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Statistics, SPEC1 2024 VCAA 6

The production of a brand of weed trimmer involves three stages, Stage 1, Stage 2 and Stage 3, which take \(W_1\) hours, \(W_2\) hours and \(W_3\) hours, respectively. Here \(W_1, W_2\) and \(W_3\) are independent random variables, which may be assumed to be normally distributed. Assume that Stage 2 starts immediately after Stage 1 ends and that Stage 3 starts immediately after Stage 2 ends.

The mean, standard deviation and cost at each stage are shown in the table below.

\begin{array}{|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Stage} & \textbf{Time (h) } & \textbf{Mean(h)} &\textbf{Standard } &  \textbf{Cost (\$/h)} \\
& & & \rule[-1ex]{0pt}{0pt} \textbf{deviation (h)}\\
\hline \rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt}& W_1 & 1.0 & 0.3 & 10 \\
\hline \rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt}& W_2 & 1.5 & 0.4 & 20 \\
\hline \rule{0pt}{2.5ex} 3 \rule[-1ex]{0pt}{0pt}& W_3 & 2.0 & 0.5 & 15 \\
\hline
\end{array}

  1. Find the mean and the variance of the total time to produce one weed trimmer.   (1 mark)

  2. Find the variance of the total cost to produce one weed trimmer.   (2 marks)
  3. If a single weed trimmer is produced, find the probability that the time spent at Stage 2 will be less than the time spent at Stage 1.
  4. Give your answer correct to two decimal places.
  5. Use  \(\operatorname{Pr}(-1<Z<1)=0.68\),  where \(Z\) is the standard normal variable with mean 0 and standard deviation 1.  (2 marks)

Show Answers Only

a.   \(\text{E}(W_1+W_2+W_3) =4.5\)

\(\text{Var}(W_1+W_2+W_3) = 0.5\)

b.   \(\text{Var}(10W_1+40W_2+15W_3) = 129.25 \)

c.   \(0.16\)

Show Worked Solution

a.   \(\text{Time to make 1 trimmer}\ = W_1+W_2+W3\)

\(\text{Using mean and std dev values from the table:}\)

\(\text{E}(W_1+W_2+W_3) = 1.0+1.5+2.0=4.5\)

\(\text{Var}(W_1+W_2+W_3) = 0.3^2+0.4^2+0.5^2 = 0.5\)
 

b.   \(\text{Cost to make 1 trimmer}\ = 10W_1+20W_2+15W3\)

\(\text{Var}(10W_1+40W_2+15W_3)\)

\(= 10^2\text{Var}(W_1)+20^2\text{Var}(W_2)+15^2\text{Var}(W_1) \)

\(= 100\text{Var}(W_1)+400\text{Var}(W_2)+225\text{Var}(W_1) \)

\(= 100\times 0.09+400 \times 0.16 +225 \times 0.25 \)

\(= 9+64+56.25 \)

\(= 129.25 \)
 

♦ Mean mark (b) 45%.

c.    \(\text{Let}\ \ X=W_1-W_2\)

\(\text{Since}\ \ W_1-W_2\ \ \text{is a linear combination of normally distributed variables}\)

\(W_1-W2\ \ \text{is also normally distributed}\)

\(\text{E}(W_1-W2)=1.0-1.5=-0.5\)

\(\text{Var}(W_1-W2)=0.3^2+0.4^2=0.25\)

\(\text{Pr}(X>0)\) \(=\text{Pr}\left( Z> \dfrac{0+0.5}{0.5}\right) \)  
  \(=\text{Pr}(Z>1)\)  
  \(=0.16\)  
♦ Mean mark (c) 50%.

Statistics, SPEC2 2022 VCAA 6

A company produces soft drinks in aluminium cans.

The company sources empty cans from an external supplier, who claims that the mass of aluminium in each can is normally distributed with a mean of 15 grams and a standard deviation of 0.25 grams.

A random sample of 64 empty cans was taken and the mean mass of the sample was found to be 14.94 grams.

Uncertain about the supplier's claim, the company will conduct a one-tailed test at the 5% level of significance. Assume that the standard deviation for the test is 0.25 grams.

  1. Write down suitable hypotheses \(H_0\) and \(H_1\) for this test.   (1 mark)

  2. Find the \(p\) value for the test, correct to three decimal places.   (1 mark)

  3. Does the mean mass of the random sample of 64 empty cans support the supplier's claim at the 5% level of significance for a one-tailed test? Justify your answer.   (1 mark)

  4. What is the smallest value of the mean mass of the sample of 64 empty cans for \(H_0\) not to be rejected? Give your answer correct to two decimal places.   (1 mark)

The equipment used to package the soft drink weighs each can after the can is filled. It is known from past experience that the masses of cans filled with the soft drink produced by the company are normally distributed with a mean of 406 grams and a standard deviation of 5 grams.

  1. What is the probability that the masses of two randomly selected cans of soft drink differ by no more than 3 grams? Give your answer correct to three decimal places.   (2 marks)

Show Answers Only

a.   \(H_0: \mu=15, \quad H_1: \mu<15\)

b.   \(p=0.027\)

c.    \(\text{Since \(\ p<0.05\), claim is not supported.}\)

d.   \(a=14.95\)

e.  \(\text{Pr}(-3<D<3)=0.329\)

Show Worked Solution

a.    \(H_0: \mu=15, \quad H_1: \mu<15\)
 

b.    \(\mu=15, \ \ \sigma=0.25\)

\(\bar{x}=14.94, \ \sigma_{\bar{x}}=\dfrac{0.25}{\sqrt{64}}=0.03125\)

\(\text{By CAS:}\)

\(p=\text{Pr}\left(\bar{X}<14.94 \mid \mu=15\right)=0.027\ \text {(3 d.p.)}\)
 

c.    \(\text{Since \(\ p<0.05\), claim is not supported.}\)

\(\text{Evidence is against \(H_0\)  at the \(5 \%\) level.}\)
 

d.    \(\text{Pr}\left(\bar{X}<a \mid \mu=15\right)>0.05\)

\(\text{Pr}\left(Z<\dfrac{a-15}{0.03125}\right)>0.05\)

\(\text{By CAS:}\ \ a=14.95\ \text{(2 d.p.)}\)
 

e.    \(\text{Let}\ \ M=\ \text{mass of one can}\)

\(M \sim N\left(406,5^2\right)\)

\(E\left(M_1\right)=E\left(M_2\right)=\mu=406\)

\(\text {Let}\ \ D=M_1-M_2\)

\(E(D)=406-406=0\)

\(\text{Var}(D)=1^2 \times \text{Var}\left(M_1\right)+(-1)^2 \times  \text{Var}\left(M_2\right)=50\)

\(\sigma_D=\sqrt{50}\)

\(D \sim N\left(0,(\sqrt{50})^2\right)\)

\(\text{By CAS: Pr\((-3<D<3)=0.329\) (3 d.p.) }\)

♦ Mean mark (e) 46%.

Statistics, SPEC2 2022 VCAA 18 MC

The time taken, \(T\) minutes, for a student to travel to school is normally distributed with a mean of 30 minutes and a standard deviation of 2.5 minutes.

Assuming that individual travel times are independent of each other, the probability, correct to four decimal places, that two consecutive travel times differ by more than 6 minutes is

  1. 0.0448
  2. 0.0897
  3. 0.1151
  4. 0.2301
  5. 0.9103
Show Answers Only

\(B\)

Show Worked Solution

\(T_d=T_1-T_2\)

\(E(T_d)=E(T_1)-E(T_2)=30-30-0\)

\(\text{Var}(T_1)=\ \text{Var}(T_2) = 2.5^2=6.25\)

\(\text{Var}(T_d)=1^2 \times \text{Var}(T_1)+(-1)^2 \times \text{Var}(T_2) = 12.5 \)

\(\sigma(T_d) = \sqrt{12.5}\)

\(1-\text{Pr}(-6 \lt T_d \lt 6)=0.0897 \ \ \text{(by CAS)}\)

\(\Rightarrow B\)

♦ Mean mark 42%.

Statistics, SPEC1 2022 VCAA 3

The time taken by a coffee machine to dispense a cup of coffee varies normally with a mean of 10 seconds and a standard deviation of 1.5 seconds.

Find the probability that more than 34 seconds is needed to dispense a total of four cups of coffee. Give your answer correct to two decimal places.   (2 marks) 

Show Answers Only

`0.98`

Show Worked Solution

`\text{1 cup:}\ \ mu=10,\ \ sigma=1.5, \ \text{Var(X)}\ =1.5^2 = 2.25`

`\text{4 cups:}\ \ mu=4 xx 10= 40,\ \text{Var(X)}\ = 4 xx 2.25=9, \ sigma = \sqrt{9}=3`

`Z\ ~\ N(0,1)`

`text{Pr}(Z>(34-40)/(3))` `= text{Pr}(Z> -2)`  
  `~~ 0.975`  
  `=0.98\ \text{(2 d.p.)}`  

♦♦ Mean mark 36%.

Statistics, SPEC1 2023 VCAA 6

Josie travels from home to work in the city. She drives a car to a train station, waits, and then rides on a train to the city. The time, \(X_c \) minutes, taken to drive to the station is normally distributed with a mean of 20 minutes \( (\mu_c=20) \) and standard deviation of 6 minutes \((\sigma_c=6) \). The waiting time, \( X_w \) minutes, for a train is normally distributed with a mean of 8 minutes \( (\mu_w=8) \) and standard deviation of \( \sqrt{3} \) minutes \( (\sigma_w=\sqrt{3}) \). The time, \( X_t \) minutes, taken to ride on a train to the city is also normally distributed with a mean of 12 minutes \( (\mu_t=12) \) and standard deviation of 5 minutes \( (\sigma_t=5) \). The three times are independent of each other.

  1. Find the mean and standard deviation of the total time, in minutes, it takes for Josie to travel from home to the city.   (2 marks)
  1. Josie's waiting time for a train on each work day is independent of her waiting time for a train on any other work day. The probability that, for 12 randomly chosen work days, Josie's average waiting time is between 7 minutes 45 seconds and 8 minutes 30 seconds is equivalent to \( \text{Pr}(a<Z<b)\), where \(Z \sim \text{N}(0,1)\) and \(a\) and \(b\) are real numbers.
  2. Find the values of \(a\) and \(b\).   (2 marks)
Show Answers Only

a.    \(E[X_c+X_w+X_t]=40\)

\(\sigma[X_c+X_w+X_t]=8\)

b.    \(a=-0.5, \ b=1\)

Show Worked Solution

a.    \(X_c ∼ N(20, 6^2), X_w ∼ N(8, (\sqrt{3})^2), X_t ∼ N(12, 5^2) \)

\(E[X_c+X_w+X_t]\) \(=E[X_c]+E[X_w]+E[X_t]\)  
  \(=20+8+12\)  
  \(=40\)  

 

\(\text{Var}[X_c+X_w+X_t]\) \(=\text{Var}[X_c]+\text{Var}[X_w]+\text{Var}[X_t]\)  
  \(=36+3+25\)  
  \(=64\)  
\(\sigma[X_c+X_w+X_t]\) \(=8\)  

 
b.    \(E(\bar X)=\mu_w=8,\ \ \sigma(\bar X)=\dfrac{\sigma_w}{\sqrt{12}}=\dfrac{\sqrt3}{\sqrt{12}}=\dfrac{1}{2} \)

\(\text{Pr}(7.75<\bar X<8.5)\) \(=\text{Pr} \Bigg{(} \dfrac{7.75-8}{\frac{1}{2}}<Z<\dfrac{8.5-8}{\frac{1}{2}}\Bigg{)} \)  
  \(=\text{Pr} \Bigg{(} \dfrac{-0.25}{\frac{1}{2}}<Z<\dfrac{0.5}{\frac{1}{2}}\Bigg{)} \)  
  \(=\text{Pr}(-0.5<Z<1)\)  

 
\(\therefore a=-0.5, \ b=1\)

Statistics, SPEC2 2021 VCAA 6

The maximum load of a lift in a chocolate company's office building is 1000 kg. The masses of the employees who use the lift are normally distributed with a mean of 75 kg and a standard deviation of 8 kg. On a particular morning there are `n` employees about to use the lift.

  1. What is the maximum possible value of `n` for there to be less than a 1% chance of the lift exceeding the maximum load?   (2 marks)

Clare, who is one of the employees, likes to have a hot drink after she exits the lift. The time taken for the drink machine to dispense a hot drink is normally distributed with a mean of 2 minutes and a standard deviation of 0.5 minutes. Times taken to dispense successive hot drinks are independent.

  1. Clare has a meeting at 9.00 am and at 8.52 am she is fourth in the queue for a hot drink. Assume that the waiting time between hot drinks dispensed is negligible and that it takes Clare 0.5 minutes to get from the drink machine to the meeting room.
  2. What is the probability, correct to four decimal places, that Clare will get to her meeting on time?   (2 marks)

Clare is a statistician for the chocolate company. The number of chocolate bars sold daily is normally distributed with a mean of 60 000 and a standard deviation of 5000. To increase sales, the company decides to run an advertising campaign. After the campaign, the mean daily sales from 14 randomly selected days was found to be 63 500.

Clare has been asked to investigate whether the advertising campaign was effective, so she decides to perform a one-sided statistical test at the 1% level of significance.

  1.   i. Write down suitable null and alternative hypotheses for this test.   (1 mark)

  2.  ii. Determine the `p` value, correct for decimal places, for this test.   (1 mark)

  3. iii. Giving a reason, state whether there is any evidence for the success of the advertising campaign.   (1 mark)

  4. Find the range of values for the mean daily sales of another 14 randomly selected days that would lead to the null hypothesis being rejected when tested at the 1% level of significance. Give your answer correct to the nearest integer.   (1 mark)

  5. The advertising campaign has been successful to the extent that the mean daily sales is now 63 000.
  6. A statistical test is applied at the 5% level of significance.
  7. Find the probability that the null hypothesis would be incorrectly accepted, based on the sales of another 14 randomly selected days and assuming a standard deviation of 5000. Give your answer correct to three decimal places.   (2 marks)

Show Answers Only
  1. `12`
  2. `0.3085`
  3.   i. `H_0: \ mu = 60 \ 000, \ H_1: \ mu > 60 \ 000`
  4.  ii.  `0.0044`
  5. iii.  `text{Successful as} \ p text{-value is below 0.01.}`
  6. `n ≥ 63\ 109`
  7. `0.274`
Show Worked Solution

a.    `text{Method 1}`

♦♦ Mean mark 22%.

`text{Let} \ \ M_i\ ~\ N (75, 8^2) \ \text{for} \ \ i = 1, 2, 3, … , n`

`W_n = W_1 + W_2 + … + W_n`

`E (W_n) = E(M_1 + … + M_n) = 75n`

`text(s.d.) (W_n) = text{s.d.} (M_1 + … M_n) = 8 sqrtn`

`W_n\ ~\ N (75n, 8^2 n)`

`text{Using} \ Z\ ~\ N (0, 1)`

`text(Pr) (W_n > 1000) = 0.01`

`text(Pr) (Z > {1000-75 n}/{8 sqrtn}) = 0.01`
 
`n = 12.5`

`:. \ text{Largest} \ n = 12`
 

`text{Method 2}`

`text{By trial and error}`
 
`text(Pr) (M_1 + … + M_11 > 1000) ≈ 0`

`text(Pr) (M_1 + … + M_12 > 1000) ≈ 0.0002`

`text(Pr) (M_1 + … + M_13 > 1000) ≈ 0.193`

`:. \ text{Largest} \ n = 12`

 

b.   `T_i\ ~\ N (2, 0.5^2) \ \ text{for} \ \ i = 1, 2, …`

Mean mark part (b) 51%.

`text{Wait time} \ (T) = T_1 + T_2 + T_3 + T_4`

`E(T) = 4 xx 2 = 8`

`text{s.d.}(T) = text{s.d.}(T_1 + T_2 + T_3 + T_4) = sqrt4 xx 0.5 = 1`
 
`T\ ~\ N (8, 1)`

`text(Pr) (T < 7.5) = 0.3085`

`text{By CAS: normCdf} (0, 7.5, 8, 1)`

 

c.i.  `H_0: \ mu = 60 \ 000`

`H_1: \ mu > 60 \ 000`
 

c.ii.  `text(Pr) (barX > 63\ 500 | mu = 60 \ 000) = 0.004407`

  `:. \ p \ text{value} = 0.0044`

 `text{By CAS: norm Cdf} (63\ 500, oo, 60\ 000, 5000/sqrt14)`
 

c.iii.  `text{S} text{ince the} \ p text{-value is below 0.01, there is strong evidence}`

   `text{the advertising was effective (against the null hypothesis).}`

 

d.   `text(Pr) (barX > n | mu = 60 \ 000) < 0.01`

♦♦♦ Mean mark part (d) 16%.

`n ≥ 63\ 109`

`text{By CAS: inv Norm} (0.99, 60 \ 000, 5000/sqrt14)`

 

e.    `text{Similar to part (d):}`

♦♦♦ Mean mark part (e) 10%.

`text(Pr) (barX > n | mu = 60 \  000) < 0.05`

`n ≤ 62198`

`text{By CAS: invNorm} \ (0.95, 60 \ 000, 5000/sqrt14)`
 
`text{If} \ \ mu = 63\ 000, text{find probability null hypothesis incorrectly accepted:}`

`text(Pr) (barX < 62\ 198 | mu = 63\ 000) = 0.274`

`text{By CAS: normCdf} (0, 62\ 198, 63\ 000, 5000/sqrt14)`


Statistics, SPEC2 2021 VCAA 20 MC

An office has two coffee machines that operate independently of each other. The time taken for each machine to produce a cup of coffee is normally distributed with a mean of 30 seconds and a standard deviation of 5 seconds. On a particular morning, a cup is produced from each machine.

The probability that the time taken by each coffee machine to produce one cup of coffee will differ by less than 3 seconds is closest to

  1. 0.164
  2. 0.236
  3. 0.329
  4. 0.451
  5. 0.671
Show Answers Only

`C`

Show Worked Solution

`T_n~N(30, 5^2), n = 1, 2\ …`

♦ Mean mark 43%.

`E(T_2-T_1) = 0`

`text(Var)(T_2-T_1)` `= text(Var)(T_2) + text(Var)(T_1)`
  `= 2 xx 5^2`
  `= 50`

 
`:. T_2-T_1~N(0, (sqrt50)^2)`

`text(Pr)(-3 < T_2-T_1 < 3) = 0.329`

`text{By CAS: normCdf}(–3,3,0,sqrt50)`

`=>\ C`

Statistics, SPEC2-NHT 2019 VCAA 20 MC

Nitrogen oxide emissions for a certain type of car are known to be normally distributed with a mean of 0.875 g/km and a standard deviation of 0.188 g/km.

For two randomly selected cars, the probability that their nitrogen oxide emissions differ by more than 0.5 g/km is closest to

  1. 0.030
  2. 0.060
  3. 0.960
  4. 0.970
  5. 0.977
Show Answers Only

`B`

Show Worked Solution

`C\ ~\ N(0.875, 0.188^2)`

`text(Let)\ \ C_1 = text(emissions of car 1,)\ C_2 = text(emissions of car 2)`

`E(C_1-C_2) = E(C_1)-E(C_2) = 0`

`text(Var)(C_1-C_2) = text(Var)(C_1) + text(Var)(C_2) = 2 xx 0.188^2`

`sigma(C_1-C_2) = sqrt2 xx 0.188`

`text(Pr)(|C_1-C_2| > 0.5) = 0.060\ \ \ text{(by CAS)}`

`=>\ B`

Statistics, SPEC2 2019 VCAA 19 MC

`X` and `Y` are independent random variables where each has a mean of 4 and a variance of 9.

If the random variable  `Z = aX + bY`  has a mean of 8 and a variance of 90, possible values of `a` and `b` are

  1. `a = 1, \ b = 1`
  2. `a = 4, \ b = −2`
  3. `a = 3, \ b = −1`
  4. `a = 1, \ b = 3`
  5. `a = −2, \ b = 4`
Show Answers Only

`C`

Show Worked Solution
`E(aX + bY)` `= aE(X) + bE(Y)`
`8` `= 4a + 4b\ \ …\ (1)`
`text(Var)(aX + bY)` `= a^2text(Var)(X) + b^2text(Var)(Y)`
`9a^2 + 9b^2` `= 90\ \ …\ (2)`

 
`text{Solve simultaneous equations (by CAS):}`

`a = 3, b = −1\ \ text(or)\ \ a = −1, b = 3`

`=>C`

Statistics, SPEC2 2016 VCAA 18 MC

Oranges grown on a citrus farm have a mean mass of 204 grams with a standard deviation of 9 grams

Lemons grown on the same farm have a mean mass of 76 grams with a standard deviation of 3 grams.

The masses of the lemons are independent of the masses of the oranges.

The mean mass and standard deviation, in grams, respectively of a set of three of these oranges and two of these lemons are

  1. `764, 3 sqrt 29`
  2. `636, 12`
  3. `764, sqrt 33`
  4. `636, 3 sqrt 10`
  5. `636, 33`
Show Answers Only

`A`

Show Worked Solution

`O~N (204, 9^2)`

`L~N (76, 3^2)`

`T = O+O+O + L + L`

`T~N (3 xx 204 + 2 xx 76, 9^2 + 9^2 + 9^2 + 3^2 + 3^2)`

`T~N (764, 261)`

`:. mu_T= 764, quad sigma_T = sqrt 261 = 3 sqrt 29`

`=>  A`

Statistics, SPEC2-NHT 2018 VCAA 6

A coffee machine dispenses coffee concentrate and hot water into a 200 mL cup to produce a long black coffee. The volume of coffee concentrate dispensed varies normally with a mean of 40 mL and a standard deviation of 1.6 mL.

Independent of the volume of coffee concentrate, the volume of water dispensed varies normally with a mean of 150 mL and a standard deviation of 6.3 mL.

  1. State the mean and the standard deviation, in millilitres, of the total volume of liquid dispensed to make a long black coffee.   (2 marks)

  2. Find the probability that a long black coffee dispensed by the machine overflows a 200 mL cup. Give your answer correct to three decimal places.   (1 mark)

  3. Suppose that the standard deviation of the volume of water dispensed by the machine can be adjusted, but that the mean volume of water dispensed and the standard deviation of the volume of coffee concentrate dispensed cannot be adjusted.
  4. Find the standard deviation of the volume of water dispensed that is needed for there to be only a 1% chance of a long black coffee overflowing a 200 mL cup. Give your answer in millilitres, correct to two decimal places.   (2 marks)

Show Answers Only
  1. `mu_V = 40 + 150 = 190\ text(mL)`

     

    `sigma_V = sqrt (1.6^2 + 6.3^2) = 6.5\ text(mL)`

  2. `0.062`
  3. `3.99\ text(mL)`
Show Worked Solution

a.   `C~N (40, 1.6^2)`

`W~N (150, 6.3^2)`

`V = C + W`

`mu_V = 40 + 150 = 190\ text(mL)`

`{:sigma^2:}_V` `= 1.6^2 + 6.3^2`  
`:.sigma_V ` `= sqrt (1.6^2 + 6.3^2)`  
  `= 6.5\ text(mL)`  

 

b.   `V~N (190, 6.5^2)`

`text(Pr)(V > 200) ~~ 0.062\ \ text{(by CAS)}`

 

c.  `C~N (40, 1.6^2)`

`W_2~N (150, sigma_w^2)`

`V_2 = C + W_2`

`V_2~N (190, 1.6^2 + sigma_w^2)`

`Z~N (0, 1)`
 

`text(Pr)(V_2 > 200) = 0.01`

`text(Pr)(Z > a) = 0.01\ \ =>\ \ a=2.326…`
  

`text(Find)\ \ sigma_w:`

`2.326…` `= (200-190)/sqrt(1.6^2 + sigma_w^2)`
`:. sigma_w` `~~ 3.99\ text(mL)\ \ \ text{(by CAS)}`

Statistics, SPEC2-NHT 2017 VCAA 18 MC

`X` is a random variable with a mean of 5 and a standard deviation of 4, and `Y` is a random variable with a mean of 3 and a standard deviation of 2.

If `X` and `Y` are independent random variables and  `Z = X-2Y`, then `Z` will have mean `mu` and standard deviation `sigma` given by

  1. `mu = -1, sigma = 0`
  2. `mu = -1, sigma = 4 sqrt 2`
  3. `mu = 2, sigma = 8`
  4. `mu = 2, sigma = 4 sqrt 2`
  5. `mu = -1, sigma = 2 sqrt 6`
Show Answers Only

`B`

Show Worked Solution

`mu_z= mu_x-2 mu_y= 5-2(3)= -1`

`sigma_z^2= sigma_x^2 + (-2)^2 sigma_y^2= 4^2 + 4(2)^2= 32`

`sigma_z= sqrt 32= 4 sqrt 2` 

`=>   B`

Statistics, SPEC1-NHT 2017 VCAA 9

The random variables `X` and `Y` are independent with  `mu_X = 4,\ text(Var)(X) = 36`  and  `mu_Y = 3,\ text(Var)(Y) = 25`.

  1. The random variable `Z` is such that  `Z = 2X + 3Y`.
  2.  i. Find `E(Z)`.   (1 mark)

  3. ii. Find the standard deviation of `Z`.   (1 mark)

  1. Researchers have reason to believe that the mean of `X` has decreased. They collect a random sample of 64 observations of `X` and find that the sample mean is  `bar X = 3.8`
  2.  i. State the null hypothesis and the alternative hypothesis that should be used to test that the mean has decreased.   (1 mark)

  3. ii. Calculate the mean and standard deviation for a distribution of sample means, `bar X`, for samples of 64 observations.   (1 mark)

Show Answers Only
  1.  i. `17`
  2. ii. `3 sqrt 41`
  3.  i. `H_0: mu = 4,\ \ H_1: mu < 4`
  4. ii. `mu_bar X = 4,\ \ sigma_bar X = 3/4`
Show Worked Solution
a.i.    `E(Z)` `= 2E(X) + 3E(Y)`
    `= 2(4) + 3(3)`
    `= 8 + 9`
    `= 17`

 

a.ii.    `text(Var)(Z)` `= 2^2 text(var)(X) + 3^2 text(var)(Y)`
    `= 4(36) + 9(25)`
    `= (4xx9xx4) + 9(25)`
    `= 9(16 + 25)`

 

`:. sigma_Z` `= sqrt(9 (41))`
  `= 3 sqrt 41`

 

b.i.    `H_0: \ mu = 4`
  `H_1: \ mu < 4`

 

b.ii.    `bar X~N(mu, (sigma^2)/n)`
  `bar X~N(4, 36/64)`

 
`E(barX) = 4`

`sigma_bar X` `= sqrt(36/64)`
  `= 3/4`

Statistics, SPEC2 2017 VCAA 18 MC

`U` and `V` are independent normally distributed random variables, where `U` has a mean of 5 and a variance of 1, and `V` has a mean of 8 and a variance of 1. The random variable `W` is defined by  `W = 4U-3V`.

In terms of the standard normal variable  `Z, Pr(W > 5)`  is equivalent to

  1. `text(Pr)(Z > (9sqrt7)/7)`
  2. `text(Pr)(Z < 1.8)`
  3. `text(Pr)(Z < (9sqrt7)/7)`
  4. `text(Pr)(Z > 0.2)`
  5. `text(Pr)(Z > 1.8)`
Show Answers Only

`E`

Show Worked Solution

`U~N(5, 1), V~N(8, 1), Z~N (0, 1)`

♦ Mean mark 42%.

`W = 4U-3V`

`:. W~N(4 xx 5-3 xx 8, 4^2 xx 1 + (-3)^2 xx 1)`

  `~N(−4, 25)`

`text(Pr)(W > 5)` `= text(Pr)(Z > (5+4)/sqrt25)`
  `= text(Pr)(Z > 9/5)`
  `= text(Pr)(Z > 1.8)`

 
`=> E`

Statistics, SPEC2-NHT 2018 VCAA 20 MC

A farm grows oranges and lemons. The oranges have a mean mass of 200 grams with a standard deviation of 5 grams and the lemons have a mean mass of 70 grams with a standard deviation of 3 grams.

Assuming masses for each type of fruit are normally distributed, what is the probability, correct to four decimal places, that a randomly selected orange will have at least three times the mass of a randomly selected lemon?

  1. 0.0062
  2. 0.0828
  3. 0.1657
  4. 0.8343
  5. 0.9172
Show Answers Only

`C`

Show Worked Solution

`O~N (200, 5^2) `

`L~N (70, 3^2)`

`text(Pr) text{(orange weighs at least 3 × lemon)}`

`=text(Pr) (O > 3L)`

`= text(Pr)(O-3L > 0)`
 

`text(Let)\ \ X = O-3L,`

`X~N (200-3 xx 70, 5^2 + (-3)^2 xx 3^2)`

`X~N (-10, 106)`

`text(Pr) (X > 0) ~~ 0.1657`

 
`=>  C`

Statistics, SPEC2 2018 VCAA 20 MC

The scores on the Mathematics and Statistics tests, expressed as percentages, in a particular year were both normally distributed. The mean and the standard deviation of the Mathematics test scores were 71 and 10 respectively, while the mean and the standard deviation of the statistics test scores were 75 and 7 respectively.

Assuming the sets of tests scores were independent of each other, the probability, correct to four decimal places, that a randomly chosen Mathematics score is higher than a randomly chosen Statistics score is

  1. 0.2877
  2. 0.3716
  3. 0.4070
  4. 0.7123
  5. 0.9088
Show Answers Only

`B`

Show Worked Solution

`M~N (71 , 10^2)`

`S~N (75, 7^2)`

`text(Pr)(M > S) = text(Pr)(M-S > 0)`

`M-S = X~N(71-75, 1^2 xx 10^2 + (-1)^2 xx 7^2)`

`X~N(-4, 149)`
 

`text(Pr)(M > S) = text(Pr)(X > 0) ~~ 0.3716`

`=>  B`

Statistics, SPEC1 2018 VCAA 4

`X` and `Y` are independent random variables. The mean and the variance of `X` are both 2, while the mean and the variance of `Y` are 2 and 4 respectively.

Given that `a` and `b` are integers, find the values of `a` and `b` if the mean and the variance of  `aX + bY`  are 10 and 44 respectively.  (4 marks)

Show Answers Only

`a = 2, qquad b = 3`

Show Worked Solution

`E(aX + bY) = aE(X) + bE(Y) = 10`

`10` `= 2a + 2b`
`5` `= a + b`
`b` `= 5-a \ \ …\ (1)`

 
`text(Var) (aX + bY) = a^2 text(Var) (X) + b^2 text(Var) (Y) = 44`

`44` `= 2a^2 + 4b^2`
`22` `= a^2 + 2b^2 \ \ …\ (2)`

 
`text{Substitute (1) into (2)}`

`22 = a^2 + 2 (5-a)^2`

`22 = a^2 + 2(25-10 a + a^2)`

`22 = 3a^2-20a + 50`

`0 = 3a^2-20a + 28`

`0 = (3a-14)(a-2)`

`a = 14/3 or a = 2`

`:. a = 2, \ b = 3,\ \ \ (a ∈ ZZ)`