SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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 2023 VCAA 19 MC

A company accountant knows that the amount owed on any individual unpaid invoice is normally distributed with a mean of $800 and a standard deviation of $200.

What is the probability, correct to three decimal places, that in a random sample of 16 unpaid invoices the total amount owed is more than $13 500?

  1. 0.087
  2. 0.191
  3. 0.413
  4. 0.587
  5. 0.809
Show Answers Only

\(B\)

Show Worked Solution

\(n=16, \ E(X)=\mu=800 \)

\(X\ \sim\ N(800, 200^2)  \)

\(\bar{X} \sim N\Bigg{(}800, \dfrac{200^2}{\sqrt{16}}\Bigg{)} \sim N(800, 50^2) \)

\(\text{Pr} (\Sigma X>13\ 500) = \text{Pr}\Bigg{(}\bar{X} > \dfrac{13\ 500}{16}\Bigg{)} = \text{Pr}(\bar{X} > 843.75)= 0.190…\)

\(\Rightarrow B\)

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 17 MC

Bottles of a particular brand of soft drink are labelled as having a volume of 1.25 L. The machines filling the bottles deliver a volume that is normally distributed with a mean of 1.26 L and a standard deviation of 0.01 L.

The probability that six bottles have a mean volume that is at least the labelled volume of 1.25 L is closest to

  1. 0.5968
  2. 0.8413
  3. 0.9750
  4. 0.9772
  5. 0.9928
Show Answers Only

`E`

Show Worked Solution

`V~N (1.26, 0.01^2)`

♦ Mean mark 50%.

`E(barV) = mu = 1.26`

`text(s.d.)(barV) = 0.01/sqrt6`

`text(Pr)(barV>= 1.25) ≈ 0.9928`

`text{By CAS: normCdf}(1.25,∞,1.26,0.01/sqrt6)`

`=> E`

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-NHT 2019 VCAA 18 MC

Consider a random variable `X` with probability density function
 

`f(x) = {(2x,, 0<= x <= 1),(0,, x < 0\ \ text(or)\ \ x > 1):}`
 

If a large number of samples, each of size 100, is taken from his distribution, then the distribution of the sample means, `barX`, will be approximately normal with mean  `E(barX) = 2/3`  and standard deviation  `text(sd)(barX)`  equal to

  1.  `sqrt2/60`
  2.  `sqrt2/6`
  3.  `1/180`
  4.  `1/18`
  5.  `sqrt2/30`
Show Answers Only

`A`

Show Worked Solution

`E(X) = E(barX) = 2/3`

`E(X^2) = int_0^1 x^2 · 2x\ dx = [(x^4)/2]_0^1 = 1/2`

`text(Var)(X)` `= E(X^2)-[E(X)]^2`
  `= 1/2-4/9`
  `= 1/18`

 
`sigma(barX) = (sqrt(1/18))/sqrt100 = 1/(10 xx 3sqrt2) = sqrt2/60`

`=>\ A`

Statistics, SPEC2 2018 VCAA 19 MC

The gestation period of cats is normally distributed with mean  `mu = 66`  days and variance  `sigma^2 = 16/9`.

The probability that a sample of five cats chosen at random has an average gestation period greater than 65 days is closest to

  1. 0.5000
  2. 0.7131
  3. 0.7734
  4. 0.8958
  5. 0.9532
Show Answers Only

`E`

Show Worked Solution

`G~N (66 , 16/9)`

`bar G~N (66, 16/(9 xx 5))`

`text(Pr)(bar G > 65) ~~ 0.9532`

`=>  E`