smc-637-45-Other probability

Probability, MET2 2023 VCAA 4

A manufacturer produces tennis balls.

The diameter of the tennis balls is a normally distributed random variable $D$, which has a mean of 6.7 cm and a standard deviation of 0.1 cm.

Find $\Pr(D>6.8)$, correct to four decimal places. (1 mark)
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Find the minimum diameter of a tennis ball that is larger than 90% of all tennis balls produced.
Give your answer in centimetres, correct to two decimal places. (1 mark)
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Tennis balls are packed and sold in cylindrical containers. A tennis ball can fit through the opening at the top of the container if its diameter is smaller than 6.95 cm.

Find the probability that a randomly selected tennis ball can fit through the opening at the top of the container.
Give your answer correct to four decimal places. (1 mark)
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In a random selection of 4 tennis balls, find the probability that at least 3 balls can fit through the opening at the top of the container.
Give your answer correct to four decimal places. (2 marks)
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A tennis ball is classed as grade A if its diameter is between 6.54 cm and 6.86 cm, otherwise it is classed as grade B.

Given that a tennis ball can fit through the opening at the top of the container, find the probability that it is classed as grade A.
Give your answer correct to four decimal places. (2 marks)
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The manufacturer would like to improve processes to ensure that more than 99% of all tennis balls produced are classed as grade A.
Assuming that the mean diameter of the tennis balls remains the same, find the required standard deviation of the diameter, in centimetres, correct to two decimal places. (2 marks)
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An inspector takes a random sample of 32 tennis balls from the manufacturer and determines a confidence interval for the population proportion of grade A balls produced.
The confidence interval is (0.7382, 0.9493), correct to four decimal places.
Find the level of confidence that the population proportion of grade A balls is within the interval, as a percentage correct to the nearest integer. (2 marks)
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A tennis coach uses both grade A and grade B balls. The serving speed, in metres per second, of a grade A ball is a continuous random variable, $V$, with the probability density function

$f(v) = \begin {cases}
\dfrac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg) &\ \ 30 \leq v \leq 3\pi^2+30 \\
0 &\ \ \text{elsewhere}
\end{cases}$

Find the probability that the serving speed of a grade A ball exceeds 50 metres per second.
Give your answer correct to four decimal places. (1 mark)
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Find the exact mean serving speed for grade A balls, in metres per second. (1 mark)
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The serving speed of a grade B ball is given by a continuous random variable, $W$, with the probability density function $g(w)$.

A transformation maps the graph of $f$ to the graph of $g$, where $g(w)=af\Bigg(\dfrac{w}{b}\Bigg)$.

If the mean serving speed for a grade B ball is $2\pi^2+8$ metres per second, find the values of $a$ and $b$. (2 marks)
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Show Answers Only

a. $0.1587$

b. $d\approx6.83$

c. $0.9938$

d. $0.9998$

e. $0,8960$

f. $0.06$

g. $90\%$

h. $0.1345$

i. $3(\pi^2+4)$

j. $a=\dfrac{3}{2}, b=\dfrac{2}{3}$

Show Worked Solution

a. $\text{Normal distribution:}\to \mu=6.7, \sigma=0.1$

$D\sim N(6.7, 0.1^2)$

$\text{Using CAS: }[\text{normCdf}(6.8, \infty, 6.7, 0.1)]$

$\Pr(D>6.8)=0.15865…\approx0.1587$

b. $\Pr(D<d)=0.90$

$\Pr\Bigg(Z<\dfrac{d-6.7}{0.1}\Bigg)=0.90$

$\text{Using CAS: }[\text{invNorm}(0.9, 6.7, 0.1)]$

$\therefore\ d=6.828155…\approx6.83\ \text{cm}$

c. $\text{Normal distribution:}\to \mu=6.7, \sigma=0.1$

$D\sim N(6.7, 0.1^2)$

$\text{Using CAS: }[\text{normCdf}(-\infty, 6.95, 6.7, 0.1)]$

$\Pr(D<6.95)=0.99379…\approx0.9938$

d. $\text{Binomial:}\to n=4, p=0.99379…$

$X=\text{number of balls}$

$X\sim \text{Bi}(4, 0.999379 …)$

$\text{Using CAS: }[\text{binomCdf}(4, 0.999379 …, 3, 4)]$

$\Pr(X\geq 3)=0.99977 …\approx 0.9998$

e.	$\Pr(\text{Grade A\|Fits})$	$=\Pr(6.54<D<6.86\|D<6.95)$
		$=\dfrac{\Pr(6.54<D<6.86)}{\Pr(D<6.95)}$
	$\text{Using CAS: }$	$=\Bigg[\dfrac{\text{normCdf}(6.54, 6.86, 6.7, 0.1)}{\text{normCdf}(-\infty, 6.95, 6.7, 0.1)}\Bigg]$
		$=\dfrac{0.89040…}{0.99977…}$
		$=0.895965…\approx 0.8960$

f. $\text{Normally distributed → symmetrical}$

$\text{Pr ball diameter outside the 99% interval}=1-0.99=0.01$

$D\sim N(6.7, \mu^2)$

$\Pr(6.54<D<6.86)>0.99$

$\therefore\ \Pr(D<6.54)<\dfrac{1-0.99}{2}$

$\text{Find z score using CAS: }\Bigg[\text{invNorm}\Bigg(\dfrac{1-0.99}{2},0,1\Bigg)\Bigg]=-2.575829…$

$\text{Then solve:} \ \dfrac{6.54-6.7}{\sigma}<-2.575829…$

$\therefore\ \sigma<0.0621\approx 0.06$

♦♦ Mean mark (f) 35%.
MARKER’S COMMENT: Incorrect responses included using $\Pr(D<6.86)=0.99$. Many answers were accepted in the range $0<\sigma\leq0.06$ as max value of SD was not asked for provided sufficient working was shown.

g. $\hat{p}=\dfrac{0.7382+0.9493}{2}=0.84375$

$\text{Solve the following simultaneous equations for }z, \hat{p}=0.84375$

$0.84375-z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}$	$=0.7382$	$(1)$
$0.84375+z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}$	$=0.9493$	$(2)$
$\text{Equation}(2)-(1)$
$2z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}$	$=0.2111$
$z=\dfrac{0.10555}{\sqrt{\dfrac{0.84375(1-0.84375)}{32}}}$	$=1.64443352$

$\text{Alternatively using CAS: Solve the following simultaneous equations for }z, \hat{p}$

$\hat{p}-z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{32}}$	$=0.7382$
$\text{and}$
$\hat{p}+z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{32}}$	$=0.9493$

$\rightarrow \ z=1.64444…\ \text{and}\ \hat{p}=0.84375$

$\therefore\ \text{Using CAS: normCdf}(-1.64443352, 1.64443352, 0 , 1)$

$\text{Level of Confidence} =0.89999133…=90\%$

♦♦ Mean mark (g) 25%.
MARKER’S COMMENT: $\hat{p}$ was often calculated incorrectly with $\hat{p}=0.8904$ frequently seen.

h. $\text{Using CAS: Evaluate }\to \displaystyle \int_{50}^{3\pi^2+30} \frac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg)\,dx=0.1345163712\approx 0.1345$

i. $\text{Using CAS: Evaluate }\to \displaystyle \int_{30}^{3\pi^2+30} v.\dfrac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg)\,dx=3(\pi^2+4)=3\pi^2+12$

j. $\text{When the function is dilated in both directions, }a\times b=1$

$\text{Method 1 : Simultaneous equations}$

$g(w) = \begin {cases}
\dfrac{b}{6\pi}\sin\Bigg(\sqrt{\dfrac{\dfrac{w}{b}-30}{3}}\Bigg) &\ \ 30b \leq v \leq b(3\pi^2+30) \\
\\ 0 &\ \ \text{elsewhere}
\end{cases}$

$\text{Using CAS: Define }g(w)\ \text{and solve the simultaneous equations below for }a, b.$

$\displaystyle \int_{30.b}^{b.(3\pi^2+30)} g(w)\,dw=1$

$\displaystyle \int_{30.b}^{b.(3\pi^2+30)} w.g(w)\,dw=2\pi^2+8$

$\therefore\ b=\dfrac{2}{3}\ \text{and}\ a=\dfrac{3}{2}$

$\text{Method 2 : Transform the mean}$

$\text{Area}$	$=1$
$\therefore\ a$	$=\dfrac{1}{b}$
$\to\ b$	$=\dfrac{E(W)}{E(V)}$
	$=\dfrac{2\pi^2+8}{3\pi^2+12}$
	$=\dfrac{2(\pi^2+4)}{3(\pi^2+4)}$
	$=\dfrac{2}{3}$
$\therefore\ a$	$=\dfrac{3}{2}$

♦♦♦ Mean mark (j) 10%.

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.

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

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

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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)
  
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2. Calculate the expected value of the extra payment for the cleaner, to the nearest dollar. (1 mark)
  
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3. Calculate the standard deviation of `C`, correct to the nearest dollar. (1 mark)
  
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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)
  
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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)
  
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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.

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

`0.0062`
`0.1056`
i.

ii. `$110 \ \ text((nearest dollar))`
iii. `$32 \ \ text(nearest dollar))`
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.)`
`2`
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)):}`

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

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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)

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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)

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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.

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)

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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)

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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)
  
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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)
  
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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)
  
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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)
  
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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)

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Show Answers Only

`10/3`
`73`
`0.2878`
`0.1512`
`10.6\ text(cm)`
1. `0.2947`
2. `7`
3. `0.0372`
4. `0.7380`
`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 2017 VCAA 3

The time Jennifer spends on her homework each day varies, but she does some homework every day.

The continuous random variable `T`, which models the time, `t,` in minutes, that Jennifer spends each day on her homework, has a probability density function `f`, where

`f(t) = {{:(1/625 (t - 20)),(1/625 (70 - t)),(0):}qquad{:(20 <= t < 45),(45 <= t <= 70),(text(elsewhere)):}:}`

Sketch the graph of `f` on the axes provided below. (3 marks)

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Find `text(Pr)(25 ≤ T ≤ 55)`. (2 marks)

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Find `text(Pr)(T ≤ 25 | T ≤ 55)`. (2 marks)

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Find `a` such that `text(Pr)(T ≥ a) = 0.7`, correct to four decimal places. (2 marks)

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The probability that Jennifer spends more than 50 minutes on her homework on any given day is `8/25`. Assume that the amount of time spent on her homework on any day is independent of the time spent on her homework on any other day.
1. Find the probability that Jennifer spends more than 50 minutes on her homework on more than three of seven randomly chosen days, correct to four decimal places. (2 marks)
  
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2. Find the probability that Jennifer spends more than 50 minutes on her homework on at least two of seven randomly chosen days, given that she spends more than 50 minutes on her homework on at least one of those days, correct to four decimal places. (2 marks)
  
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Let `p` be the probability that on any given day Jennifer spends more than `d` minutes on her homework.

Let `q` be the probability that on two or three days out of seven randomly chosen days she spends more than `d` minutes on her homework.

Express `q` as a polynomial in terms of `p`. (2 marks)

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1. Find the maximum value of `q`, correct to four decimal places, and the value of `p` for which this maximum occurs, correct to four decimal places. (2 marks)
  
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2. Find the value of `d` for which the maximum found in part g.i. occurs, correct to the nearest minute. (2 marks)
  
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Show Answers Only

`4/5`
`1/41`
`39.3649`
1. `0.1534`
2. `0.7626`
`q =7p^2(1-p)^4(2p+3)`
1. `p = 0.3539quadtext(and)quadq = 0.5665`
2. `49\ text(min)`

Show Worked Solution

MARKER’S COMMENT: Many did not draw graph along `t`-axis between 0 and 20 and for `t>70`.

b. `text(Pr)(25 <= T <= 55)`

`= int_25^45 1/625 (t – 20)\ dt + int_45^55 1/625 (70 – t)\ dt`

`= 4/5`

c. `text(Pr)(T ≤ 25 | T ≤ 55)`

`=(text(Pr)(T <= 25))/(text(Pr)( T <= 55))`

`= (int_20^25 1/625(t – 20)\ dt)/(1 – int_55^70 1/625(70 – t)\ dt)`

`= (1/50)/(1 – 9/50)`

`= 1/41`

d. `text(Pr)(T ≥ a) = 0.7`

♦ Mean mark part (d) 36%.

`=>\ text(Pr)(T <= a) = 0.3`

`text(Solve:)`

`int_20^a 1/625(t – 20)\ dt`

`= 0.3quadtext(for)quada ∈ (20, 45)`

`:. a == 39.3649`

e.i. `text(Let)\ X =\ text(Number of days Jenn studies more than 50 min)`

`X ~\ text(Bi) (7, 8/25)`

`text(Pr)(X >= 4) = 0.1534`

e.ii.	`text(Pr)(X >= 2 \| X >= 1)`	`= (text(Pr)(X >= 2))/(text(Pr)(X >= 1))`
		`= (0.7113…)/(0.9327…)`
		`= 0.7626\ \ text{(to 4 d.p.)}`

f. `text(Let)\ Y =\ text(Number of days Jenn spends more than)\ d\ text(min)`

`Y ~\ text(Bi)(7,p)`

♦ Mean mark part (f) 36%.

`q`	`= text(Pr)(Y = 2) + text(Pr)(Y = 3)`
	`= ((7),(2))p^2(1 – p)^5 + ((7),(3))p^3(1 – p)^4`
	`= 21p^2(1 – p)^5 + 35p^3(1 – p)^4`
	`=7p^2(1-p)^4[3(1-p)+5p]`
	`=7p^2(1-p)^4(2p+3)`

g.i. `text(Solve)\ \ q′(p) = 0,`

♦♦ Mean mark part (g)(i) 30%.

`p`	`=0.35388…`
	`=0.3539\ \ text{(to 4 d.p.)}`

`:. q_text(max)= 0.5665\ \ text{(to 4 d.p.)}`

g.ii. `text(Pr)(T > d) = p= 0.35388…`

♦♦♦ Mean mark part (g)(ii) 8%.

`text(Solve:)`

`int_d^70 (1/625(70 – t))dt`

`= 0.35388… quadtext(for)quadd ∈ (45,70)`

`:. d`	`=48.967…`
	`=49\ text(mins)`

Probability, MET1 2012 VCAA 8b

The probability density function `f` of a random variable `X` is given by

`qquad qquad f(x) = {((x + 1)/12, 0 <= x <= 4), (\ \ 0, text{otherwise}):}`

Find the value of `b` such that `text(Pr) (X <= b) = 5/8.` (3 marks)

Show Answers Only

`3`

Show Worked Solution

`1/12 int_0^b (x + 1)\ dx`	`= 5/8`
`[1/2 x^2 + x]_0^b`	`= 15/2`
`1/2b^2 + b`	`= 15/2`
`b^2 + 2b – 15`	`= 0`
`(b + 5) (b – 3)`	`= 0`

`:. b = 3,\ \ \ b in (0, 4)`

Probability, MET1 2016 VCAA 8*

Let `X` be a continuous random variable with probability density function

`f(x) = {(−4xlog_e(x),0<x<=1),(0,text(elsewhere)):}`

Part of the graph of `f` is shown below. The graph has a turning point at `x = 1/e`.

Show by differentiation that `(x^k)/(k^2)(k log_e(x)-1)` is an antiderivative of `x^(k – 1) log_e(x)`, where `k` is a positive real number. (2 marks)

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Calculate `text(Pr)(X > 1/e)`. (2 marks)

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Show Answers Only

`text(See Worked Solutions)`
`1-3/(e^2)`

Show Worked Solution

a. `text(Using Product Rule:)`

♦♦ Mean mark part (a) 28%.
MARKER’S COMMENT: Students who expanded before differentiating tended to score more highly.

`d/(dx) ((x^k)/(k^2)(klog_e(x)-1))`

`=d/(dx)((x^k)/k log_e(x)-(x^k)/(k^2))`

`= x^(k-1) log_e(x) + 1/k x^(k-1)-1/k x^(k-1)`

`= x^(k-1) log_e(x)`

`:. intx^(k-1) log_e(x)\ dx = (x^k)/(k^2)(klog_e(x)-1)`

b. `text(Pr)(x > 1/e)`

♦♦♦ Mean mark 16%.

`= −4 int_(1/e)^1 (xlog_e(x))\ dx,\ text(where)\ k = 2`

`= −4[(x^2)/4(2log_e(x)-1)]_(1/e)^1`

`= −4[1/4(0 -1)-1/(4e^2)(2log_e(e^(−1))-1)]`

`= −4[−1/4 + 1/(4e^2) + 1/(2e^2)]`

`= 1-3/(e^2)`

Probability, MET2 2009 VCAA 11 MC

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)):}`

The value of `a` such that `text(Pr) (X > a) = 0.2` is closest to

`0.26`
`0.30`
`0.32`
`0.35`
`0.40`

Show Answers Only

`D`

Show Worked Solution

`text(Solve)\ \ int_a^(1/2) f (x)\ dx`	`= 0.2,\ \ a in [0, 1/2]`
`:. a`	`~~ 0.35`

`=> D`

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):}`

Find correct to four decimal places
1. `text(Pr)(3 <= X <= 5)` (1 mark)
  
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2. `text(Pr)(3 <= Y <= 5)` (3 marks)
  
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Find the mean of `Y`, correct to three decimal places. (3 marks)

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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)

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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.

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

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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.

`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 2010 VCAA 11 MC

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

`f(x) = {(cos(2x), if (3 pi)/4 < x < (5 pi)/4), (qquad qquad quad 0,\ \ \ text(elsewhere)):}`

The value of `a` such that `text(Pr) (X < a) = 0.25` is closest to

`2.25`
`2.75`
`2.88`
`3.06`
`3.41`

Show Answers Only

`C`

Show Worked Solution

`text(Solve)\ \ int_((3 pi)/4)^a cos (2x)\ dx = 1/4\ \ text(for)\ \ a in ((3 pi)/4, (5 pi)/4)`

`:. a`	`= (11 pi)/12\ \ \ text([by CAS.])`
	`~~ 2.88`

`=> C`

Probability, MET2 2016 VCAA 18 MC

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

`qquad f(x) = {(1/4 cos (x/2), 3 pi <= x <= 5 pi), (0, text{elsewhere}):}`

The value of `a` such that `text(Pr) (X < a) = (sqrt 3 + 2)/4` is

`(19 pi)/6`
`(14 pi)/3`
`(10 pi)/3`
`(29 pi)/6`
`(17 pi)/3`

Show Answers Only

`B`

Show Worked Solution

`int_(3 pi)^a f(x)\ dx`	`= (sqrt 3 + 2)/4`
`[1/2 sin (x/2)]_(3pi)^a`	`= (sqrt 3 + 2)/4`
`1/2 sin (a/2) +1/2`	`= (sqrt 3 + 2)/4`
`sin (a/2)`	`=sqrt3/2`
`a/2`	`=pi/3, (2pi)/3, (4pi)/3, (5pi)/3, (7pi)/3, …`
`:. a`	`= (14 pi)/3\ \ text(for)\ a in (3 pi, 5 pi)`

`=> B`

Probability, MET1 2006 VCAA 6

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

`f(x) = {(x/12,\ \ 1 <= x <= 5), (\ 0,\ \ text(otherwise)):}`

Find `text(Pr) (X < 3)` (2 marks)

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If `text(Pr) (X >= a) = 5/8`, find the value of `a`. (2 marks)

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Show Answers Only

`1/3`
`sqrt 10`

Show Worked Solution

a. `text(Pr) (X < 3)`

`= int_1^3 1/12x\ dx`

`= 1/12 [1/2 x^2]_1^3`

`= 1/24 [3^2 – 1^2]`

`= 1/3`

b. `text(Pr) (X >= a) = 5/8`

♦ Mean mark 47%.

`int_a^5 x/12\ dx`	`= 5/8`
`1/24 [x^2]_a^5`	`= 5/8`
`(5^2 – a^2)`	`= 15`
`a^2`	`= 10`
`:. a`	`= sqrt 10,\ \ \ a in (1, 5)`

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.

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

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

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Show Answers Only

`6/125`
`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 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)
  
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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)
  
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Find the mean diameter of medium oranges, in centimetres. (1 mark)

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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.

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)

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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)
  
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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)
  
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Show Answers Only

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

e.	\(\Pr(\text{Grade A\|Fits})\)	\(=\Pr(6.54<D<6.86\|D<6.95)\)
		\(=\dfrac{\Pr(6.54<D<6.86)}{\Pr(D<6.95)}\)
	\(\text{Using CAS: }\)	\(=\Bigg[\dfrac{\text{normCdf}(6.54, 6.86, 6.7, 0.1)}{\text{normCdf}(-\infty, 6.95, 6.7, 0.1)}\Bigg]\)
		\(=\dfrac{0.89040…}{0.99977…}\)
		\(=0.895965…\approx 0.8960\)

\(0.84375-z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}\)	\(=0.7382\)	\((1)\)
\(0.84375+z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}\)	\(=0.9493\)	\((2)\)
\(\text{Equation}(2)-(1)\)
\(2z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}\)	\(=0.2111\)
\(z=\dfrac{0.10555}{\sqrt{\dfrac{0.84375(1-0.84375)}{32}}}\)	\(=1.64443352\)

\(\hat{p}-z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{32}}\)	\(=0.7382\)
\(\text{and}\)
\(\hat{p}+z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{32}}\)	\(=0.9493\)

\(\text{Area}\)	\(=1\)
\(\therefore\ a\)	\(=\dfrac{1}{b}\)
\(\to\ b\)	\(=\dfrac{E(W)}{E(V)}\)
	\(=\dfrac{2\pi^2+8}{3\pi^2+12}\)
	\(=\dfrac{2(\pi^2+4)}{3(\pi^2+4)}\)
	\(=\dfrac{2}{3}\)
\(\therefore\ a\)	\(=\dfrac{3}{2}\)