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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 2021 VCAA 4

A teacher coaches their school's table tennis team.

The teacher has an adjustable ball machine that they use to help the players practise.

The speed, measured in metres per second, of the balls shot by the ball machine is a normally distributed random variable `W`.

The teacher sets the ball machine with a mean speed of 10 metres per second and standard deviation of 0.8 metres per second.

  1. Determine  `text(Pr) (W ≥11)`, correct to three decimal places.   (1 mark)

  2. Find the value of `k`, in metres per second, which 80% of ball speeds are below. Give your answer in metres per second, correct to one decimal place.   (1 mark) 

The teacher adjusts the height setting for the ball machine. The machine now shoots balls high above the table tennis table.

Unfortunately, with the new height setting, 8% of balls do not land on the table.

Let  `overset^P`  be the random variable representing the sample proportion of the balls that do not land on the table in random samples of 25 balls.

  1. Find the mean and the standard deviation of  `overset^P`.   (2 marks)

  2. Use the binomial distribution to find  `text(Pr) (overset^P > 0.1)`, correct to three decimal places.   (2 marks) 

The teacher can also adjust the spin setting on the ball machine.

The spin, measured in revolutions per second, is a continuous random variable  `X` with the probability density function
 

       `f(x) = {(x/500, 0 <= x < 20), ({50-x}/{750}, 20 <= x <= 50), (\ 0, text(elsewhere)):}`
 

  1. Find the maximum possible spin applied by the ball machine, in revolutions per second.   (1 mark)

Show Answers Only

  1. `0.106`
  2. `0.8`
  3. `sqrt46/125`
  4. `0.323`
  5. `50 \ text{revolutions per second}`
  6. This content is no longer in the Study Design.
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Show Worked Solution

a.   `W\ ~\ N (10,0.8^2)`

`text{By CAS: norm Cdf} \ (11, ∞, 10, 0.8)`

`text(Pr) (W >= 11) = 0.106`
 

b.    `text(Pr) (W < k) = 0.8 \ \ \ text{By CAS: inv Norm} \ (0.8, 10, 0.8)`
 

c.    `E(overset^P) =  0.08 = 2/25`

♦ Mean mark part (c) 45%.

`text(s.d.) (overset^P) = sqrt{{0.08 xx 0.92}/{25}} = sqrt46/125`
 

d.    `X\ ~\ text(Bi) (25, 0.08)`

♦ Mean mark part (d) 49%.

`text{By CAS: binomCdf} \ (25, 0.08, 3, 25)`

`text(Pr) (overset^P > 0.1)` `= text(Pr) (X > 0.1 xx 25)`
  `= text(Pr) (X > 2.5)`
  `= text(Pr) (X >= 3)`
  `= 0.323`

 

e.    `text{Maximum spin = 50 revolutions per second}`

♦♦ Mean mark part (e) 21%.

 

f.    This content is no longer in the Study Design.

g.    This content is no longer in the Study Design.

h.    This content is no longer in the Study Design.

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`