (1 mark)
  2. Find  \(\text{E}(T)\).   (2 marks)
  3. What is the probability that a person has to queue for more than two minutes, given that they have already queued for one minute?   (3 marks)
Show Answers Only
a.    \(\displaystyle \int_{0}^{4} (16-t^2)\,dt\) \(=1\)
  \(k\left[8t^2-\dfrac{t^4}{4}\right]_0^4\) \(=1\)
  \(k\Bigg(8\times16-\dfrac{16\times16}{4}\Bigg)\) \(=1\)
  \(64k\) \(=1\)
  \(\therefore\ k\) \(=\dfrac{1}{64}\)

b.    \(E(T)=\dfrac{32}{15}\)

c.    \(\dfrac{16}{25}=0.64\)

Show Worked Solution
a.    \(\displaystyle \int_{0}^{4} kt(16-t^2)\,dt\) \(=1\)
  \(k\left[8t^2-\dfrac{t^4}{4}\right]_0^4\) \(=1\)
  \(k\Bigg(8\times16-\dfrac{16\times16}{4}\Bigg)\) \(=1\)
  \(64k\) \(=1\)
  \(\therefore\ k\) \(=\dfrac{1}{64}\)

♦ Mean mark (a) 43%.
b.    \(E(T)\) \(=\dfrac{1}{64}\displaystyle \int_{0}^{4} (16t^2-t^4)\,dt\)
    \(=\dfrac{1}{64}\left[\dfrac{16t^3}{3}-\dfrac{t^5}{5}\right]_0^4\)
    \(=\dfrac{1}{64}\Bigg(\dfrac{1024}{3}-\dfrac{1024}{5}-0\Bigg)\)
    \(=\dfrac{1}{64}\times\dfrac{2048}{15}\)
    \(=\dfrac{32}{15}\)

♦♦ Mean mark (b) 38%.
c.    \(\text{Pr}(2<T<4|T>1)\) \(=\dfrac{\text{Pr}(2<T<4)}{\text{Pr}(T>1)}\)
    \(=\dfrac{\dfrac{1}{64}\displaystyle \int_{2}^{4} (16t-t^3)\,dt}{\dfrac{1}{64}\displaystyle \int_{1}^{4} (16t-t^3)\,dt}\)
    \(=\dfrac{\left[8t^2-\dfrac{t^4}{4}\right]_2^4}{\left[8t^2-\dfrac{t^4}{4}\right]_1^4}\)
    \(=\dfrac{(64-(32-4))}{\bigg(64-\bigg(8-\dfrac{1}{4}\bigg)\bigg)}\)
    \(=\dfrac{36}{\bigg(\dfrac{225}{4}\bigg)}=\dfrac{144}{225}=\dfrac{16}{25}=0.64\)

♦♦♦ Mean mark (c) 24%.
MARKER’S COMMENT: Simplifying fractions caused problems. Cancelling factors will assist with calculations.

Probability, MET1 2021 VCAA 7

A random variable  `X`  has the probability density function  `f`  given by

`f(x) = {{:(k/(x^2)),(0):}\ \ \ \ {:(1 <= x <= 2),(text{elsewhere}):}:}`

where  `k`  is a positive real number.

  1. Show that  `k = 2`.  (1 mark)

  2. Find  `E(X)`.  (2 marks)

Show Answers Only

  1. `text(See Worked Solution)`
  2. `2log_e2`

Show Worked Solution

♦ Mean mark part (a) 48%.

a.    `int_1^2 k/x^2\ dx` `=1`
  `k[- 1/x]_1^2` `=1`
  `k(-1/2+1)` `=1`
  `k/2` `=1`
  `:.k` `=2\ \ text{… as required}`

 

♦ Mean mark part (b) 44%.
MARKER’S COMMENT: Common error not recognising  `log_e1=0`

b.    `E(X)` `=int_1^2 x * 2/x^2\ dx`
    `=int_1^2 2/x\ dx`
    `=[2log_e x]_1^2`
    `=2log_e2-2log_e1`
    `=2log_e2`

Probability, MET2-NHT 2019 VCAA 10 MC

Let  `f`  be the probability density function  `f : [ 0, (2)/(3)] → R, \ f(x) = kx(2x + 1)(3x -2)(3x + 2)`.
The value of  `k`  is

  1.     `(308)/(405)`
  2.  `–(308)/(405)`
  3.  `–(405)/(308)`
  4.     `(405)/(308)`
  5.     `(960)/(133)`
Show Answers Only

`C`

Show Worked Solution

`text(Solve for)\ k :`

`k int_0^((2)/(3)) x (2x + 1)(3x – 2)(3x + 2)\ dx = 1`

`k = -(405)/(308)`
 

`=> \ C`

Probability, MET2 2019 VCAA 18 MC

The distribution of a continuous random variable, `X`, is defined by the probability density function  `f`, where
 

`f(x) = {(p(x)), (0):} qquad {:(-a <= x <= b), (text(otherwise)):}`
 

and  `a, b in RR^+`

The graph of the function  `p`  is shown below.
 


 

It is known that the average value of  `p`  over the interval  `[-a, b]`  is  `3/4`.

`text(Pr)(X > 0)`  is

A.   `2/3`

B.   `3/4`

C.   `4/5`

D.   `7/9`

E.   `5/6`

Show Answers Only

`D`

Show Worked Solution
`text(Area)` `= 1/2 (a xx 2a) + 1/2 b(2a + b)`
`1` `= a^2 + ab + (b^2)/2\ …\ (1)`

 

`text(Average value)` `= 1/(b + a) (a^2 + ab + (b^2)/2)`
`3/4` `= 1/(b + a)\ …\ (2)`

 
`text(Solve for) \ \ a\ \ text{(by CAS):}`

`a = – sqrt 2/3`

`text(Pr)(X > 0)` `= 1 – text(Pr)(X < 0)`
  `= 1 – a^2`
  `= 7/9`

 
`=>   D`

Probability, MET2 2017 VCAA 19 MC

A probability density function  `f` is given by

`f(x) = {{:(cos(x) + 1),(0):}qquad{:(k < x < (k + 1)),(text(elsewhere)):}:}`

where `0 < k < 2`.

The value of `k` is

  1. `1`
  2. `(3pi - 1)/2`
  3. `pi - 1`
  4. `(pi - 1)/2`
  5. `pi/2`
Show Answers Only

`D`

Show Worked Solution
`int_k^(k+1)(cos(x) + 1)\ dx` `=[sinx +x]_k^(k+1)`
`sin(k+1)+(k+1) -sink – k` `=1`
`sin(k+1)-sink` `=0`
`sin(k+1)-sin(pi-k)` `=0`
   
`k+1` `=pi-k`
`2k` `=pi-1`
`:.k` `=(pi-1)/2`

`=> D`

Probability, MET1 2008 VCAA 4

The function
 

`f(x) = {{:(k),(0):}{:(sin(pix)qquadtext(if)qquadx ∈ [0,1]),(qquadqquadqquadqquadquadtext(otherwise)):}`
 

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

  1. Show that  `k = pi/2`.  (2 marks)

  2. Find  `text(Pr)(X <= 1/4 | X <= 1/2)`.  (3 marks)

Show Answers Only

  1. `text(See Worked Solutions)`
  2. `1 – (sqrt2)/2`

Show Worked Solution

a.   `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`

 

b.   `text(Conditional Probability:)`

♦ Mean mark 47%.
MARKER’S COMMENT: Few students used the symmetry of the probability density function to calculate the denominator.

`text(Pr)(X <= 1/4 | X <= 1/2)`

`= (text(Pr)(X <= 1/4))/(text(Pr)(X <= 1/2))`
`= (pi/2 int_0^(1/4) sin(pix)dx)/(1/2)`
`= -pi/pi [cos(pix)]_0^(1/4)`
`= -1 [cos(pi/4) – cos(0)]`
`= -1 [1/(sqrt2) – 1]`
`= 1 – (sqrt2)/2`

Probability, MET1 2010 VCAA 7

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  `text(Pr) (X < 3)`  as a  definite integral. (Do not evaluate the definite integral.)  (1 mark)

Show Answers Only

  1. `6/125`
  2. `int_0^3 f(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.   `text(Pr) (X < 3) = int_0^3 f(x)\ dx`

Probability, MET2 2015 VCAA 13 MC

The function `f` is a probability density function with rule
 

`f(x) = {(ae^x, 0 <= x <= 1), (ae, 1 < x <= 2), (\ 0, text(otherwise)):}`
 

The value of `a` is

  1. `1`
  2. `e`
  3. `1/e`
  4. `1/(2e)`
  5. `1/(2e - 1)`
Show Answers Only

`E`

Show Worked Solution

`text(Total area) = 1`

`int_0^1 ae^x\ dx + int_1^2 ae\ dx` `= 1`
`[ae^x]_0^1 + [ae*x]_1^2` `=1`
`[ae-a] + [2ae-ae]` `=1`
`a(2e-1)` `=1`
`:. a` `= 1/(2e – 1)`

`=>   E`

Probability, MET2 2015 VCAA 9 MC

The graph of the probability density function of a continuous random variable, `X`, is shown below.

VCAA 2015 9mc

If  `a > 2`, then `text(E)(X)` is equal to

  1. `8`
  2. `5`
  3. `4`
  4. `3`
  5. `2`
Show Answers Only

`B`

Show Worked Solution
`text(Area under rectangle)` `= 1`
`1/6(a – 2)` `= 1`
`:. a` `= 8`
♦ Mean mark 37%.
MARKER’S COMMENT: Once `a=8` is found, the expected value can also be quickly found halfway between 2 and 8, due to the uniform distribution.

 

`text(E)(X)` `= int_2^8 x (1/6)\ dx`
  `=[x^2/12]_2^8`
  `=64/12 – 4/12`
  `=5`

 
`=>   B`