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Probability, MET1 2023 VCAA SM-Bank 7

The duration of telemarketing calls to mobile phone users is a continuous random variable \(T\) minutes, with probability density function

\(f(t)= \begin{cases} \dfrac{2}{5} e^{-\frac{2}{5} t} & t \geq 0 \\ \ 0 & \text {elsewhere }\end{cases}\)

Find the value of \(k\) such that 90% of telemarketing calls last less than \(k\) minutes. Express your answer in the form  \(\dfrac{a}{b} \,\log _e(c)\),  where \(a, b\) and \(c\) are positive integers.   (3 marks)

Show Answers Only

\(k=\dfrac{5}{2}\log_{e}{10}\)

Show Worked Solution

\(\text{Pr}(T<k)=\displaystyle\int_{0}^{k}\dfrac{2}{5} e^{-\frac{2}{5} t}\,dt=0.9\)

\(\dfrac{2}{5}\Bigg[-\dfrac{5}{2}e^{-\frac{2t}{5}}\displaystyle\Bigg]_{0}^{k}\) \(=0.9\)
\(-e^{-\frac{2k}{5}}+1\) \(=0.9\)
\(e^{-\frac{2k}{5}}\) \(=0.1=\dfrac{1}{10}\)
\(e^{\frac{2k}{5}}\) \(=10\)
\(\dfrac{2k}{5}\) \(=\log_{3}{10}\)
\(\therefore\ k\) \(=\dfrac{5}{2}\log_{3}{10}\)

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`.

  1. 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)

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

Show Answers Only

  1. `text(See Worked Solutions)`
  2.  `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 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 2014 VCAA 8

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

`f(x) = {{:(1/5e^(−x/5),x >= 0),(0, x < 0):}`
 

The median of `X` is  `m`.

  1. Determine the value of  `m`.  (2 marks)
  2. The value of `m` is a number greater than 1.

     

    Find `text(Pr)(X < 1 | X <= m)`.  (2 marks)

Show Answers Only
  1. `−5log_e(1/2)\ \ text(or)\ \ 5log_e(2)\ \ text(or)\ \ log_e 32`
  2. `2(1 – e^(−1/5))`
Show Worked Solution
a.    `1/5 int_0^m e^(−x/5)dx` `= 1/2`
  `1/5 xx (−5)[e^(−x/5)]_0^m` `= 1/2`
  `[-e^(- x/5)]_0^m` `= 1/2`
  `-e^(−m/5) + 1` `= 1/2`
  `e^(−m/5)` `= 1/2`
  `- m/5` `= log_e(1/2)`

 

`:. m = −5log_e(1/2)\ \ \ (text(or)\ \ 5log_e(2),\ text(or)\ \ log_e 32)`

 

b.   `text(Using Conditional Probability:)`

♦ Part (b) mean mark 38%.
MARKER’S COMMENT: A common error was assuming `m` obtained in part (a) was equivalent to `text(Pr)(X<=m)`.
`text(Pr)(X < 1 | X <= m)` `= (text(Pr)(X < 1))/(text(Pr)(X <= m))`
  `= (1/5 int_0^1 e^(−x/5)dx)/(1/2)`
  `= (1/5(−5)[e^(−x/5)]_0^1)/(1/2)`
  `= −2[(e^(−1/5)) – e^0]`
  `= 2(1 – e^(−1/5))`

Probability, MET2 2015 VCAA 13 MC

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

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

The value of `a` is

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

`E`

Show Worked Solution

`text(Total area) = 1`

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

`=>   E`