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Statistics, 2ADV S2 2024 HSC 16

Flowers were planted in two gardens (Garden A and Garden B).

On a particular day, 25 flowers were randomly selected from each garden and their heights measured in millimetres.

The data are represented in parallel box-plots.
 

Compare the two datasets by examining the skewness of the distributions, and the measures of central tendency and spread.   (3 marks)

Show Answers Only

\(\text{Skewness comparison:}\)

\(\rightarrow\ \text{Garden A is negatively skewed while Garden B is positively skewed.}\)
  

\(\text{Measures of central tendency comparison:}\)

\(\rightarrow\ \text{IQR (A)}\ \approx 17\ \text{ is greater than IQR (B)}\ \approx 9\)

\(\rightarrow\ \text{This means the middle 50% of data points of B are more closely clustered} \)
     \(\text{around the median than A.}\)

\(\text{Spread comparison:}\)

\(\rightarrow\ \text{Range (A)}\ \approx 39\ \text{is less than range (B)}\ \approx 28\)

\(\rightarrow\ \text{This means the spread of data points of A is wider than the spread of data points of B.}\)

\(\text{Overall, flowers from Garden B will tend to be higher than Garden A flowers as well as exhibiting}\)
\(\text{a larger range of heights.}\)

Show Worked Solution

\(\text{Skewness comparison:}\)

\(\rightarrow\ \text{Garden A is negatively skewed while Garden B is positively skewed.}\)
  

\(\text{Measures of central tendency comparison:}\)

\(\rightarrow\ \text{IQR (A)}\ \approx 17\ \text{ is greater than IQR (B)}\ \approx 9\)

\(\rightarrow\ \text{This means the middle 50% of data points of B are more closely clustered}\)
    \(\text{around the median than A.}\)
 

\(\text{Spread comparison:}\)

\(\rightarrow\ \text{Range (A)}\ \approx 39\ \text{is less than range (B)}\ \approx 28\)

\(\rightarrow\ \text{This means the spread of data points of A is wider than the spread of data points of B.}\)
 

\(\text{Overall, flowers from Garden B will tend to be higher than Garden A flowers as well as exhibiting}\)
\(\text{a larger range of heights.}\)

L&E, 2ADV E1 2024 HSC 13

The graph shows the populations of two different animals, \(W\) and \(K\), in a conservation park over time. The \(y\)-axis is the size of the population and the \(x\)-axis is the number of years since 1985 .

Population \(W\) is modelled by the equation  \(y=A \times(1.055)^x\).

Population \(K\) is modelled by the equation  \(y=B \times(0.97)^x\).
 

Complete the table using the information provided.   (3 marks)

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=\ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} & & \\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}& & 61 \\
\hline
\end{array}

Show Answers Only

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

Show Worked Solution

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

\(\text{Calculations:}\)

\(\text{Population } W: \ (1.055)^x \Rightarrow \text { increase of 5.5% each year}\)

\(\text {Population } K: \ (0.97)^x \Rightarrow 1-0.97=0.03 \Rightarrow \text { decrease of 3.0% each year}\)

\(\text {Predicted population }(W)=34 \times(1.055)^{50}=494\)

Algebra, STD2 A4 2024 HSC 22

The graph shows the populations of two different animals, \(W\) and \(K\), in a conservation park over time. The \(y\)-axis is the size of the population and the \(x\)-axis is the number of years since 1985 .

Population \(W\) is modelled by the equation  \(y=A \times(1.055)^x\).

Population \(K\) is modelled by the equation  \(y=B \times(0.97)^x\).
 

Complete the table using the information provided.   (3 marks)

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=\ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} & & \\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}& & 61 \\
\hline
\end{array}

Show Answers Only

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

Show Worked Solution

\begin{array}{|l|c|c|}
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}&\text {Population } W & \text {Population } K \\
\hline
\rule{0pt}{2.5ex}\text { Population in 1985 }\rule[-1ex]{0pt}{0pt} & A=34 & B=280 \\
\hline
\rule{0pt}{2.5ex}\text { Percentage yearly change in the population }\rule[-1ex]{0pt}{0pt} &+5.5\% & -3.0\%\\
\hline
\rule{0pt}{2.5ex}\text { Predicted population when } x=50 \rule[-1ex]{0pt}{0pt}&494 & 61 \\
\hline
\end{array}

\(\text{Calculations:}\)

\(\text{Population } W: \ (1.055)^x \Rightarrow \text { increase of 5.5% each year}\)

\(\text {Population } K: \ (0.97)^x \Rightarrow 1-0.97=0.03 \Rightarrow \text { decrease of 3.0% each year}\)

\(\text {Predicted population }(W)=34 \times(1.055)^{50}=494\)

Financial Maths, 2ADV M1 2024 HSC 26

Twenty-five years ago, Phoenix deposited a single sum of money into a new bank account, earning 2.4% interest per annum compounding monthly.

Present value interest factors for an annuity of $1 for various interest rates \((r)\) and numbers of periods \((n)\) are given in the table.

Phoenix made the following withdrawals from this account.

  • $2000 at the end of each month for the first 15 years, starting at the end of the first month.
  • $1200 at the end of each month for the next 10 years, starting at the end of the 181st month after the account was opened.

Calculate the minimum sum that Phoenix could have deposited in order to make these withdrawals.   (4 marks)

Show Answers Only

\(\text{Minimum deposit}\ = $391\,344.80\)

Show Worked Solution

\(\text{1st Annuity}\)

\(\text{Find PVA for \$2000 paid monthly for 1st 15 years:}\)

\(r= \dfrac{2.4%}{12} = 0.2\% = 0.002\)

\(\text{Total payments (to Phoenix)}\ = 15 \times 12 = 180\)

\(\text{PVA factor (from table)}\ = 151.036\)

\(\text{PVA (1st annuity)}\ = 2000 \times 151.036 = $302\,072\)
 

\(\text{2nd Annuity}\)

\(\text{Find PVA for \$1200 paid monthly from year 16 to 25:}\)

\(\text{PVA (2nd annuity) = PVA (25 years) }-\text{ PVA (15 years)}\)

\(r= \dfrac{2.4%}{12} = 0.2\% = 0.002\)

\(\text{Total payments (25 years)}\ = 25 \times 12 = 300\)

\(\text{PVA factors (from table): 225.430 (25 years), 151.036 (15 years)}\)

\(\text{PVA (2nd annuity)}\) \(=(1200 \times 225.430)-(1200 \times 151.036)\)  
  \(=$89\,272.80\)  

 
\(\therefore\ \text{Minimum deposit}\ = 302\,072+89\,272.80 = $391\,344.80\)

Financial Maths, STD2 F5 2024 HSC 41

Twenty-five years ago, Phoenix deposited a single sum of money into a new bank account, earning 2.4% interest per annum compounding monthly.

Present value interest factors for an annuity of $1 for various interest rates \((r)\) and numbers of periods \((n)\) are given in the table.

Phoenix made the following withdrawals from this account.

  • $2000 at the end of each month for the first 15 years, starting at the end of the first month.
  • $1200 at the end of each month for the next 10 years, starting at the end of the 181st month after the account was opened.

Calculate the minimum sum that Phoenix could have deposited in order to make these withdrawals.   (4 marks)

Show Answers Only

\(\text{Minimum deposit}\ = $391\,344.80\)

Show Worked Solution

\(\text{1st Annuity}\)

\(\text{Find PVA for \$2000 paid monthly for 1st 15 years:}\)

\(r= \dfrac{2.4%}{12} = 0.2\% = 0.002\)

\(\text{Total payments (to Phoenix)}\ = 15 \times 12 = 180\)

\(\text{PVA factor (from table)}\ = 151.036\)

\(\text{PVA (1st annuity)}\ = 2000 \times 151.036 = $302\,072\)
 

\(\text{2nd Annuity}\)

\(\text{Find PVA for \$1200 paid monthly from year 16 to 25:}\)

\(\text{PVA (2nd annuity) = PVA (25 years) }-\text{ PVA (15 years)}\)

\(r= \dfrac{2.4%}{12} = 0.2\% = 0.002\)

\(\text{Total payments (25 years)}\ = 25 \times 12 = 300\)

\(\text{PVA factors (from table): 225.430 (25 years), 151.036 (15 years)}\)

\(\text{PVA (2nd annuity)}\) \(=(1200 \times 225.430)-(1200 \times 151.036)\)  
  \(=$89\,272.80\)  

 
\(\therefore\ \text{Minimum deposit}\ = 302\,072+89\,272.80 = $391\,344.80\)

Statistics, 2ADV S2 2024 HSC 21

A researcher is studying anacondas (a type of snake).

A dataset recording the age (in years) and length (in cm) of female and male anacondas is displayed on the graph.

Anacondas reach maturity at about 4 years of age.
 

Write THREE observations about anacondas that may be made from the scatterplot. (Note: No calculations are required.)   (3 marks)

Show Answers Only

\(\text{Answers could include any three of the following:}\)

\(→\ \text{Female anacondas are longer than males of the equivalent age.}\)

\(→\ \text{Female anacondas grow more quickly than male anacondas from birth until maturity which can be seen}\)
\(\ \ \ \ \text{by the steeper gradient of the LOBF for each dataset over this period.}\)

\(→\ \text{Female anacondas continue to grow to at least 10 years of age, well past their age of maturity at 4 years of age.}\)

\(→\ \text{Male anacondas’ growth slows noticeably and flattens out once they hit their age of maturity at 4 years old.}\)

Show Worked Solution

\(\text{Answers could include any three of the following:}\)

\(→\ \text{Female anacondas are longer than males of the equivalent age.}\)

\(→\ \text{Female anacondas grow more quickly than male anacondas from birth until maturity which can be seen}\)
\(\ \ \ \ \text{by the steeper gradient of the LOBF for each dataset over this period.}\)

\(→\ \text{Female anacondas continue to grow to at least 10 years of age, well past their age of maturity at 4 years of age.}\)

\(→\ \text{Male anacondas’ growth slows noticeably and flattens out once they hit their age of maturity at 4 years old.}\)

Statistics, 2ADV S3 2024 HSC 23

A random variable is normally distributed with mean 0 and standard deviation 1. The table gives the probability that this random variable is less than \(z\).

\begin{array} {|c|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} z \rule[-1ex]{0pt}{0pt} & 0.6 & 0.7 & 0.8 & 0.9 & 1.0 & 1.1 & 1.2 & 1.3 & 1.4 \\
\hline
\rule{0pt}{2.5ex} \textit{Probability} \rule[-1ex]{0pt}{0pt} & 0.7257 & 0.7580 & 0.7881 & 0.8159 & 0.8413 & 0.8643 & 0.8849 & 0.9032 & 0.9192 \\
\hline
\end{array}

The probability values given in the table for different values of \(z\) are represented by the shaded area in the following diagram.
 

The scores in a university examination with a large number of candidates are normally distributed with mean 58 and standard deviation 15.

  1. By calculating a \(z\)-score, find the percentage of scores that are between 58 and 70.   (2 marks)
  2. Explain why the percentage of scores between 46 and 70 is twice your answer to part (a).   (1 mark)
  3. By using the values in the table above, find an approximate minimum score that a candidate would need to be placed in the top 10% of the candidates.   (2 marks)
Show Answers Only

a.   \(28.81%\)

b.   \(\text{Normal distribution is symmetrical about the mean (58).}\)

\(\text{Since 70 and 46 are both the same distance (12) from the mean, the percentage}\)

\(\text{of scores in this range will be twice the answer in part (a).}\)

c.   \(\text{Approx minimum score = 78%}\)

Show Worked Solution

a.   \(z\text{-score (58)}\ =\dfrac{x-\mu}{\sigma} = \dfrac{58-58}{15}=0\)

\(z\text{-score (70)}\ = \dfrac{70-58}{15}=0.8\)

\(\text{Using table:}\)

\(\text{% between 58–70}\ =0.7881-0.5=0.2881=28.81%\)
 

b.   \(\text{Normal distribution is symmetrical about the mean (58).}\)

\(\text{Since 70 and 46 are both the same distance (12) from the mean, the percentage}\)

\(\text{of scores in this range will be twice the answer in part (a).}\)
 

c.   \(z\text{-score 1.3 has a table value 0.9032}\)

\(1-0.9032=0.0968\ \Rightarrow\ \text{i.e. 9.68% of students score higher.}\)

\(\text{Find}\ x\ \text{for a}\ z\text{-score of 1.3:}\)

\(1.3\) \(=\dfrac{x-58}{15}\)  
\(x\) \(=1.3 \times 15 +58\)   
  \(=77.5\)  

 
\(\therefore\ \text{Approx minimum score = 78%}\)

Statistics, STD2 S5 2024 HSC 35

A random variable is normally distributed with mean 0 and standard deviation 1. The table gives the probability that this random variable is less than \(z\).

\begin{array} {|c|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} z \rule[-1ex]{0pt}{0pt} & 0.6 & 0.7 & 0.8 & 0.9 & 1.0 & 1.1 & 1.2 & 1.3 & 1.4 \\
\hline
\rule{0pt}{2.5ex} \textit{Probability} \rule[-1ex]{0pt}{0pt} & 0.7257 & 0.7580 & 0.7881 & 0.8159 & 0.8413 & 0.8643 & 0.8849 & 0.9032 & 0.9192 \\
\hline
\end{array}

The probability values given in the table for different values of \(z\) are represented by the shaded area in the following diagram.
 

The scores in a university examination with a large number of candidates are normally distributed with mean 58 and standard deviation 15.

  1. By calculating a \(z\)-score, find the percentage of scores that are between 58 and 70.   (2 marks)
  2. Explain why the percentage of scores between 46 and 70 is twice your answer to part (a).   (1 mark)
  3. By using the values in the table above, find an approximate minimum score that a candidate would need to be placed in the top 10% of the candidates.   (2 marks)
Show Answers Only

a.   \(28.81%\)

b.   \(\text{Normal distribution is symmetrical about the mean (58).}\)

\(\text{Since 70 and 46 are both the same distance (12) from the mean, the percentage}\)

\(\text{of scores in this range will be twice the answer in part (a).}\)

c.   \(\text{Approx minimum score = 78%}\)

Show Worked Solution

a.   \(z\text{-score (58)}\ =\dfrac{x-\mu}{\sigma} = \dfrac{58-58}{15}=0\)

\(z\text{-score (70)}\ = \dfrac{70-58}{15}=0.8\)

\(\text{Using table:}\)

\(\text{% between 58–70}\ =0.7881-0.5=0.2881=28.81%\)
 

b.   \(\text{Normal distribution is symmetrical about the mean (58).}\)

\(\text{Since 70 and 46 are both the same distance (12) from the mean, the percentage}\)

\(\text{of scores in this range will be twice the answer in part (a).}\)
 

c.   \(z\text{-score 1.3 has a table value 0.9032}\)

\(1-0.9032=0.0968\ \Rightarrow\ \text{i.e. 9.68% of students score higher.}\)

\(\text{Find}\ x\ \text{for a}\ z\text{-score of 1.3:}\)

\(1.3\) \(=\dfrac{x-58}{15}\)  
\(x\) \(=1.3 \times 15 +58\)   
  \(=77.5\)  

 
\(\therefore\ \text{Approx minimum score = 78%}\)

Statistics, STD2 S4 2024 HSC 30

A researcher is studying anacondas (a type of snake).

A dataset recording the age (in years) and length (in cm) of female and male anacondas is displayed on the graph.

Anacondas reach maturity at about 4 years of age.
 

Write THREE observations about anacondas that may be made from the scatterplot. (Note: No calculations are required.)   (3 marks)

Show Answers Only

\(\text{Answers could include three of the following:}\)

\(→\ \text{Female anacondas are longer than males of the equivalent age.}\)

\(→\ \text{Female anacondas grow more quickly than male anacondas from birth until maturity which can be seen by}\)
\(\ \ \ \ \text{the steeper gradient of the LOBF for each dataset over this period.}\)

\(→\ \text{Female anacondas continue to grow to at least 10 years of age, well past their age of maturity at 4 years of age.}\)

\(→\ \text{Male anacondas’ growth slows noticeably and flattens out once they hit their age of maturity at 4 years old.}\)

Show Worked Solution

\(\text{Answers could include three of the following:}\)

\(→\ \text{Female anacondas are longer than males of the equivalent age.}\)

\(→\ \text{Female anacondas grow more quickly than male anacondas from birth until maturity which can be seen by}\)
\(\ \ \ \ \text{the steeper gradient of the LOBF for each dataset over this period.}\)

\(→\ \text{Female anacondas continue to grow to at least 10 years of age, well past their age of maturity at 4 years of age.}\)

\(→\ \text{Male anacondas’ growth slows noticeably and flattens out once they hit their age of maturity at 4 years old.}\)

Statistics, 2ADV S2 2024 HSC 8 MC

Some data are used to create a box plot shown.
 

A histogram is created from the same set of data.

Which of these histograms is NOT possible for the given box plot?
 


 

Show Answers Only

\(D\)

Show Worked Solution

\(\text{By inspection of box plot,}\ \ IQR=\dfrac{2}{3}\times \text{range}\)

\(\text{ln options A, B and C (given 16 data points):}\)

\(Q_1\ \text{(4th data point)}\ \rightarrow \text{2nd column}\)

\(Q_3\ \text{(13th data point)}\  \rightarrow \text{6th column}\)

\(\text{Since}\ IQR=\dfrac{2}{3}\times \text{range}\ \Rightarrow\ \text{histograms are possible.}\)
 

\(\text{ln option D:}\)

\(Q_1  \rightarrow \text{3rd column}\)

\(Q_3 \rightarrow \text{5th column}\)

\(\text{Since}\ IQR=\dfrac{1}{3} \times \text{range}\ \Rightarrow\ \text{not possible.}\)

\(\Rightarrow D\)

Statistics, STD2 S1 2024 HSC 15 MC

Some data are used to create a box plot shown.
 

A histogram is created from the same set of data.

Which of these histograms is NOT possible for the given box plot?
 


 

Show Answers Only

\(D\)

Show Worked Solution

\(\text{By inspection of box plot,}\ \ IQR=\dfrac{2}{3}\times \text{range}\)

\(\text{ln options A, B and C (given 16 data points):}\)

\(Q_1\ \text{(4th data point)}\ \rightarrow \text{2nd column}\)

\(Q_3\ \text{(13th data point)}\  \rightarrow \text{6th column}\)

\(\text{Since}\ IQR=\dfrac{2}{3}\times \text{range}\ \Rightarrow\ \text{histograms are possible.}\)
 

\(\text{ln option D:}\)

\(Q_1  \rightarrow \text{3rd column}\)

\(Q_3 \rightarrow \text{5th column}\)

\(\text{Since}\ IQR=\dfrac{1}{3} \times \text{range}\ \Rightarrow\ \text{not possible.}\)

\(\Rightarrow D\)

Statistics, STD2 S1 2024 HSC 28

Flowers were planted in two gardens (Garden A and Garden B).

On a particular day, 25 flowers were randomly selected from each garden and their heights measured in millimetres.

The data are represented in parallel box-plots.
 

Compare the two datasets by examining the skewness of the distributions, and the measures of central tendency and spread.   (3 marks)

Show Answers Only

\(\text{Skewness comparison:}\)

\(\rightarrow\ \text{Garden A is negatively skewed while Garden B is positively skewed.}\)
  

\(\text{Measures of central tendency comparison:}\)

\(\rightarrow\ \text{IQR (A)}\ \approx 17\ \text{ is greater than IQR (B)}\ \approx 9\)

\(\rightarrow\ \text{This means the middle 50% of data points of B are more closely clustered around the median than A.}\)
 

\(\text{Spread comparison:}\)

\(\rightarrow\ \text{Range (A)}\ \approx 39\ \text{is greater than range (B)}\ \approx 28\)

\(\rightarrow\ \text{This means the spread of data points of A is wider than the spread of data points of B.}\)
 

\(\text{Overall, flowers from Garden A will tend to be higher than Garden B flowers as well as exhibiting a larger range of heights.}\)

Show Worked Solution

\(\text{Skewness comparison:}\)

\(\rightarrow\ \text{Garden A is negatively skewed while Garden B is positively skewed.}\)
  

\(\text{Measures of central tendency comparison:}\)

\(\rightarrow\ \text{IQR (A)}\ \approx 17\ \text{ is greater than IQR (B)}\ \approx 9\)

\(\rightarrow\ \text{This means the middle 50% of data points of B are more closely clustered around the median than A.}\)
 

\(\text{Spread comparison:}\)

\(\rightarrow\ \text{Range (A)}\ \approx 39\ \text{is greater than range (B)}\ \approx 28\)

\(\rightarrow\ \text{This means the spread of data points of A is wider than the spread of data points of B.}\)
 

\(\text{Overall, flowers from Garden B will tend to be higher than Garden A flowers as well as exhibiting a larger range of heights.}\)

Statistics, 2ADV S3 2024 HSC 3 MC

Pia's marks in Year 10 assessments are shown. The scores for each subject were normally distributed.

\begin{array}{|l|c|c|c|}
\hline & \textit {Pia's mark} & \textit {Year 10 mean} & \textit {Year 10 standard} \\
&&&\textit {deviation}\\
\hline \text {English} & 78 & 66 & 6 \\
\hline \text {Mathematics} & 80 & 71 & 10 \\
\hline \text {Science} & 77 & 70 & 15 \\
\hline \text {History} & 85 & 72 & 9 \\
\hline
\end{array}

In which subject did Pia perform best in comparison with the rest of Year 10?

  1. English
  2. Mathematics
  3. Science
  4. History
Show Answers Only

\(A\)

Show Worked Solution

\(\text {Consider the z-score of each option:}\)

\(z \text {-score (English)}=\dfrac{78-66}{6}=2\)

\(z \text {-score (Maths)}=\dfrac{80-71}{10}=0.9\)

\(z \text {-score (Science) }=\dfrac{77-70}{15}=0.46 \ldots\)

\(z \text {-score (History})=\dfrac{85-72}{9}=1.4 \ldots\)

\(\Rightarrow A\)

Statistics, STD2 S5 2024 HSC 5 MC

Pia's marks in Year 10 assessments are shown. The scores for each subject were normally distributed.

\begin{array}{|l|c|c|c|}
\hline & \textit {Pia's mark} & \textit {Year 10 mean} & \textit {Year 10 standard} \\
&&&\textit {deviation}\\
\hline \text {English} & 78 & 66 & 6 \\
\hline \text {Mathematics} & 80 & 71 & 10 \\
\hline \text {Science} & 77 & 70 & 15 \\
\hline \text {History} & 85 & 72 & 9 \\
\hline
\end{array}

In which subject did Pia perform best in comparison with the rest of Year 10?

  1. English
  2. Mathematics
  3. Science
  4. History
Show Answers Only

\(A\)

Show Worked Solution

\(\text {Consider the z-score of each option:}\)

\(z \text {-score (English)}=\dfrac{78-66}{6}=2\)

\(z \text {-score (Maths)}=\dfrac{80-71}{10}=0.9\)

\(z \text {-score (Science) }=\dfrac{77-70}{15}=0.46 \ldots\)

\(z \text {-score (History})=\dfrac{85-72}{9}=1.4 \ldots\)

\(\Rightarrow A\)

Statistics, 2ADV S3 2023 HSC 23

A random variable is normally distributed with a mean of 0 and a standard deviation of 1 . The table gives the probability that this random variable lies below `z` for some positive values of `z`.

The probability values given in the table are represented by the shaded area in the following diagram.
 

The weights of adult male koalas form a normal distribution with mean `mu` = 10.40 kg, and standard deviation `sigma` = 1.15 kg.

In a group of 400 adult male koalas, how many would be expected to weigh more than 11.93 kg?  (4 marks)

Show Answers Only

`37\ text{koalas*}`

`text{*36 or 36.72 koalas would also receive full marks}`

Show Worked Solution
`ztext{-score (11.93)}` `=(x-mu)/sigma`  
  `=(11.93-10.4)/1.15`  
  `=1.330`  

 
`Ptext{(Koala weighs > 11.93 kg)}\ = P(z>1.330)`

`text{Using the table:}`

`P(z>1.33)` `=1-0.9082`  
  `=0.0918`  

 

`:.\ text{Expected koalas > 11.93 kg}` `=0.0918 xx 400`  
  `=36.72`  
  `=37\ text{koalas*}`  

 
`text{*36 or 36.72 koalas would also receive full marks}`

Statistics, 2ADV S2 2023 HSC 18

A university uses gas to heat its buildings. Over a period of 10 weekdays during winter, the gas used each day was measured in megawatts (MW) and the average outside temperature each day was recorded in degrees Celsius (°C).

Using `x` as the average daily outside temperature and `y` as the total daily gas usage, the equation of the least-squares regression line was found.

The equation of the regression line predicts that when the temperature is 0°C, the daily gas usage is 236 MW.

The ten temperatures measured were: 0°, 0°, 0°, 2°, 5°, 7°, 8°, 9°, 9°, 10°,

The total gas usage for the ten weekdays was 1840 MW.

In any bivariate dataset, the least-squares regression line passes through the point `(bar x,bar y)`, where `bar x` is the sample mean of the `x`-values and `bary` is the sample mean of the `y`-values.

  1. Using the information provided, plot the point `(bar x,bar y)` and the `y`-intercept of the least-squares regression line on the grid.  (3 marks)
     

 

  1. What is the equation of the regression line?  (2 marks)

  2. In the context of the dataset, identify ONE problem with using the regression line to predict gas usage when the average outside temperature is 23°C.  (1 mark)

Show Answers Only

a.    
       

b.    `y=-10.4x+236`

c.    `text{Answers could include one of the following:}`

`text{→ 23°C is outside the range of the dataset and requires the trend}`

`text{to be extrapolated.}`

`text{→ At 23°C, the equation predicts negative daily gas usage.}`

Show Worked Solution

a.    `barx=(0+0+0+2+5+7+8+9+9+10)/10=5^@text{C}`

`bary=1840/10=184`

`text{Regression line passes through:}\ (0,236) and (5,184)`
 

 

b.    `m=(y_2-y_1)/(x_2-x_1)=(184-236)/(5-0)=-10.4`

`text{Equation of line}\ m=-10.4\ text{passing through}\ (0,236):`

`(y-y_1)` `=m(x-x_1)`  
`y-236` `=-10.4(x-0)`  
`y` `=-10.4x+236`  

 
c.    `text{Answers could include one of the following:}`

`text{→ 23°C is outside the range of the dataset and requires the trend}`

`text{to be extrapolated.}`

`text{→ At 23°C, the equation predicts negative daily gas usage.}`

Trigonometry, 2ADV T1 2023 HSC 16

The diagram shows a shape `APQBCD`. The shape consists of a rectangle `ABCD` with an arc `PQ` on side `AB` and with side lengths `BC` = 3.6 m and `CD` = 8.0 m.

The arc `PQ` is an arc of a circle with centre `O` and radius 2.1 m and `∠POQ=110°`.

 

What is the perimeter of the shape `APQBCD`? Give your answer correct to one decimal place.  (4 marks)

Show Answers Only

`23.8\ text{m}`

Show Worked Solution
`text{Arc}\ PQ` `=110/360 xx pi xx 2 xx 2.1`  
  `=4.03171… \ text{m}`  

 
`text{Consider}\ ΔOPQ:`
 
 

`sin 55^@` `=x/2.1`  
`x` `=2.1 xx sin 55^@`  
  `=1.7202…`  

 
`PQ=2x=3.440\ text{m}`

`:.\ text{Perimeter}` `=8+(2xx3.6)+4.031+(8-3.440)`  
  `=23.79…\ text{m}`  
  `=23.8\ text{m  (to 1 d.p.)}`  

Financial Maths, 2ADV M1 2023 HSC 15

A table of future value interest factors for an annuity of $1 is shown.
 

  1. Micky wants to save $450 000 over the next 10 years.
  2. If the interest rate is 6% per annum compounding annually, how much should Micky contribute each year? Give your answer to the nearest dollar.  (2 marks)

  3. Instead, Micky decides to contribute  $8535 every three months for 10 years to an annuity paying 6% per annum, compounding quarterly.
  4. How much will Micky have at the end of 10 years?  (3 marks)

Show Answers Only

  1. `$34\ 140`
  2. `$463\ 177.38`

Show Worked Solution

a.    `text{Applicable interest rate}\ =6%`

`text{Compounding periods}\ =10xx1=10`

`=>\ text{Factor}\ = 13.181`

`:.\ text{Contribution (annual)}` `=(450\ 000)/13.181`  
  `=$34\ 140`  

 
b.    `text{Applicable interest rate}\ =(6%)/4=1.5%\ text{per quarter}`

`text{Compounding periods}\ =10xx4=40`

`=>\ text{Factor}\ = 54.268`

`text{Total (after 10 years)}` `=8535 xx 54.268`  
  `=$463\ 177.38`  

Statistics, 2ADV S2 2023 HSC 1 MC

The number of bees leaving a hive was observed and recorded over 14 days at different times of the day.
 

Which Pearson's correlation coefficient best describes the observations?

  1. – 0.8
  2. – 0.2
  3. 0.2
  4. 0.8
Show Answers Only

`D`

Show Worked Solution

`text{Correlation is positive and strong.}`

`text{Best option:}\ r=0.8`

`=>D`

NOTE: Inputting all data points into a calculator is unnecessary and time consuming here.

Statistics, STD2 S5 2023 HSC 38

A random variable is normally distributed with a mean of 0 and a standard deviation of 1 . The table gives the probability that this random variable lies below `z` for some positive values of `z`.

The probability values given in the table are represented by the shaded area in the following diagram.
 

The weights of adult male koalas form a normal distribution with mean `mu` = 10.40 kg, and standard deviation `sigma` = 1.15 kg.

In a group of 400 adult male koalas, how many would be expected to weigh more than 11.93 kg?  (4 marks)

Show Answers Only

`37\ text{koalas*}`

`text{*36 or 36.72 koalas would also receive full marks}`

Show Worked Solution
`ztext{-score (11.93)}` `=(x-mu)/sigma`  
  `=(11.93-10.4)/1.15`  
  `=1.330`  

 
`Ptext{(Koala weighs > 11.93 kg)}\ = P(z>1.330)`

`text{Using the table:}`

`P(z>1.33)` `=1-0.9082`  
  `=0.0918`  
♦ Mean mark 39%.
`:.\ text{Expected koalas > 11.93 kg}` `=0.0918 xx 400`  
  `=36.72`  
  `=37\ text{koalas*}`  

 
`text{*36 or 36.72 koalas would also receive full marks}`

Statistics, STD2 S4 2023 HSC 34

A university uses gas to heat its buildings. Over a period of 10 weekdays during winter, the gas used each day was measured in megawatts (MW) and the average outside temperature each day was recorded in degrees Celsius (°C).

Using `x` as the average daily outside temperature and `y` as the total daily gas usage, the equation of the least-squares regression line was found.

The equation of the regression line predicts that when the temperature is 0°C, the daily gas usage is 236 MW.

The ten temperatures measured were: 0°, 0°, 0°, 2°, 5°, 7°, 8°, 9°, 9°, 10°,

The total gas usage for the ten weekdays was 1840 MW.

In any bivariate dataset, the least-squares regression line passes through the point `(bar x,bar y)`, where `bar x` is the sample mean of the `x`-values and `bary` is the sample mean of the `y`-values.

  1. Using the information provided, plot the point `(bar x,bar y)` and the `y`-intercept of the least-squares regression line on the grid.  (3 marks)
     

 

  1. What is the equation of the regression line?  (2 marks)

  2. In the context of the dataset, identify ONE problem with using the regression line to predict gas usage when the average outside temperature is 23°C.  (1 mark)

Show Answers Only

a.    
         

b.    `y=-10.4x+236`

c.    `text{Answers could include one of the following:}`

`text{→ 23°C is outside the range of the dataset and requires the trend}`

`text{to be extrapolated.}`

`text{→ At 23°C, the equation predicts negative daily gas usage.}`

Show Worked Solution

a.    `barx=(0+0+0+2+5+7+8+9+9+10)/10=5^@text{C}`

`bary=1840/10=184`

`text{Regression line passes through:}\ (0,236) and (5,184)`
 

♦ Mean mark (a) 44%.

b.    `m=(y_2-y_1)/(x_2-x_1)=(184-236)/(5-0)=-10.4`

`text{Equation of line}\ m=-10.4\ text{passing through}\ (0,236):`

`(y-y_1)` `=m(x-x_1)`  
`y-236` `=-10.4(x-0)`  
`y` `=-10.4x+236`  
♦♦♦ Mean mark (b) 21%.

 
c.    `text{Answers could include one of the following:}`

`text{→ 23°C is outside the range of the dataset and requires the trend}`

`text{to be extrapolated.}`

`text{→ At 23°C, the equation predicts negative daily gas usage.}`

♦♦♦ Mean mark 23%.

Measurement, STD2 M6 2023 HSC 33

The diagram shows a shape `APQBCD`. The shape consists of a rectangle `ABCD` with an arc `PQ` on side `AB` and with side lengths `BC` = 3.6 m and `CD` = 8.0 m.

The arc `PQ` is an arc of a circle with centre `O` and radius 2.1 m and `∠POQ=110°`.

 

What is the perimeter of the shape `APQBCD`? Give your answer correct to one decimal place.  (4 marks)

Show Answers Only

`23.8\ text{m}`

Show Worked Solution
`text{Arc}\ PQ` `=110/360 xx pi xx 2.1^2`  
  `=4.03171… \ text{m}`  

 
`text{Consider}\ ΔOPQ:`
 

♦ Mean mark 42%.
`sin 55^@` `=x/2.1`  
`x` `=2.1 xx sin 55^@`  
  `=1.7202…`  

 
`PQ=2x=3.440\ text{m}`

`:.\ text{Perimeter}` `=8+(2xx3.6)+4.031+(8-3.440)`  
  `=23.79…`  
  `=23.8\ text{m  (to 1 d.p.)}`  

Financial Maths, STD2 F5 2023 HSC 25

A table of future value interest factors for an annuity of $1 is shown.
 

  1. Micky wants to save $450 000 over the next 10 years.
  2. If the interest rate is 6% per annum compounding annually, how much should Micky contribute each year? Give your answer to the nearest dollar.  (2 marks)

  3. Instead, Micky decides to contribute  $8535 every three months for 10 years to an annuity paying 6% per annum, compounding quarterly.
  4. How much will Micky have at the end of 10 years?  (3 marks)

Show Answers Only

  1. `$34\ 140`
  2. `$463\ 177.38`

Show Worked Solution

a.    `text{Applicable interest rate}\ =6%`

`text{Compounding periods}\ =10xx1=10`

`=>\ text{Factor}\ = 13.181`

`:.\ text{Contribution (annual)}` `=(450\ 000)/13.181`  
  `=$34\ 140`  

 
b.    `text{Applicable interest rate}\ =(6%)/4=1.5%\ text{per quarter}`

`text{Compounding periods}\ =10xx4=40`

`=>\ text{Factor}\ = 54.268`

`text{Total (after 10 years)}` `=8535 xx 54.268`  
  `=$463\ 177.38`  
Mean mark (b) 53%.
 

Statistics, STD2 S4 2023 HSC 3 MC

The number of bees leaving a hive was observed and recorded over 14 days at different times of the day.
 

Which Pearson's correlation coefficient best describes the observations?

  1. `-0.8`
  2. `-0.2`
  3. `0.2`
  4. `0.8`
Show Answers Only

`D`

Show Worked Solution

`text{Correlation is positive and strong.}`

`text{Best option:}\ r=0.8`

`=>D`

NOTE: Inputting all data points into a calculator is unnecessary and time consuming here.

Functions, 2ADV F1 2022 HSC 12

A student believes that the time it takes for an ice cube to melt (`M` minutes) varies inversely with the room temperature `(T^@ text{C})`. The student observes that at a room temperature of `15^@text{C}` it takes 12 minutes for an ice cube to melt.

  1. Find the equation relating `M` and `T`.    (2 marks)

  2. By first completing this table of values, graph the relationship between temperature and time from `T=5^@C` to `T=30^@ text{C}`.   (2 marks)
     

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M &  &  &  \\
\hline \end{array}

 
                   

Show Answers Only

a.    `M=180/T`

 b.    

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M & 36 & 12 & 6 \\
\hline \end{array}       

 

Show Worked Solution
a.    `M` `prop 1/T`
  `M` `=k/T`
  `12` `=k/15`
  `k` `=15 xx 12`
    `=180`

 
`:.M=180/T`
 


♦ Mean mark (a) 49%.

b.   

\begin{array} {|c|c|c|c|}
\hline  \ \ T\ \  & \ \ 5\ \  & \ 15\  & \ 30\  \\
\hline M & 36 & 12 & 6 \\
\hline \end{array}

Algebra, STD2 A4 2022 HSC 24

A student believes that the time it takes for an ice cube to melt (`M` minutes) varies inversely with the room temperature `(T^@ text{C})`. The student observes that at a room temperature of `15^@text{C}` it takes 12 minutes for an ice cube to melt.

  1. Find the equation relating `M` and `T`.   (2 marks)

  2. By first completing this table of values, graph the relationship between temperature and time from `T=5^@C` to `T=30^@ text{C}.`   (2 marks)

\begin{array} {|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ T\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 5\ \ \  & \ \ 15\ \ \  & \ \ \ 30\ \ \  \\
\hline
\rule{0pt}{2.5ex} \ \ M\ \ \rule[-1ex]{0pt}{0pt} & & & \\
\hline
\end{array}

 
           

Show Answers Only
a.    `M` `prop 1/T`
  `M` `=k/T`
  `12` `=k/15`
  `k` `=15 xx 12`
    `=180`

 
`:.M=180/T`

b.   

\begin{array} {|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ T\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 5\ \ \  & \ \ 15\ \ \  & \ \ 30\ \  \\
\hline
\rule{0pt}{2.5ex} \ \ M\ \ \rule[-1ex]{0pt}{0pt} & 36 & 12 & 6 \\
\hline
\end{array}

 

     

Show Worked Solution
a.    `M` `prop 1/T`
  `M` `=k/T`
  `12` `=k/15`
  `k` `=15 xx 12`
    `=180`

 
`:.M=180/T`


♦♦ Mean mark part (a) 29%.

b.   

\begin{array} {|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \ \ T\ \  \rule[-1ex]{0pt}{0pt} & \ \ \ 5\ \ \  & \ \ 15\ \ \  & \ \ 30\ \  \\
\hline
\rule{0pt}{2.5ex} \ \ M\ \ \rule[-1ex]{0pt}{0pt} & 36 & 12 & 6 \\
\hline
\end{array}

 

     


♦ Mean mark 44%.

Statistics, 2ADV S3 2022 HSC 26

The life span of batteries from a particular factory is normally distributed with a mean of 840 hours and a standard deviation of 80 hours.

It is known from statistical tables that for this distribution approximately 60% of the batteries have a life span of less than 860 hours.

What is the approximate percentage of batteries with a life span between 820 and 920 hours?  (3 marks)

Show Answers Only

`44text{%}`

Show Worked Solution

`mu=840, \ sigma=80`

`ztext{-score (860)}\ = (x-mu)/sigma=(860-840)/80=0.25` 

`ztext{-score (820)}\ =(820-840)/80=-0.25` 

`ztext{-score (920)}\ =(920-840)/80=1`
 

`text{50% of batteries have a life span below 840 hours (by definintion)}`

`=>\ text{10% lie between 840 and 860 hours}`

`=>\ text{By symmetry, 10% lie between 820 and 840 hours}`

`=> P(-0.25<=z<=0)=10text{%}`
 

`:.\ text{Percentage between 820 and 920}`

`=P(-0.25<=z<=1)`

`=P(-0.25<=z<=0) + P(0<=z<=1)`

`=10+34`

`=44text{%}`

Statistics, 2ADV S2 2022 HSC 24

Jo is researching the relationship between the ages of teenage characters in television series and the ages of actors playing these characters.

After collecting the data, Jo finds that the correlation coefficient is 0.4564.

A scatterplot showing the data is drawn. The line of best fit with equation  `y=-7.51+1.85 x`, is also drawn.
 


 

Describe and interpret the data and other information provided, with reference to the context given.  (4 marks)

Show Answers Only

`text{Correlation coefficient}\ (r) = 0.4564`

    • `text{Moderate and positive correlation}`

`text{Gradient of LOBF}\ = 1.85`

    • `text{On average, each extra year of a character’s age results}`
      `text{in the actor being 1.85 years older.}`

`text{Mode of data set = 15 years}`
  

`text{Limitations}`

    • `text{Data set is very restricted with just a 4 year range of}`
      `text{character ages.}`
    • `text{LOBF not useful when extrapolated to the left as it drops}`
      `text{below zero (on y-axis).}`
    • `text{Relationship describes correlation only, not causation.}`
Show Worked Solution

`text{Correlation coefficient}\ (r) = 0.4564`

    • `text{Moderate and positive correlation}`

`text{Gradient of LOBF}\ = 1.85`

    • `text{On average, each extra year of a character’s age results}`
      `text{in the actor being 1.85 years older.}`

`text{Mode of data set = 15 years}`
  

`text{Limitations}`

    • `text{Data set is very restricted with just a 4 year range of}`
      `text{character ages.}`
    • `text{LOBF not useful when extrapolated to the left as it drops}`
      `text{below zero (on y-axis).}`
    • `text{Relationship describes correlation only, not causation.}`

Financial Maths, 2ADV M1 2022 HSC 21

Eli is choosing between two investment options.

A table of future value interest factors for an annuity of $1 is shown.

  1. What is the value of Eli's investment after 10 years using Option 1 ?  (2 marks)

  2. What is the difference between the future values after 10 years using Option 1 and Option 2?  (2 marks)

Show Answers Only
  1. `$45\ 097.17`
  2. `$40.87`
Show Worked Solution

a.   `text{Monthly r/i}\ = 1.2/12=0.1text{%}\ \ =>\ \ r= 0.001`

`text{Compounding periods}\ (n)=12xx10=120`

`FV` `=PV(1+r)^n`  
  `=40\ 000(1+0.001)^120`  
  `=$45\ 097.17`  

 

b.   `text{Quarterly r/i}\ = 2.4/4=0.6text{%}\ \ =>\ \ r= 0.006`

`text{Compounding periods}\ (N) =4xx10=40`

`text{Annuity factor (from table) = 45.05630}`

`FV` `=1000xx45.05630`  
  `=45\ 056.30`  

 

`text{Difference}` `=45\ 097.17-45\ 056.30`  
  `=$40.87`  

Statistics, 2ADV S2 2022 HSC 11

The table shows the types of customer complaints received by an online business in a month.

  1. What are the values of `A` and `B`?  (2 marks)

  2. The data from the table are shown in the following Pareto chart.
     
     
  3. The manager will address 80% of the complaints.
  4. Which types of complaints will the manager address?  (1 mark)

Show Answers Only
  1. `A=160, \ B=96`
  2. `text{Stock shortages and delivery fees.}`
Show Worked Solution

a.   `A=98+62=160`

`text{% Damaged items}\ = 8/200 xx 100 = 4text{%}`

`text{Cumulative % after damaged items = 96%}`

`B = 92+4=96`
 

b.   `text{The right hand side cumulative frequency percentage}`

`text{shows that 80% of all complaints received concern}`

`text{stock shortages and delivery fees.}`

`:.\ text{The manager will address stock shortages and delivery fees.}`

Statistics, STD2 S5 2022 HSC 37

The life span of batteries from a particular factory is normally distributed with a mean of 840 hours and a standard deviation of 80 hours.

It is known from statistical tables that for this distribution approximately 60% of the batteries have a life span of less than 860 hours.

What is the approximate percentage of batteries with a life span between 820 and 920 hours?  (3 marks)

Show Answers Only

`44text{%}`

Show Worked Solution

`mu=840, \ sigma=80`

`ztext{-score (860)}\ = (x-mu)/sigma=(860-840)/80=0.25` 

`ztext{-score (820)}\ =(820-840)/80=-0.25` 

`ztext{-score (920)}\ =(920-840)/80=1`
 

`text{50% of batteries have a life span below 840 hours (by definition)}`

`=>\ text{10% lie between 840 and 860 hours}`

`=>\ text{By symmetry, 10% lie between 820 and 840 hours}`

`=> P(-0.25<=z<=0)=10text{%}`
 

`:.\ text{Percentage between 820 and 920}`

`=P(-0.25<=z<=1)`

`=P(-0.25<=z<=0) + P(0<=z<=1)`

`=10+34`

`=44text{%}`


♦♦ Mean mark 28%.

Statistics, STD2 S4 2022 HSC 35

Jo is researching the relationship between the ages of teenage characters in television series and the ages of actors playing these characters.

After collecting the data, Jo finds that the correlation coefficient is 0.4564.

A scatterplot showing the data is drawn. The line of best fit with equation  `y=-7.51+1.85 x`, is also drawn.
 


 

Describe and interpret the data and other information provided, with reference to the context given.  (4 marks)

Show Answers Only

`text{Correlation coefficient}\ (r) = 0.4564`

    • `text{Moderate and positive correlation}`

`text{Gradient of LOBF}\ = 1.85`

    • `text{On average, each extra year of a character’s age results}`
      `text{in the actor being 1.85 years older.}`

`text{Mode of data set = 15 years}`
  

`text{Limitations}`

    • `text{Data set is very restricted with just a 4 year range of}`
      `text{character ages.}`
    • `text{LOBF not useful when extrapolated to the left as it drops}`
      `text{below zero (on y-axis).}`
    • `text{Relationship describes correlation only, not causation.}`
Show Worked Solution

`text{Correlation coefficient}\ (r) = 0.4564`

    • `text{Moderate and positive correlation}`

`text{Gradient of LOBF}\ = 1.85`

    • `text{On average, each extra year of a character’s age results}`
      `text{in the actor being 1.85 years older.}`

`text{Mode of data set = 15 years}`
  

`text{Limitations}`

    • `text{Data set is very restricted with just a 4 year range of}`
      `text{character ages.}`
    • `text{LOBF not useful when extrapolated to the left as it drops}`
      `text{below zero (on y-axis).}`
    • `text{Relationship describes correlation only, not causation.}`

♦♦ Mean mark 30%.

Financial Maths, STD2 F5 2022 HSC 30

Eli is choosing between two investment options.

A table of future value interest factors for an annuity of $1 is shown.

  1. What is the value of Eli's investment after 10 years using Option 1 ?  (2 marks)

  2. What is the difference between the future values after 10 years using Option 1 and Option 2?  (2 marks)

Show Answers Only
  1. `$45\ 097.17`
  2. `$40.87`
Show Worked Solution

a.   `text{Monthly r/i}\ = 1.2/12=0.1text{%}\ \ =>\ \ r= 0.001`

`text{Compounding periods}\ (n)=12xx10=120`

`FV` `=PV(1+r)^n`  
  `=40\ 000(1+0.001)^120`  
  `=$45\ 097.17`  

 


♦ Mean mark 48%.

b.   `text{Quarterly r/i}\ = 2.4/4=0.6text{%}\ \ =>\ \ r= 0.006`

`text{Compounding periods}\ (N) =4xx10=40`

`text{Annuity factor (from table) = 45.05630}`

`FV` `=1000xx45.05630`  
  `=45\ 056.30`  

 

`text{Difference}` `=45\ 097.17-45\ 056.30`  
  `=$40.87`  

♦ Mean mark 43%.

Statistics, STD2 S1 2022 HSC 19

The table shows the types of customer complaints received by an online business in a month.
 

  1. What are the values of `A` and `B`?  (2 marks)

  2. The data from the table are shown in the following Pareto chart.
     
     
  3. The manager will address 80% of the complaints.
  4. Which types of complaints will the manager address?  (1 mark)

Show Answers Only
  1. `A=160, \ B=96`
  2. `text{Stock shortages and delivery fees.}`
Show Worked Solution

a.   `A=98+62=160`

`text{% Damaged items}\ = 8/200 xx 100 = 4text{%}`

`text{Cumulative % after damaged items = 96%}`

`B = 92+4=96`
 

b.   `text{The right hand side cumulative frequency percentage}`

`text{shows that 80% of all complaints received concern}`

`text{stock shortages and delivery fees.}`

`:.\ text{The manager will address stock shortages and delivery fees.}`


♦ Mean mark part (b) 41%.

Statistics, STD2 S5 2021 HSC 41

In a particular city, the heights of adult females and the heights of adult males are each normally distributed.

Information relating to two females from that city is given in Table 1.
 

The means and standard deviations of adult females and males, in centimetres, are given in Table 2.
 


 

A selected male is taller than 84% of the population of adult males in this city.

By first labelling the normal distribution curve below with the heights of the two females given in Table 1, calculate the height of the selected male, in centimetres, correct to two decimal places.  (4 marks)

 

Show Answers Only

`178.95 \ text{cm}`

Show Worked Solution

 

`z text{-score (175 cm, female)} = 2`

♦♦♦ Mean mark 16%.

`z text{-score (160.6 cm, female)} = -1`
 

`text{Find} \ mu \ text{of female heights:}`

`mu – sigma` `= 160.6`  
`mu + 2sigma` `= 175`  
`3 sigma` `= 175 – 160.6`  
`sigma` `= 14.4/3`  
  `= 4.8 \ text{cm}`  
`:. \ mu` `= 165.4 \ text{cm}`  

 

`text{Selected male’s height has} \ z text{-score} = 1`

`mu text{(male)} = 1.05 times 165.4 = 173.67`

`sigma \ text{(male)} = 1.1 times 4.8 = 5.28`

 

`:. \ text{Actual male height}` `= 173.67 + 5.28`  
  `= 178.95 \ text{cm}`  

Statistics, 2ADV S2 2021 HSC 17

For a sample of 17 inland towns in Australia, the height above sea level, `x` (metres), and the average maximum daily temperature, `y` (°C), were recorded.

The graph shows the data as well as a regression line.
 

     
 

The equation of the regression line is  `y = 29.2 − 0.011x`.

The correlation coefficient is  `r = –0.494`.

  1. i.  By using the equation of the regression line, predict the average maximum daily temperature, in degrees Celsius, for a town that is 540 m above sea level. Give your answer correct to one decimal place.  (1 mark)

  2. ii. The gradient of the regression line is −0.011. Interpret the value of this gradient in the given context.  (2 marks)

  3. The graph below shows the relationship between the latitude, `x` (degrees south), and the average maximum daily temperature, `y` (°C), for the same 17 towns, as well as a regression line.
     
     
         
     
    The equation of the regression line is  `y = 45.6 − 0.683x`.
  4. The correlation coefficient is  `r = − 0.897`.
  5. Another inland town in Australia is 540 m above sea level. Its latitude is 28 degrees south.
  6. Which measurement, height above sea level or latitude, would be better to use to predict this town’s average maximum daily temperature? Give a reason for your answer.  (1 mark)

Show Answers Only
  1. i.  `23.3°text(C)`
  2. ii. `text(See Worked Solutions)`
  3. `text(Latitude. Correlation coefficient shows a stronger relationship.)`
Show Worked Solution
a.i.    `y` `=29.2 – 0.011(540)`
    `=23.26`
    `=23.3°text{C  (1 d.p.)`
 

a.ii.  `text(On average, the average maximum daily temperature of)`

`text(inland towns drops by 0.011 of a degree for every metre)`

`text(above sea level the town is situated.)`
 

b.  `text(The correlation co-efficient of the regression line using)`

`text(latitude is significantly stronger than the equivalent)`

`text(co-efficient for the regression line using height above sea)`

`text(level.)`

`:.\ text(The equation using latitude is preferred.)`

Statistics, STD2 S4 2021 HSC 33

For a sample of 17 inland towns in Australia, the height above sea level, `x` (metres), and the average maximum daily temperature, `y` (°C), were recorded.

The graph shows the data as well as a regression line.
 

     
 

The equation of the regression line is  `y = 29.2 − 0.011x`.

The correlation coefficient is  `r = –0.494`.

  1. i.  By using the equation of the regression line, predict the average maximum daily temperature, in degrees Celsius, for a town that is 540 m above sea level. Give your answer correct to one decimal place.  (1 mark)

  2. ii. The gradient of the regression line is −0.011. Interpret the value of this gradient in the given context.  (2 marks)

  3. The graph below shows the relationship between the latitude, `x` (degrees south), and the average maximum daily temperature, `y` (°C), for the same 17 towns, as well as a regression line.
     
     
         
     
    The equation of the regression line is  `y = 45.6 − 0.683x`.
  4. The correlation coefficient is  `r = − 0.897`.
  5. Another inland town in Australia is 540 m above sea level. Its latitude is 28 degrees south.
  6. Which measurement, height above sea level or latitude, would be better to use to predict this town’s average maximum daily temperature? Give a reason for your answer.  (1 mark)

Show Answers Only
  1. i.  `23.3°text(C)`
  2. ii. `text(See Worked Solutions)`
  3. `text(Latitude. Correlation coefficient shows a stronger relationship.)`
Show Worked Solution
a.i.    `y` `=29.2 – 0.011(540)`
    `=23.26`
    `=23.3°text{C  (1 d.p.)`
♦♦ Mean mark part a.ii. 28%.

 
a.ii.  `text(On average, the average maximum daily temperature of)`

`text(inland towns drops by 0.011 of a degree for every metre)`

`text(above sea level the town is situated.)`
 

♦♦♦ Mean mark part b 18%.

b.  `text(The correlation co-efficient of the regression line using)`

`text(latitude is significantly stronger than the equivalent)`

`text(co-efficient for the regression line using height above sea)`

`text(level.)`

`:.\ text(The equation using latitude is preferred.)`

Trigonometry, 2ADV T1 2021 HSC 12

A right-angled triangle  `XYZ`  is cut out from a semicircle with centre `O`. The length of the diameter  `XZ`  is 16 cm and  `angle YXZ`  = 30°, as shown on the diagram.
 


 

  1. Find the length of  `XY`  in centimetres, correct to two decimal places.  (2 marks)

  2. Hence, find the area of the shaded region in square centimetres, correct to one decimal place.  (3 marks)

Show Answers Only
  1. `13.86 \ text{cm}`
  2. `45.1 \ text{cm}^2`
Show Worked Solution

 

a.    `cos 30^@` `=(XY)/16`
  `XY` `= 16 \ cos 30^@`
    `= 13.8564`
    `= 13.86 \ text{cm (2 d.p.)}`

 

b.    `text{Area of semi-circle}` `= 1/2 times pi r^2`
    `= 1/2 pi times 8^2`
    `= 100.531 \ text{cm}^2`

 

`text{Area of} \ Δ XYZ` `= 1/2 a b sin C`  
  `= 1/2 xx 16 xx 13.856 xx sin 30^@`  
  `= 55.42 \ text{cm}^2`  

 

`:. \ text{Shaded Area}` `= 100.531 – 55.42`  
  `= 45.111`  
  `= 45.1 \ text{cm}^2 \ text{(1 d.p.)}`  

Measurement, STD2 M6 2021 HSC 32

A right-angled triangle  `XYZ`  is cut out from a semicircle with centre `O`. The length of the diameter  `XZ`  is 16 cm and  `angle YXZ`  = 30°, as shown on the diagram.
 


 

  1. Find the length of  `XY`  in centimetres, correct to two decimal places.  (2 marks)

  2. Hence, find the area of the shaded region in square centimetres, correct to one decimal place.  (3 marks)

Show Answers Only
  1. `13.86 \ text{cm}`
  2. `45.1 \ text{cm}^2`
Show Worked Solution

 

a.    `cos 30^@` `=(XY)/16`
  `XY` `= 16 \ cos 30^@`
    `= 13.8564`
    `= 13.86 \ text{cm (2 d.p.)}`

 

b.    `text{Area of semi-circle}` `= 1/2 times pi r^2`
    `= 1/2 pi times 8^2`
    `= 100.531 \ text{cm}^2`

♦ Mean mark part (b) 36%.
`text{Area of} \ Δ XYZ` `= 1/2 ab\ sin C`  
  `= 1/2 xx 16 xx 13.856 xx sin 30^@`  
  `= 55.42 \ text{cm}^2`  

 

`:. \ text{Shaded Area}` `= 100.531-55.42`  
  `= 45.111`  
  `= 45.1 \ text{cm}^2 \ \ text{(1 d.p.)}`  

Trigonometry, 2ADV T1 2021 HSC 18

The diagram shows a triangle `ABC` where `AC` = 25 cm, `BC` = 16 cm, `angle BAC` = 28° and angle `ABC` is obtuse.
 


 

Find the size of the obtuse angle `ABC` correct to the nearest degree.  (3 marks)

Show Answers Only

`133°`

Show Worked Solution

`text(Using the sine rule:)`

`sin theta/25` `= (sin 28°)/16`
`sin theta` `= (25 xx sin 28°)/16`
`sin theta` `= 0.73355`
`theta` `= 47°`
 
`:. angleABC` `= 180-47`
  `= 133°`

Measurement, STD2 M6 2021 HSC 37

The diagram shows a triangle `ABC` where `AC` = 25 cm, `BC` = 16 cm, `angle BAC` = 28° and angle `ABC` is obtuse.
 


 

Find the size of the obtuse angle `ABC` correct to the nearest degree.  (3 marks)

Show Answers Only

`133°`

Show Worked Solution

`text(Using the sine rule:)`

♦♦ Mean mark 31%.
`sin theta/25` `= (sin 28°)/16`
`sin theta` `= (25 xx sin 28°)/16`
`sin theta` `= 0.73355`
`theta` `= 47°`
 
`:. angleABC` `= 180-47`
  `= 133°`

Statistics, 2ADV S3 2021 HSC 22

A random variable is normally distributed with mean 0 and standard deviation 1. The table gives the probability that this random variable lies between 0 and `z` for different values of `z`.
 

   

The probability values given in the table for different values of `z` are represented by the shaded area in the following diagram.
 

  1. Using the table, find the probability that a value from a random variable that is normally distributed with a mean of 0 and standard deviation 1 lies between 0.1 and 0.5.  (1 mark)

  2. Birth weights are normally distributed with a mean of 3300 grams and a standard deviation of 570 grams. By first calculating a `z`-score, find how many babies, out of 1000 born, are expected to have a birth weight greater than 3528 grams.  (3 marks)

Show Answers Only
  1. `0.1517`
  2. `345\ text(babies)`
Show Worked Solution

♦♦ Mean mark part (a) 29%.
COMMENT: Note the Advanced and Std2 questions varied slightly but used the same table and graph.
a.    `P(0.1 < x < 0.5)` `= 0.1915 – 0.0398`
    `= 0.1517`
 

b.   `mu = 330, sigma = 570`

♦ Mean mark part (b) 48%.
`ztext(-score)\ (3528)` `= (x – mu)/sigma`
  `= (3528 – 3300)/570`
  `= 0.4`

 

`P(ztext(-score) > 0.4)`  `= 0.5 – 0.1554`
  `= 0.3446`

 
`:.\ text(Expected babies > 3528 grams)`

`= 1000 xx 0.3446`

`= 344.6`

`~~ 345\ text(babies)`

Probability, 2ADV S1 2021 HSC 6 MC

There are 8 chocolates in a box. Three have peppermint centres (P) and five have caramel centres (C).

Kim randomly chooses a chocolate from the box and eats it. Sam then randomly chooses and eats one of the remaining chocolates.

A partially completed probability tree is shown.
 

What is the probability that Kim and Sam choose chocolates with different centres?

  1. `\frac{15}{64}`
  2. `\frac{15}{56}`
  3. `\frac{15}{32}`
  4. `\frac{15}{28}`
Show Answers Only

`D`

Show Worked Solution

 

`Ptext{(different centres)}` `= P text{(PC)} + P text{(CP)}`
  `=\frac{3}{8} · \frac{5}{7} + \frac{5}{8} · \frac{3}{7}`
  `= \frac{15}{56} + \frac{15}{56}`
  `= \frac{15}{28}`

 
`=> D`

Probability, STD2 S2 2021 HSC 11 MC

There are 8 chocolates in a box. Three have peppermint centres (P) and five have caramel centres (C).

Kim randomly chooses a chocolate from the box and eats it. Sam then randomly chooses and eats one of the remaining chocolates.

A partially completed probability tree is shown.
 

What is the probability that Kim and Sam choose chocolates with different centres?

  1. `\frac{15}{64}`
  2. `\frac{15}{56}`
  3. `\frac{15}{32}`
  4. `\frac{15}{28}`
Show Answers Only

`D`

Show Worked Solution

♦♦ Mean mark 35%.

 

`Ptext{(different centres)}` `= P text{(PC)} + P text{(CP)}`
  `=\frac{3}{8} · \frac{5}{7} + \frac{5}{8} · \frac{3}{7}`
  `= \frac{15}{56} + \frac{15}{56}`
  `= \frac{15}{28}`

 
`=> D`

Functions, 2ADV F2 2021 HSC 5 MC

Which of the following best represents the graph of  `y = 10 (0.8)^x`?
 

Show Answers Only

`A`

Show Worked Solution

`\text{By elimination:}`

`\text{When} \ x = 0 \ , \ y = 10(0.8) ^0 = 10`

`-> \ text{Eliminate B and D}`

`text(As)\ \ x→oo, \ y→0`

`-> \ text{Eliminate C}`

`=> A`

Algebra, STD2 A4 2021 HSC 10 MC

Which of the following best represents the graph of  `y = 10 (0.8)^x`?
 

Show Answers Only

`A`

Show Worked Solution

`\text{By elimination:}`

♦ Mean mark 41%.

`\text{When} \ x = 0 \ , \ y = 10(0.8) ^0 = 10`

`-> \ text{Eliminate B and D}`

`text(As)\ \ x→oo, \ y→0`

`-> \ text{Eliminate C}`

`=> A`

Statistics, 2ADV S2 2021 HSC 4 MC

The number of downloads of a song on each of twenty consecutive days is shown in the following graph.
 

Which of the following graphs best shows the cumulative number of downloads up to and including each day?

Show Answers Only

`C`

Show Worked Solution

`text{The gradient of the cumulative frequency histogram}`

`text{will increase gradually, be steepest at day 10 then}`

`text{decrease gradually.}`

`=> C`

Statistics, STD2 S1 2021 HSC 7 MC

The number of downloads of a song on each of twenty consecutive days is shown in the following graph.
 

Which of the following graphs best shows the cumulative number of downloads up to and including each day?

Show Answers Only

`C`

Show Worked Solution

`text{The gradient of the cumulative frequency histogram}`

♦ Mean mark 50%.

`text{will increase gradually, be steepest at day 10 then}`

`text{decrease gradually.}`

`=> C`

Statistics, STD2 S4 2020 HSC 36

A cricket is an insect. The male cricket produces a chirping sound.

A scientist wants to explore the relationship between the temperature in degrees Celsius and the number of cricket chirps heard in a 15-second time interval.

Once a day for 20 days, the scientist collects data. Based on the 20 data points, the scientist provides the information below.

  • A box-plot of the temperature data is shown.
     
       
  • The mean temperature in the dataset is 0.525°C below the median temperature in the dataset.
  • A total of 684 chirps was counted when collecting the 20 data points.

The scientist fits a least-squares regression line using the data `(x, y)`, where `x` is the temperature in degrees Celsius and `y` is the number of chirps heard in a 15-second time interval. The equation of this line is

`y = −10.6063 + bx`,

where `b` is the slope of the regression line.

The least-squares regression line passes through the point  `(barx, bary)`, where  `barx`  is the sample mean of the temperature data and  `bary`  is the sample mean of the chirp data.

Calculate the number of chirps expected in a 15-second interval when the temperature is 19° Celsius. Give your answer correct to the nearest whole number.  (5 marks)

Show Answers Only

`29\ text(chirps)`

Show Worked Solution

`y = −10.6063 + bx`

♦♦♦ Mean mark 18%.

`text(Find)\ b:`

`text(Line passes through)\ \ (barx, bary)`

`barx` `= 22 – 0.525`
  `= 21.475`
`bary` `= text(total chirps)/text(number of data points)`
  `= 684/20`
  `= 34.2`

 

`34.2` `= −10.6063 + b(21.475)`
`:.b` `= 44.8063/21.475`
  `~~ 2.0864`

 
`text(If)\ \ x = 19,`

`y` `= −10.6063 + 2.0864 xx 19`
  `= 29.03`
  `= 29\ text(chirps)`

Measurement, STD2 M6 2020 HSC 32

The diagram shows a regular decagon (ten-sided shape with all sides equal and all interior angles equal). The decagon has centre `O`.
 

The perimeter of the shape is 80 cm.

By considering triangle `OAB`, calculate the area of the ten-sided shape. Give your answer in square centimetres correct to one decimal place.  (4 marks)

Show Answers Only

`492.4\ text(cm²)`

Show Worked Solution

`angle AOB = 360/10 = 36^@`

♦♦ Mean mark 33%.

`AB = 80/10 = 8\ text(cm)`

`DeltaAOB\ text(is made up of 2 identical right-angled triangles)`
 

`tan 18^@` `= 4/x`
`x` `= 4/(tan 18^@)`

 

`:.\ text(Area of decagon)` `= 20 xx 1/2 xx 4/(tan 18^@) xx 4`
  `= 492.429…`
  `= 492.4\ text(cm²  (to 1 d.p.))`