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Statistics, MET2 2025 VCAA 3

The time taken for a driver to travel to work each day, in minutes, is modelled by a continuous random variable \(T\) with probability density function

\(f(t)=\left\{\begin{array}{cl}
\dfrac{1}{1\,215\,000}(t-29)(59-t)^3 & 29 \leq t \leq 59 \\
0 & \text {otherwise}
\end{array}\right.\)

    1. Find the mean time taken, in minutes, for the driver to travel to work each day.   (1 mark)

    2. Find the standard deviation of the time taken, in minutes, for the driver to travel to work each day.   (2 marks)

  2. The driver allows \(k\) minutes to travel to work each day. If the journey takes longer than \(k\) minutes, the driver will be late. Whether the driver is late on a particular day is independent of whether they are late on any other day.
    1. If \(k=47\), write a definite integral to show that the probability of the driver being late is 0.08704    (1 mark)
    2. If \(k=47\), find the probability that the driver will be late on at least one day in a five-day working week.
    3. Give your answer correct to four decimal places.   (2 marks)

    4. For \(k=47\), let \(\hat{P}\) be the proportion of days the driver is late in any five-day working week. Find \(\operatorname{Pr}(0.4 \leq \hat{P} \leq 0.6)\) correct to four decimal places.   (2 marks)

    5. Find the integer \(k\) such that the probability, correct to one decimal place, of the driver being late at least once in any five-day working week is 0.2    (2 marks)

  3. At a given traffic light, the wait time is modelled by a normal distribution with a mean of 2.5 minutes and a standard deviation of \(\sigma\) minutes.
    1. If \(\sigma=0.6\), find the probability that the wait time will be less than 3.5 minutes.
    2. Give your answer correct to two decimal places.   (1 mark)

    3. Find the value of \(\sigma\) such that there is a 2% chance of a wait time longer than 3.5 minutes.
    4. Give your answer correct to two decimal places.   (1 mark)

  4. The driver passes through three traffic lights \((A, B\) and \(C)\) on their journey to work. The probability of each traffic light being red is shown in the table below.
  5. \begin{array}{|l|c|c|c|}
    \hline \rule{0pt}{2.5ex}\text {Traffic light} \rule[-1ex]{0pt}{0pt}& \quad A \quad & \quad B \quad & \quad C  \quad\\
    \hline \rule{0pt}{2.5ex} \text {Probability that the traffic light is red} \quad  \rule[-1ex]{0pt}{0pt}& 0.2 & 0.3 & 0.1 \\
    \hline
    \end{array}
  6. Let \(Y\) be the random variable representing the number of traffic lights that are red on the driver's journey to work. Assume that each traffic light being red is independent of any other traffic light being red.
  7. Complete the following table for the probability distribution of \(Y\).    (2 marks)

  8. \begin{array}{|c|c|c|c|c|}
    \hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& \ \ \quad 0 \ \ \quad & \ \ \quad 1 \ \ \quad & \ \ \quad 2 \ \ \quad & \ \ \quad 3 \ \ \quad \\
    \hline \rule{0pt}{2.5ex}\operatorname{Pr}(Y=y) \rule[-1ex]{0pt}{0pt}& & & & \\
    \hline
    \end{array}
Show Answers Only

a.i.   \(\displaystyle \int_{29}^{59}(t \times f(t)) d t=39\)
 

a.ii. \(\text{Strategy 1}\)

\(\text{sd}(T)=\sqrt{\displaystyle\int_{29}^{59} t^2 f(t) d t-39^2}=\dfrac{10 \sqrt{14}}{7}\)

\(\text{Strategy 2}\)

\(\text{sd}(T)=\sqrt{\displaystyle\int_{29}^{59}(t-39)^2 f(t)\, d t}=\dfrac{10 \sqrt{14}}{7}\)
 

b.i.  \(P\text{(driver being late)}=\displaystyle\int_{47}^{59} f(t)\, d t=0.08704\)
 

b.ii.  \(P(\text{(driver late at least 1 day in week)}=0.3658 \ \ \text{(4 d.p.)}\)
 

b.iii. \(0.0631 \ \text{(4 d.p.)}\)
 

b.iv. \(k=49\)
 

c.i.  \(P(W<3.5)=0.95\)
 

c.ii   \(\sigma=0.49\)
 

d.

\begin{array}{|c|c|c|c|c|}
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad& \quad 2 \quad & \quad 3 \quad \\
\hline \rule{0pt}{2.5ex}\operatorname{Pr}(Y=y) \rule[-1ex]{0pt}{0pt}& \frac{63}{125}=0.504& \frac{199}{500}=0.398 & \frac{23}{250}=0.092 & \frac{3}{500}=0.006 \\
\hline
\end{array}

Show Worked Solution

a.i.   \(\text{Calculate (by CAS):}\)

\(\displaystyle \int_{29}^{59}(t \times f(t)) d t=39\)
 

a.ii. \(\text{Strategy 1}\)

\(\text{sd}(T)=\sqrt{\displaystyle\int_{29}^{59} t^2 f(t) d t-39^2}=\dfrac{10 \sqrt{14}}{7}\)

\(\text{Strategy 2}\)

\(\text{sd}(T)=\sqrt{\displaystyle\int_{29}^{59}(t-39)^2 f(t)\, d t}=\dfrac{10 \sqrt{14}}{7}\)
 

b.i.  \(P\text{(driver being late)}\)

\(=\displaystyle\int_{47}^{59} f(t)\, d t=0.08704\)
 

b.ii.  \(P(\text{(driver late at least 1 day in week)}\)

\(=1-P(\text{never late in 5 days})\)

\(=1-(0.08704)^5\)

\(=0.3658 \ \ \text{(4 d.p.)}\)
 

b.iii. \(\text{Let} \ \ Y \sim \text{Bi}(5,0.08704)\)

\(\operatorname{Pr}(0.4 \leqslant \hat{P} \leqslant 0.6)=\operatorname{Pr}(2 \leqslant Y \leqslant 3)=0.0631 \ \text{(4 d.p.)}\)
 

b.iv. \(\text{Solve for} \ k:\)

\(1-\left(1-\displaystyle \int_k^{59} f(t) d t\right)^5=0.2\)

\(k=49\)
 

c.i.  \(W \sim N\left(\mu, \sigma^2\right) \sim\left(2.5,0.6^2\right)\)

\(\text{Solve (by CAS):}\)

\(P(W<3.5)=0.95\)
 

c.ii   \(\text{Find \(z\)-score when \(P(W>3.5)=0.02\)}\)

\(z \text {-score }=2.0537 \ldots\)

\(\text{Solve for} \ \sigma :\)

\(\dfrac{3.5-2.5}{\sigma}=2.0537 \ldots \ \Rightarrow \ \sigma=0.49\ \text{(2 d.p.)}\)
 

d.

\begin{array}{|c|c|c|c|c|}
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& \quad 0 \quad & \quad 1 \quad& \quad 2 \quad & \quad 3 \quad \\
\hline \rule{0pt}{2.5ex}\operatorname{Pr}(Y=y) \rule[-1ex]{0pt}{0pt}& \frac{63}{125}=0.504& \frac{199}{500}=0.398 & \frac{23}{250}=0.092 & \frac{3}{500}=0.006 \\
\hline
\end{array}

Probability, MET2 2022 VCAA 14 MC

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

\(f(x)=\begin{cases}\dfrac{2}{9}xe^{-\frac{1}{9}x^2}         &\ \ x\geq 0 \\ \\ 0       &\ \ x<0 \\ \end{cases}\)
 

The expected value of \(X\), correct to three decimal places, is

  1. 1.000
  2. 2.659
  3. 3.730
  4. 6.341
  5. 9.000
Show Answers Only

\(B\)

Show Worked Solution
\(\text{E}(X) = \displaystyle\int_0^{\infty} \dfrac{2}{9} x^2 e^{-\frac{1}{9} x^2}\,dx\) \(\approx 2.65868\dots\)    [by CAS]  
   \(\approx 2.659\)  

  
\(\Rightarrow B\)

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`

Statistics, MET1-NHT 2018 VCAA 6

The discrete random variable `X` has the probability mass function
 

`text(Pr)(X = x) = {(kx), (k), (0):} qquad {:(x∈{1, 4, 6}), (x = 3), (text(otherwise)):}`
 

  1. Show that  `k = (1)/(12)`.  (2 marks)

  2. Find  `text(E)(X)`.  (1 mark)

  3. Evaluate  `text(Pr)(X ≥ 3 | X ≥ 2)`.  (1 mark)

Show Answers Only

  1. `text(Proof (See Worked Solution))`
  2. `(14)/(3)`
  3. `1`

Show Worked Solution

a.     `qquad X` `qquad 1 qquad` `qquad 3 qquad` `qquad 4 qquad` `qquad 6 qquad`
  `qquad Pr(X = x) qquad` `k` `k` `4k` `6k`

 

`k + k+4k + 6k` `= 1`
`12k` `= 1`
`k` `= (1)/(12)`

 

b.    `E(X)` `= ∑ x text(Pr)(X = x)`
  `= k + 3k + 16k + 36k`
  `= 56k`
  `= (56)/(12)`
  `= (14)/(3)`

 

c.    `text(Pr) (X ≥ 3 | X ≥ 2)` `= (text(Pr) (X ≥ 3 ∩ X ≥ 2))/(text(Pr) (X ≥ 2))`
  `= (text(Pr) (X ≥ 3))/(text(Pr) (X ≥ 2))`
  `= (k + 4k + 6k)/(k + 4k + 6k)`
  `= 1`

Statistics, MET2-NHT 2019 VCAA 3

Concerts at the Mathsland Concert Hall begin `L` minutes after the scheduled starting time. `L` is a random variable that is normally distributed with a mean of 10 minutes and a standard deviation of four minutes.

  1. What proportion of concerts begin before the scheduled starting time, correct to four decimal places?   (1 mark)

  2. Find the probability that a concert begins more than 15 minutes after the scheduled starting time, correct to four decimal places.   (1 mark)

If a concert begins more than 15 minutes after the scheduled starting time, the cleaner is given an extra payment of $200. If a concert begins up to 15 minutes after the scheduled starting time, the cleaner is given an extra payment of $100. If a concert begins at or before the scheduled starting time, there is no extra payment for the cleaner.

Let `C` be the random variable that represents the extra payment for the cleaner, in dollars.

    1. Using your responses from part a. and part b., complete the following table, correct to three decimal places.   (1 mark)

       

    2. Calculate the expected value of the extra payment for the cleaner, to the nearest dollar.   (1 mark)

    3. Calculate the standard deviation of `C`, correct to the nearest dollar.   (1 mark)

The owners of the Mathsland Concert Hall decide to review their operation. They study information from 1000 concerts at other similar venues, collected as a simple random sample. The sample value for the number of concerts that start more than 15 minutes after the scheduled starting time is 43.

    1. Find the 95% confidence interval for the proportion of the concerts that begin more than 15 minutes after the scheduled starting time. Give values correct to three decimal places.   (1 mark)

    2. Explain why this confidence interval suggests that the proportion of concerts that begin more than 15 minutes after the scheduled starting time at the Mathsland Concert Hall is different from the proportion at the venue in the sample.   (1 mark)

The owners of the Mathsland Concert Hall decide that concerts must not begin before the scheduled starting time. They also make changes to reduce the number of concerts that begin after the scheduled starting time. Following these changes, `M` is the random variable that represents the number of minutes after the scheduled starting time that concerts begin. The probability density function for `M` is
 

`qquad qquad f(x) = {(8/(x + 2)^3), (0):} qquad {:(x ≥ 0), (x < 0):}`
 

where `x` is the the time, in minutes, after the scheduled starting time.

  1. Calculate the expected value of `M`.   (2 marks) 
    1. Find the probability that a concert now begins more than 15 minutes after the scheduled starting time.   (1 mark)

    2. Find the probability that each of the next nine concerts begins more than 15 minutes after the scheduled starting time and the 10th concert begins more than 15 minutes after the scheduled starting time. Give your answer correct to four decimal places.   (2 marks)

    3. Find the probability that a concert begins up to 20 minutes after the scheduled starting time, given that it begins more than 15 minutes after the scheduled starting time. Give your answer correct to three decimal places.   (2 marks)

Show Answers Only

  1. `0.0062`
  2. `0.1056`
  3. i.   

     ii.   `$110 \ \ text((nearest dollar))`
     iii.  `$32 \ \ text(nearest dollar))`
  4. i.   `(0.030, 0.056)`
    ii.   `text(The proportion of concerts that begin more than 15 minutes)` 

     

    `qquad text(late is not within the sample 95% confidence interval.)`

  5.  `2`
  6. i.   `(4)/(289)`
    ii.   `0.0122 \ \ text((to 4 decimal places))`
    iii.  `0.403 \ \ text((to 3 decimal places))`

Show Worked Solution

a.    `L\ ~\ N (10, 4^2)`

`text(Pr) (L < 0)` `= P(z < –2.5)`
  `= 0.0062`

 

b.    `text(Pr) (L > 15)` `= text(Pr) ( z > 1.25)`
  `= 0.1056`

 

c.i.

ii.  `E(C)` `= 0.8882 xx 100 + 0.1056 xx 200`
  `= 109.94`
  `= $110 \ \ text((nearest dollar))`

 

iii.  `E(C^2)` `= 100^2 xx 0.8882 + 200^2 xx 0.1056`
  `= 13\ 106`

 

`text(s.d.) (C)` `= sqrt(E(C^2) – [E(C)]^2)`
  `= sqrt(13\ 106 – (109.94)^2)`
  `= sqrt(1019.1 …)`
  `= 31.92 …`
  `= $32 \ \ text((nearest dollar))`

 

 
d.i.  `overset^p = (43)/(1000) \ \ , \ n = 1000`

`95% \ text(C. I.)` `= overset^p ± 1.96 sqrt((overset^p(1 – overset^p))/(n))`
  `= (0.030, 0.056)`

 
ii.   `text(The proportion of concerts that begin more than 15 minutes)`

`text(late is not within the sample 95% confidence interval.)`

 

e.    `E(M)` `= int_0^∞ x ((8)/((x +2)^3))\ dx`
  `= 2`

 

f.i.    `text(Pr)(M > 15)` `=int_0^∞ x ((8)/((x +2)^3))\ dx`
  `= (4)/(289)`

 
ii.  `text(Pr)(M > 15) = (4)/(289) \ \ , \ \ text(Pr)(M ≤ 15) = (285)/(289)`
 

`:. \ text(Pr) text{(9 concerts}\ \ M ≤ 15 ,\ text{10th concert}\ \ M > 15)`

`= ((205)/(289))^9 xx (4)/(289)`

`= 0.0122 \ \ text((to 4 decimal places))`

 
iii.  `text(Pr)(15 < M < 20)= int_15^20 (8)/((x + 2)^3)\ dx = (195)/(34\ 969)`

 `text(Pr)(M > 15)= int_15^oo (8)/((x + 2)^3)\ dx = (4)/(289)`
 

`text(Pr)(M < 20 | M>15)` `= (text(Pr)(15 < M < 20))/(text(Pr)(M > 15))`
  `= ((195)/(34\ 969))/((4)/(289))`
  `= (195)/(484)`
  `= 0.403 \ \ text((to 3 decimal places))`

Statistics, MET2 2019 VCAA 4

The Lorenz birdwing is the largest butterfly in Town A.

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

`f(x) = {(4/625 (5x^3-x^4), quad 0 <= x <= 5),(0, quad text(elsewhere)):}`
 

  1. Find the mean life span of the Lorenz birdwing butterfly.  (2 marks)

  2. 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)

  3. What is the probability that a Lorenz birdwing butterfly lives for at least four weeks, given that it lives for at least two weeks, correct to four decimal places?  (2 marks)

The wingspans of Lorenz birdwing butterflies in Town A are normally distributed with a mean of 14.1 cm and a standard deviation of 2.1 cm.

  1. Find the probability that a randomly selected Lorenz birdwing butterfly in Town A has a wingspan between 16 cm and 18 cm, correct to four decimal places.  (1 mark)

  2. A Lorenz birdwing butterfly is considered to be very small if its wingspan is in the smallest 5% of all the Lorenz birdwing butterflies in Town A.

     

    Find the greatest possible wingspan, in centimetres, for a very small Lorenz birdwing butterfly in Town A, correct to one decimal place.  (1 mark)

Each year, a detailed study is conducted on a random sample of 36 Lorenz birdwing butterflies in Town A.

A Lorenz birdwing butterfly is considered to be very large if its wingspan is greater than 17.5 cm. The probability that the wingspan of any Lorenz birdwing butterfly in Town A is greater than 17.5 cm is 0.0527, correct to four decimal places.

    1. Find the probability that three or more of the butterflies, in a random sample of 36 Lorenz birdwing butterflies from Town A, are very large, correct to four decimal places.  (1 mark)

    2. The probability that `n` or more butterflies, in a random sample of 36 Lorenz birdwing butterflies from Town A, are very large is less than 1%.

       

      Find the smallest value of `n`, where `n` is an integer.  (2 marks)

    3. For random samples of 36 Lorenz birdwing butterflies in Town A, `hat p` is the random variable that represents the proportion of butterflies that are very large.
    4. Find the expected value and the standard deviation of `hat p`, correct to four decimal places.  (2 marks)

    5. What is the probability that a sample proportion of butterflies that are very large lies within one standard deviation of 0.0527, correct to four decimal places? Do not use a normal approximation.  (2 marks)

  2. The Lorenz birdwing butterfly also lives in Town B.

     

    In a particular sample of Lorenz birdwing butterflies from Town B, an approximate 95% confidence interval for the proportion of butterflies that are very large was calculated to be (0.0234, 0.0866), correct to four decimal places.

     

    Determine the sample size used in the calculation of this confidence interval.  (2 marks)

Show Answers Only

  1. `10/3`
  2. `73`
  3. `0.2878`
  4. `0.1512`
  5. `10.6\ text(cm)`
    1. `0.2947`
    2. `7`
    3. `0.0372`
    4. `0.7380`
  7. `200`

Show Worked Solution

a.    `mu` `= 4/625 int_0^5 x(5x^3-x^4)\ dx`
    `= 4/625[x^5-1/6 x^6]_0^5`
    `= 10/3\ \ \ text{(by CAS)}`

 

b.    `text(Pr)(X > 2)` `= 4/625 int_2^5 5x^3-x^4\ dx`
    `= 4/625[5/4x^4-x^5/5]_2^5`
    `= 0.9129…`

 

`:.\ text(Expected number)` `= 80 xx 0.9129…`
  `~~ 73.03`
  `~~ 73`

 

c.    `text(Pr)(X = 4|X> 2)` `= (text(Pr)(X >= 4))/(text(Pr)(X >= 2))`
    `= 0.26272/0.91296`
    `= 0.2878`

 

d.   `W\ ~\ N (14.1, 2.1^2)`

`text(Pr)(16 < W < 18) = 0.1512\ \ \ text{(by CAS)}`

 

e.   `text(Solution 1:)`

`text(Pr)(W < w) = 0.05`

`text(Pr)(Z < z) = 0.05\ \ =>\ \ z = -1.6449\ \ text{(by CAS)}`

`(w-14.1)/2.1` `= -1.6449`
`w` `= 10.6\ text(cm)`

 
`text(Solution 2:)`

`text(invNorm)`

`text(Tail setting: left)`

`text(prob: 0.05)`

`sigma: 2.1`

`mu: 14.1`

`=> 10.6\ \ \ text{cm (by CAS)}`

 

f.i.   `L\ ~\ text(Bi)(n, p)\ ~\ text(Bi) (36, 0.0527)`

`text(Pr)(L >= 3) ~~ 0.2947`

 

f.ii.    `text(Pr)(L >= n) < 0.01`
 

`text(CAS: binomialCdf) (x, 36, 36, 0.0527)`

`text(Pr)(L >= 0) = 0.011 > 0.01`

`text(Pr)(L >= 7) = 0.002 < 0.01`

`:.\ text(Smallest)\ n = 7`

 

f.iii.    `E(hat p)` `= p = 0.0527`
  `sigma(hat p)` `= sqrt((p(1-p))/n) = sqrt((0.0527(1-0.0527))/36) ~~ 0.0372`

 

f.iv.        `hat p +- 1 sigma: (0.0527-0.0372, 0.0527 + 0.0372) = (0.0155, 0.0899)`

`text(Pr)(0.0155 < hat p < 0.0899)`

`= text(Pr)(36 xx 0.0155 < L < 36 xx 0.0899)`

`= text(Pr)(0.56 < L < 3.24)`

`= text(Pr)(1 <= L <= 3)`

`~~ 0.7380`

 

g.   `0.0234 = hat p-1.96 sqrt((hat p(1-hat p))/n) qquad text{… (1)}`

`0.0866 = hat p + 1.96 sqrt((hat p(1-hat p))/n) qquad text{… (2)}`

`text(Solve simultaneous equations:)`

`hat p ~~ 0.055, quad n ~~ 199.96`

`:.\ text(Sample size) = 200`

Probability, MET2 2011 VCAA 2*

In a chocolate factory the material for making each chocolate is sent to one of two machines, machine A or machine B.

The time, `X` seconds, taken to produce a chocolate by machine A, is normally distributed with mean 3 and standard deviation 0.8.

The time, `Y` seconds, taken to produce a chocolate by machine B, has the following probability density function
 

`f(y) = {{:(0,y < 0),(y/16,0 <= y <= 4),(0.25e^(−0.5(y-4)),y > 4):}`
 

  1. Find correct to four decimal places

    1. `text(Pr)(3 <= X <= 5)`   (1 mark)

    2. `text(Pr)(3 <= Y <= 5)`   (3 marks)

  2. Find the mean of `Y`, correct to three decimal places.   (3 marks)

  3. It can be shown that  `text(Pr)(Y <= 3) = 9/32`. A random sample of 10 chocolates produced by machine B is chosen. Find the probability, correct to four decimal places, that exactly 4 of these 10 chocolate took 3 or less seconds to produce.   (2 marks)

All of the chocolates produced by machine A and machine B are stored in a large bin. There is an equal number of chocolates from each machine in the bin.

It is found that if a chocolate, produced by either machine, takes longer than 3 seconds to produce then it can easily be identified by its darker colour.

  1. A chocolate is selected at random from the bin. It is found to have taken longer than 3 seconds to produce.
  2. Find, correct to four decimal places, the probability that it was produced by machine A.   (3 marks)

Show Answers Only

a.i.  `0.4938`

a.ii. `0.4155`

b.    `4.333`

c.    `0.1812`

d.    `0.4103`

Show Worked Solution

a.i.   `X ∼\ N(3,0.8^2)`

`text(Pr)(3 <= X <= 5) = 0.4938\ \ text{(4 d.p.)}`

 

a.ii.    `text(Pr)(3 <= Y <= 5)` `= text(Pr)(3 <= Y <= 4) + (4 < Y <= 5)`
    `= int_3^4 (y/16)dy + int_4^5(1/4 e^(−1/2(y-4)))dy`
    `= 0.4155\ \ text{(4 d.p.)}`

 

b.    `text(E)(Y)` `= int_0^4 y(y/16)dy + int_4^∞ 0.25ye^(−1/2(y-4))dy`
    `= 4.333\ \ text{(3 d.p.)}`

 

c.   `text(Solution 1)`

`text(Let)\ \ W = #\ text(chocolates from)\ B\ text(that take less)`

`text(than 3 seconds)`

`W ∼\ text(Bi)(10, 9/32)`

`text(Using CAS: binomPdf)(10, 9/32,4)`

`text(Pr)(W = 4) = 0.1812\ \ text{(4 d.p.)}`
 

`text(Solution 2)`

`text(Pr)(W = 4)`  `=((10),(4)) (9/32)^4 (23/32)^6` 
  `=0.1812`

♦♦♦ Mean mark part (e) 19%.
MARKER’S COMMENT: Students who used tree diagrams were the most successful.

 

d.   
`text(Pr)(A | L)` `= (text(Pr)(AL))/(text(Pr)(L))`
  `= (0.5 xx 0.5)/(0.5 xx 0.5 + 0.5 xx 23/32)`
  `= 0.4103\ \ text{(4 d.p.)}`

Probability, MET2 2016 VCAA 3*

A school has a class set of 22 new laptops kept in a recharging trolley. Provided each laptop is correctly plugged into the trolley after use, its battery recharges.

On a particular day, a class of 22 students uses the laptops. All laptop batteries are fully charged at the start of the lesson. Each student uses and returns exactly one laptop. The probability that a student does not correctly plug their laptop into the trolley at the end of the lesson is 10%. The correctness of any student’s plugging-in is independent of any other student’s correctness.

  1. Determine the probability that at least one of the laptops is not correctly plugged into the trolley at the end of the lesson. Give your answer correct to four decimal places.   (2 marks)

  2. A teacher observes that at least one of the returned laptops is not correctly plugged into the trolley.
  3. Given this, find the probability that fewer than five laptops are not correctly plugged in. Give your answer correct to four decimal places.   (2 marks)

The time for which a laptop will work without recharging (the battery life) is normally distributed, with a mean of three hours and 10 minutes and standard deviation of six minutes. Suppose that the laptops remain out of the recharging trolley for three hours.

  1. For any one laptop, find the probability that it will stop working by the end of these three hours. Give your answer correct to four decimal places.   (2 marks)

A supplier of laptops decides to take a sample of 100 new laptops from a number of different schools. For samples of size 100 from the population of laptops with a mean battery life of three hours and 10 minutes and standard deviation of six minutes, `hat P` is the random variable of the distribution of sample proportions of laptops with a battery life of less than three hours.

  1. Find the probability that `text(Pr) (hat P >= 0.06 | hat P >= 0.05)`. Give your answer correct to three decimal places. Do not use a normal approximation.   (3 marks)

It is known that when laptops have been used regularly in a school for six months, their battery life is still normally distributed but the mean battery life drops to three hours. It is also known that only 12% of such laptops work for more than three hours and 10 minutes.

  1. Find the standard deviation for the normal distribution that applies to the battery life of laptops that have been used regularly in a school for six months, correct to four decimal places.   (2 marks)

The laptop supplier collects a sample of 100 laptops that have been used for six months from a number of different schools and tests their battery life. The laptop supplier wishes to estimate the proportion of such laptops with a battery life of less than three hours.

  1. Suppose the supplier tests the battery life of the laptops one at a time.
  2. Find the probability that the first laptop found to have a battery life of less than three hours is the third one.   (1 mark)

The laptop supplier finds that, in a particular sample of 100 laptops, six of them have a battery life of less than three hours.

  1. Determine the 95% confidence interval for the supplier’s estimate of the proportion of interest. Give values correct to two decimal places.   (1 mark)

  2. The supplier also provides laptops to businesses. The probability density function for battery life, `x` (in minutes), of a laptop after six months of use in a business is
     

     

    `qquad qquad f(x) = {(((210-x)e^((x-210)/20))/400, 0 <= x <= 210), (0, text{elsewhere}):}`
     

  3. Find the mean battery life, in minutes, of a laptop with six months of business use, correct to two decimal places.   (1 mark)

Show Answers Only

  1. `0.9015`
  2. `0.9311`
  3. `0.0478`
  4. `0.658`
  5. `8.5107`
  6. `1/8`
  7. `p in (0.01, 0.11)`
  8. `170.01\ text(min)`

Show Worked Solution

a.   `text(Solution 1)`

`text(Let)\ \ X = text(number not correctly plugged),`

`X ~ text(Bi) (22, .1)`

`text(Pr) (X >= 1) = 0.9015\ \ [text(CAS: binomCdf)\ (22, .1, 1, 22)]`

 

`text(Solution 2)`

`text(Pr) (X>=1)` `=1-text(Pr) (X=0)`
  `=1-0.9^22`
  `=0.9015\ \ text{(4 d.p.)}`

 

 b.   `text(Pr) (X < 5 | X >= 1)`

MARKER’S COMMENT: Early rounding was a common mistake, producing 0.9312.

`= (text{Pr} (1 <= X <= 4))/(text{Pr} (X >= 1))`

`= (0.83938…)/(0.9015…)\ \ [text(CAS: binomCdf)\ (22, .1, 1,4)]`

`= 0.9311\ \ text{(4 d.p.)}`

 

c.   `text(Let)\ \ Y = text(battery life in minutes)`

MARKER’S COMMENT: Some working must be shown for full marks in questions worth more than 1 mark.

`Y ~ N (190, 6^2)`

`text(Pr) (Y <= 180)= 0.0478\ \ text{(4 d.p.)}`

`[text(CAS: normCdf)\ (−oo, 180, 190,6)]`

 

d.   `text(Let)\ \ W = text(number with battery life less than 3 hours)`

♦ Mean mark part (d) 33%.

`W ~ Bi (100, .04779…)`

`text(Pr) (hat P >= .06 | hat P >= .05)` `= text(Pr) (X_2 >= 6 | X_2 >= 5)`
  `= (text{Pr} (X_2 >= 6))/(text{Pr} (X_2 >= 5))`
  `= (0.3443…)/(0.5234…)`
  `= 0.658\ \ text{(3 d.p.)}`

 

e.   `text(Let)\ \ B = text(battery life), B ~ N (180, sigma^2)`

`text(Pr) (B > 190)` `= .12`
`text(Pr) (Z < a)` `= 0.88`
`a` `dot = 1.17499…\ \ [text(CAS: invNorm)\ (0.88, 0, 1)]`
`-> 1.17499` `= (190-180)/sigma\ \ [text(Using)\ Z = (X-u)/sigma]`
`:. sigma` `dot = 8.5107`

 

f.    `text(Pr) (MML)` `= 1/2 xx 1/2 xx 1/2`
    `= 1/8`

 

g.   `text(95% confidence int:) qquad quad [(text(CAS:) qquad qquad 1-text(Prop)\ \ z\ \ text(Interval)), (x = 6), (n = 100)]`

`p in (0.01, 0.11)`

 

h.    `mu` `= int_0^210 (x* f(x)) dx`
  `:. mu` `dot = 170.01\ text(min)`

Probability, MET1 2013 VCAA 8

A continuous random variable, `X`, has a probability density function
 

`f(x) = { (pi/4 cos ((pi x)/4),\ \ \ text(if)\ x in [0, 2]), (\ \ \ 0,\ \ \ text(otherwise)):}`
 

Given that  `d/(dx) (x sin ((pi x)/4)) = (pi x)/4 cos ((pi x)/4) + sin ((pi x)/4)`,  find  `text(E)(X).`  (3 marks)

Show Answers Only

`2 – 4/pi`

Show Worked Solution

`text(Find)\ \ int (pix)/4 cos ((pi x)/4)\ dx`

`text(Integrating the given equation:)`

`int (pi x)/4 cos ((pi x)/4) + sin ((pi x)/4)\ dx` `= x sin ((pi x)/4)`
`int (pi x)/4 cos ((pi x)/4)\ dx – 4/pi cos ((pi x)/4)` `= x sin ((pi x)/4)`
`:. int (pi x)/4 cos ((pi x)/4)\ dx` `= x sin ((pi x)/4) + 4/pi cos ((pi x)/4)`
♦♦ Mean mark 32%.

 

`:.  text(E)(X)` `= int_0^2 x (pi/4 cos (pi/4 x)) dx`
  `= [x sin ((pi x)/4) + 4/pi cos ((pi x)/4)]_0^2`
  `= (2 sin (pi/2) + 4/pi cos (pi/2)) – (0 + 4/pi cos (0))`
  `= (2 + 0) – (0 +4/pi)`
  `= 2 – 4/pi`

Probability, MET2 2014 VCAA 4*

Patricia is a gardener and she owns a garden nursery. She grows and sells basil plants and coriander plants.

The heights, in centimetres, of the basil plants that Patricia is selling are distributed normally with a mean of 14 cm and a standard deviation of 4 cm. There are 2000 basil plants in the nursery.

  1. Patricia classifies the tallest 10 per cent of her basil plants as super.
  2. What is the minimum height of a super basil plant, correct to the nearest millimetre?   (1 mark)

Patricia decides that some of her basil plants are not growing quickly enough, so she plans to move them to a special greenhouse. She will move the basil plants that are less than 9 cm in height.

  1. How many basil plants will Patricia move to the greenhouse, correct to the nearest whole number?   (2 marks)

The heights of the coriander plants, `x` centimetres, follow the probability density function  `h(x)`,

`h(x) = {(pi/100 sin ((pi x)/50), 0 < x < 50), (\ \ \ \ \ \0, text(otherwise)):}`

  1. State the mean height of the coriander plants.   (1 mark)

Patricia thinks that the smallest 15 per cent of her coriander plants should be given a new type of plant food

  1. Find the maximum height, correct to the nearest millimetre, of a coriander plant if it is to be given the new type of plant food.   (2 marks)

Patricia also grows and sells tomato plants that she classifies as either tall or regular. She finds that 20 per cent of her tomato plants are tall.

A customer, Jack, selects `n` tomato plants at random.

  1. Let `q` be the probability that at least one of Jack’s `n` tomato plants is tall.
  2. Find the minimum value of `n` so that `q` is greater than 0.95.   (2 marks)

Show Answers Only

  1. `191\ text(mm)`
  2. `211\ text(plants)`
  3. `25\ text(cm)`
  4. `127\ text(mm)`
  5. `14\ text(plants)`

Show Worked Solution

a.   `text(Let)\ \ X = text(plant height,)`

♦ Mean mark 43%.

`X ∼\ text(N)(14,4^2)`

`text(Pr)(X > a)` `= 0.1`
`a` `= 19.1\ text(cm)quadtext([CAS: invNorm)\ (.9,14,4)]`
  `=191\ text{mm  (nearest mm)}`

 

`:.\ text(Min super plant height is 191 mm.)`

 

b.   `text(Pr)(X < 9) = 0.10565…\ qquadtext([CAS: normCdf)\ (−∞,9,14,4)]`

`:.\ text(Number moved to greenhouse)`

`= 0.10565… xx 2000`

`= 211\ text(plants)`

 

c.    `text(E)(X)` `= int_0^50  (x xx pi/100 sin((pix)/50))dx`
    `= 25\ text(cm)`

 

d.    `text(Solve:)\ \ int_0^a h(x)\ dx` `= 0.15\ \ text(for)\ \ a ∈ (0,50)`

♦♦ Mean mark 33%.

`:.a` `=12.659…\ text(cm)`
  `=127\ text{mm  (nearest mm)}`

 

e.   `text(Let)\ \ Y =\ text(Number of tall plants,)`

♦♦ Mean mark 30%.

`Y ∼\ text(Bi) (n,0.2)`

`text(Pr)(Y >= 1)` `> 0.95`
`1-text(Pr)(Y = 0)` `> 0.95`
`0.05` `> 0.8^n`
`n` `> 13.4\ \ text([by CAS])`

 

`:. n_text(min) = 14\ text(plants)`

Probability, MET2 2015 VCAA 3

Mani is a fruit grower. After his oranges have been picked, they are sorted by a machine, according to size. Oranges classified as medium are sold to fruit shops and the remainder are made into orange juice.

The distribution of the diameter, in centimetres, of medium oranges is modelled by a continuous random variable, `X`, with probability density function
 

`f(x) = {(3/4(x-6)^2(8-x), 6 <= x <= 8), (\ \ \ \ \ \ \ 0, text(otherwise)):}`
 

    1. Find the probability that a randomly selected medium orange has a diameter greater than 7 cm.   (2 marks)

    2. Mani randomly selects three medium oranges.
    3. Find the probability that exactly one of the oranges has a diameter greater than 7 cm.
    4. Express the answer in the form `a/b`, where `a` and `b` are positive integers.   (2 marks)

  2. Find the mean diameter of medium oranges, in centimetres.   (1 mark)

For oranges classified as large, the quantity of juice obtained from each orange is a normally distributed random variable with a mean of 74 mL and a standard deviation of 9 mL.

  1. What is the probability, correct to three decimal places, that a randomly selected large orange produces less than 85 mL of juice, given that it produces more than 74 mL of juice?  (2 marks)

Mani also grows lemons, which are sold to a food factory. When a truckload of lemons arrives at the food factory, the manager randomly selects and weighs four lemons from the load. If one or more of these lemons is underweight, the load is rejected. Otherwise it is accepted.

It is known that 3% of Mani’s lemons are underweight.

    1. Find the probability that a particular load of lemons will be rejected. Express the answer correct to four decimal places.   (2 marks)

    2. Suppose that instead of selecting only four lemons, `n` lemons are selected at random from a particular load.
    3. Find the smallest integer value of `n` such that the probability of at least one lemon being underweight exceeds 0.5  (2 marks)

Show Answers Only

    1. `11/16`
    2. `825/4096`
  2. `36/5\ text(cm)`
  3. `0.778`
    1. `0.1147`
    2. `23`

Show Worked Solution

a.i.    `text(Pr)(X > 7)` `= int_7^8 f(x)\ dx`
    `= int_7^8 (3/4(x-6)^2(8-x))\ dx`
    `= 11/16`

 

a.ii.   `text(Let)\ \ Y =\ text(number with diameter > 7cm)`

  `Y ∼\ text(Bi)(3,11/6)`

`text(Pr)(Y = 1)` `= ((3),(1))(11/16)^1 xx (5/16)^2`
  `= 825/4096`

 

b.   `text{E(X)}` `= int_6^8 (x xx f(x))\ dx`
    `= 36/5`
    `=7.2\ text(cm)`

 

c.   `text(Let)\ \ L = text(Large juice quantity)`

`L ∼\ N(74,9^2)`

`text(Pr)(L < 85 | L > 74)` `= (text(Pr)(L < 85 ∩ L>74))/(text(Pr)(L > 74))`
  `= (text(Pr)(74 < L < 85))/(text(Pr)(L > 74))`
  `= (0.3891…)/0.5`
  `= 0.7783…` 
  `=0.778\ \ text{(3 d.p.)}`

 

d.i.  `text{Solution 1 [by CAS]}`

`text(Let)\ \ W =\ text(number of lemons under weight)`

 `W ∼\ text(Bi)(4,0.03)`

`text(Pr)(W >= 1) = 0.1147qquad[text(CAS: binomCdf)\ (4,0.03,1,4)]`

 

`text(Solution 2)`

`text(Pr)(W>=1)` `=1-text(Pr)(W=0)`
  `=1-(0.97)^4`
  `=0.11470…`
  `=0.1147\ \ text{(4 d.p.)}`

 

d.ii.  `W ∼\ text(Bi)(n,0.03)`

♦ Mean mark 44%.

`text(Pr)(W >= 1)` `> 1/2`
`1-text(Pr)(W = 0)` `> 1/2`
`1/2` `> (0.97)^n`
`n` `> 22.8`
`:. n_text(min)` `= 23`

Probability, MET2 2013 VCAA 2

FullyFit is an international company that owns and operates many fitness centres (gyms) in several countries. At every one of FullyFit’s gyms, each member agrees to have his or her fitness assessed every month by undertaking a set of exercises called `S`. There is a five-minute time limit on any attempt to complete `S` and if someone completes `S` in less than three minutes, they are considered fit.

  1. At FullyFit’s Melbourne gym, it has been found that the probability that any member will complete `S` in less than three minutes is `5/8.` This is independent of any member.
  2. In a particular week, 20 members of this gym attemp `S.`
    1. Find the probability, correct to four decimal places, that at least 10 of these 20 members will complete `S` in less than three minutes (2 marks)

    2. Given that at least 10 of these 20 members complete `S` in less than three minutes, what is the probability, correct to three decimal places, that more than 15 of them complete `S` in less than three minutes?  (3 marks)

  3. When FullyFit surveyed all its gyms throughout the world, it was found that the time taken by members to complete `S` is a continuous random variable `X`, with a probability density function `g`, as defined below.
     
    `qquad qquad qquad g(x) = {(((x-3)^3 + 64)/256 ,  1 <= x <= 3),((x + 29)/128 ,  3 < x <= 5),(0 ,  text(elsewhere)):}`

    1. Find `text(E)(X)`, correct to four decimal places.  (2 marks)

    2. In a random sample of 200 FullyFit members, how many members would be expected to take more than four minutes to complete `S?` Give your answer to the nearest integer.  (2 marks)

Show Answers Only

a.i.   `0.9153`

a.ii.   `0.086`

b.i.   `3.0458`

b.ii.  `52\ text(people)`

Show Worked Solution

a.i.   `text(Let)\ \ y =\ text(number who complete in less than 3 min)`

`Y ∼\ text(Bi)(20, 5/8)`

`text(Pr)(Y >= 10)` `= 0.91529…`
  `= 0.9153\ \ text{(4 d.p.)}`

 

a.ii.    `text(Pr)(Y> 15 | Y >= 10)` `= (text(Pr)(Y > 15))/(text(Pr)(Y >= 10))`
    `= (0.079041…)/(0.915292…)`
    `= 0.086\ \ text{(3 d.p.)}`

 

b.i.    `text(E)(X)` `=int_-oo^oo (x xx g(x))\ dx`
    `= int_1^3 x[((x-3)^3 + 64)/256]dx + int_3^5 x((x + 29)/128) dx`
    `= 3.04583…`
    `= 3.0458\ \ text{(4 d.p.)}`

 

b.ii.  `text(Let)\ \ W =\ text(number who take more than 4 min)`

♦ Mean mark 48%.

`W ∼\ text(Bi)(200, int_4^5 (x + 29)/128\ dx)`

`W ∼\ text(Bi)(200, 67/256)`

`text(E)(W)` `= np`
  `= 200 xx 67/256`
  `= 1675/32`
  `= 52.3437…`
  `= 52\ text(people)`

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`

Probability, MET2 2014 VCAA 16 MC

The continuous random variable `X`, with probability density function  `p(x)`, has mean 2 and variance 5.

The value of  `int_-oo^oo x^2 p(x)\ dx`  is

  1. `1`
  2. `7`
  3. `9`
  4. `21`
  5. `29`
Show Answers Only

`C`

Show Worked Solution
♦ Mean mark 46%.
`text(VAR)(X)` `= text(E)(X^2) – [text(E)(X)]^2`
`5` `= int_(−∞)^∞ x^2 p(x)\ dx – (2)^2`
`:. 9` `= int_(−∞)^∞ x^2 p(x)\ dx`

 
`=>   C`