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Statistics, SPEC2 2024 VCAA 20 MC

The masses of avocados in a crop may be assumed to be normally distributed, with a mean of 200 grams and a standard deviation of 7.5 grams.

After an avocado of mass \(M\) grams is peeled and the stone is removed, the mass of edible flesh \(F\) grams is given by \(F=0.70 M\). Four avocados are randomly selected from the crop.

What is the probability, correct to four decimal places, that a total of more than 570 grams of edible flesh is obtained?

  1. \(0.0868\)
  2. \(0.1705\)
  3. \(0.2128\)
  4. \(0.3170\)
Show Answers Only

\(B\)

Show Worked Solution

\(M \sim N\left(200,7.5^2\right)\)

\(F=0.7 M\)

\(F \sim N \left(200 \times 0.7, 7.5^2 \times 0.7\right) \sim N\left(140, 5.25^2\right)\)

♦ Mean mark 48%.

\(\text{Mass of flesh in 4 avocados}\)

\(F_1+F_2+F_3+F_4 \sim N\left(4 \times 140,4 \times 5.25^2\right) \sim N\left(560,10.5^2\right)\)

\(\operatorname{Pr}\left(F_1+F_2+F_3, F_4>570\right)=0.17045 \ \ \text{(by CAS)}\)

\(\Rightarrow B\)

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, SPEC2 2021 VCAA 19 MC

The mean unscaled score for a certain assessment task is 25 and the variance is 36. The scores scaled so that the mean score is 30 and the variance is 49. Let `S` be the scaled scores, to the nearest integer, and let `X` be the unscaled scores.

If the scaling function takes the form  `S = mX+n`, where  `m ∈ R^+` and `n ∈ R`, then a score of 32 would be scaled to

  1. 22
  2. 34
  3. 36
  4. 38
  5. 40
Show Answers Only

`D`

Show Worked Solution
`E(S)` `= E(mX + n)`
  `= mE(X) + n`
  `= 25m + n`

 

♦ Mean mark 51%.
`text(Var)(S)` `= text(Var)(mX + n)`
  `= m^2text(Var)(X)`
  `= 36m^2`

 
`text{Solve (by CAS)}:`

`25m + n` `= 30\ …\ (1)`
`36m^2` `= 49\ …\ (2)`

 
`m = 7/6, n = 5/6`
 

`:.\ text(Adjusted score of 32)`

`= 7/6 xx 32 + 5/6`

`~~ 38`

`=>\ D`

Statistics, SPEC1 2019 VCAA 3

A machine produces chocolate in the form of a continuous cylinder of radius 0.5 cm. Smaller cylindrical pieces are cut parallel to its end, as shown in the diagram below.

The lengths of the pieces vary with a mean of 3 cm and a standard deviation of 0.1 cm.
 


 

  1. Find the expected volume of a piece of chocolate in cm³.   (1 mark)

  2. Find the variance of the volume of a piece of chocolate in cm6.   (1 mark)

  3. Find the expected surface area of a piece of chocolate in cm².   (1 mark)

Show Answers Only
  1. `0.75pi\ text(cm³)`
  2. `0.000625pi^2\ text(cm)^6`
  3. `3.5pi\ text(cm²)`
Show Worked Solution
a.    `E(V)` `= pir^2 xx E(h)`
    `= pi xx 0.5^2 xx 3`
    `= 0.75pi\ text(cm³)`

 

b.    `text(Var)(V)` `= text(Var)(pir^2h)`
    `= pi^2 xx 0.5^4 xx text(Var)(h)`
    `= 0.0625pi^2 xx 0.1^2`
    `= 0.000625pi^2\ text(cm)^6`

 

c.    `E text{(Surface Area)}` `= 2pir^2 + 2pir xx E(h)`
    `= 2pir(r +E(h))`
    `= pi(0.5 + 3)`
    `= 3.5pi\ text(cm²)`

Statistics, SPEC2 2017 VCAA 6

A dairy factory produces milk in bottles with a nominal volume of 2 L per bottle. To ensure most bottles contain at least the nominal volume, the machine that fills the bottles dispenses volumes that are normally distributed with a mean of 2005 mL and a standard deviation of 6 mL.

  1. Find the percentage of bottles that contain at least the nominal volume of milk, correct to one decimal place.   (1 marks)

Bottles of milk are packed in crates of 10 bottles, where the nominal total volume per crate is 20 L.

  1. Show that the total volume of milk contained in each crate varies with a mean of 20 050 mL and a standard deviation of  `6sqrt10`  mL.   (2 marks)

  2. Find the percentage, correct to one decimal place, of crates that contain at least the nominal volume of 20 L.   (1 mark)

  3. Regulations require at least 99.9% of crates to contain at least the nominal volume of 20 L.
  4. Assuming the mean volume dispensed by the machine remains 2005 mL, find the maximum allowable standard deviation of the bottle-filling machine needed to achieve this outcome. Give your answer in millilitres, correct to one decimal place.   (3 marks)

  5. A nearby dairy factory claims the milk dispensed into its 2 L bottles varies normally with a mean of 2005 mL and a standard deviation of 2 mL.
  6. When authorities visit the nearby dairy factory and check a random sample of 10 bottles of milk, they find the mean volume to be 2004 mL.
  7. Assuming that the standard deviation of 2 mL is correct, carry out a one-sided statistical test and determine, stating a reason, whether the nearby dairy’s claim should be accepted at the 5% level of significance.   (2 marks)

Show Answers Only
  1. `79.8text(%)`
  2. `text(See Worked Solutions)`
  3. `99.6text(%)`
  4. `text(max) (σ_x) ~~ 5.1`
  5. `text(The claim should be accepted at a 5% significance level.)`
Show Worked Solution

a.   `V_B~N(2005, 6^2)`

`text(Pr)(V_B > 2000)` `~~ 0.797672\ \ \ text{(by CAS)}`
  `~~ 79.8text(%)`

 

b.   `text(Let)\ \ V_C=\ text(Volume of a crate)`

`mu_(V_C)` `= 10 xx mu_(V_B)=20\ 050`  
     
  `σ_(V_C)^2` `= σ_(V_B)^2 xx 10`
    `= 360`
  `:. σ_(V_C)` `= 6sqrt(10)`

 

c.  `V_C~N(20\ 050, (6sqrt10)^2)`

`text(Pr)(V_C > 20\ 000)` `~~ 0.995796`
  `~~ 99.6text(%)`

 

d.   `text(Let)\ \ V_A =\ text(New distribution)`

`V_A~N(20\ 050, (σ_x  sqrt10)^2)`

`text(Pr)(V_A > 20\ 000)` `>= 0.999`  
`text(Pr)(Z=a)` `=0.999`  
`:.a` `=-3.0902`  

 

`(20\ 000-20\ 050)/(σ_x sqrt10)` `=-3.0902`
`σ_x` `=5.116…`

 
`:.\ text(max) (σ_x) ~~ 5.1`

 

e.   `H_0: mu=2005`

`H_1: mu<2005`

`D~N(2005, 2^2)\ \ =>\ \ barD~N(2005, (2^2)/10)`

`p` `= text(Pr)(barD < 2004)`
  `~~ 0.056923`

  

`:.\ text(S)text(ince)\ \ p>0.05,\ text(the claim should be accepted)`

`text(at a 5% significance level.)`

Statistics, SPEC2-NHT 2018 VCAA 19 MC

A local supermarket sells apples in bags that have negligible mass. The stated mass of a bag of apples is 1 kg.

The mass of this particular type of apple is known to be normally distributed with a mean of 115 grams and a standard deviation of 7 grams. A particular bag contains nine randomly selected apples.

The probability that the nine apples in this bag have a total mass of less than 1 kg is

  1. 0.0478
  2. 0.1132
  3. 0.4265
  4. 0.5373
  5. 0.9522
Show Answers Only

`A`

Show Worked Solution

`M~N (115, 7^2)`

`T = M + M + M + M + M + M + M + M + M`

`T~N (9 xx 115, 9 xx 7^2)~N(1035, 21^2)`

`\text{Pr} (T < 1000) ~~ 0.0478`

 
`=>  A`