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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.

  1. Find  \(\Pr(D>6.8)\), correct to four decimal places.   (1 mark)
  2. Find the minimum diameter of a tennis ball that is larger than 90% of all tennis balls produced.
  3. Give your answer in centimetres, correct to two decimal places.   (1 mark)

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.

  1. Find the probability that a randomly selected tennis ball can fit through the opening at the top of the container.
  2. Give your answer correct to four decimal places.   (1 mark)
  3. 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.
  4. Give your answer correct to four decimal places.   (2 marks)

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.

  1. 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.
  2. Give your answer correct to four decimal places.   (2 marks)
  3. The manufacturer would like to improve processes to ensure that more than 99% of all tennis balls produced are classed as grade A.
  4. 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)
  5. 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.
  6. The confidence interval is (0.7382, 0.9493), correct to four decimal places.
  7. 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)

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
 

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

  1. Find the probability that the serving speed of a grade A ball exceeds 50 metres per second.
  2. Give your answer correct to four decimal places.   (1 mark)
  3. Find the exact mean serving speed for grade A balls, in metres per second.   (1 mark)

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 \(g(w)=af\Bigg(\dfrac{w}{b}\Bigg)\).

  1. 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)
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%.

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, 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

  1. `0.26`
  2. `0.30`
  3. `0.32`
  4. `0.35`
  5. `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 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

  1. `2.25`
  2. `2.75`
  3. `2.88`
  4. `3.06`
  5. `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

  1. `(19 pi)/6`
  2. `(14 pi)/3`
  3. `(10 pi)/3`
  4. `(29 pi)/6`
  5. `(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 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 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)`