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Data Analysis, GEN2 2024 NHT 1

Data was collected to investigate the behaviour of tides in Sydney Harbour.

There are usually two high tides and two low tides each day.

The variables in this study were:

  • Day: the day number in the sample
  • LLT: the height of the lowest low tide for that day (in metres)
  • HHT: the height of the highest high tide for that day (in metres)

Table 1 displays the data collected for a sample of 14 consecutive days in February 2021.

Table 1

\begin{array}{|c|c|c|}
\hline
 \rule{0pt}{2.5ex}\ \ \ \textit{Day}\ \ \ \rule[-1ex]{0pt}{0pt}& \textit{LLT (m)} & \textit{HHT (m)}\\
\hline \rule{0pt}{2.5ex}1 \rule[-1ex]{0pt}{0pt}& 0.43 & 1.65 \\
\hline \rule{0pt}{2.5ex}2 \rule[-1ex]{0pt}{0pt}& 0.49 & 1.55 \\
\hline \rule{0pt}{2.5ex}3 \rule[-1ex]{0pt}{0pt}& 0.55 & 1.44 \\
\hline \rule{0pt}{2.5ex}4 \rule[-1ex]{0pt}{0pt}& 0.61 & 1.42 \\
\hline \rule{0pt}{2.5ex}5 \rule[-1ex]{0pt}{0pt}& 0.68 & 1.42 \\
\hline \rule{0pt}{2.5ex}6 \rule[-1ex]{0pt}{0pt}& 0.73 & 1.42 \\
\hline \rule{0pt}{2.5ex}7 \rule[-1ex]{0pt}{0pt}& 0.72 & 1.42 \\
\hline \rule{0pt}{2.5ex}8 \rule[-1ex]{0pt}{0pt}& 0.65 & 1.47 \\
\hline \rule{0pt}{2.5ex}9 \rule[-1ex]{0pt}{0pt}& 0.57 & 1.55 \\
\hline \rule{0pt}{2.5ex}10 \rule[-1ex]{0pt}{0pt}& 0.48 & 1.64 \\
\hline \rule{0pt}{2.5ex}11 \rule[-1ex]{0pt}{0pt}& 0.39 & 1.74 \\
\hline \rule{0pt}{2.5ex}12 \rule[-1ex]{0pt}{0pt}& 0.30 & 1.83 \\
\hline \rule{0pt}{2.5ex}13 \rule[-1ex]{0pt}{0pt}& 0.25 & 1.90 \\
\hline \rule{0pt}{2.5ex}14 \rule[-1ex]{0pt}{0pt}& 0.22 & 1.92 \\
\hline
\end{array}

  1. For the \(H H T\) values in Table 1:
    1. Calculate the mean, in metres.
    2. Round your answer to one decimal place.   (1 mark)

    3. Calculate the standard deviation, in metres.
    4. Round your answer to three decimal places.    (1 mark)

  2. Use the \(HHT\) data from Table 1 to construct a boxplot on the grid below.    (2 marks)

     

  1. The five-number summary of the \(L L T\) data is shown in Table 2 below.
  2. Table 2

\begin{array}{|c|c|c|c|c|}
\hline \rule{0pt}{2.5ex}\textbf{Minimum} \rule[-1ex]{0pt}{0pt}& \ \ \textbf{Q1} \ \ & \textbf{Median} & \ \  \textbf{Q3} \ \ & \textbf{Maximum} \\
\hline \rule{0pt}{2.5ex}0.22 \rule[-1ex]{0pt}{0pt}& 0.39 & 0.52 & 0.65 & 0.73 \\
\hline
\end{array}

  1. Show that the minimum \(L L T\) value of 0.22 m is not an outlier.    (2 marks)

  2. A least squares line can be used to model the association between \(L L T\) and \(H H T\). In this model, \(H H T\) is the response variable.
  3. Use the data from Table 1 to determine the equation of this least squares line.
  4. Round the values of the intercept and slope to four significant figures.
  5. Write your answers in the boxes provided.    (2 marks)

Show Answers Only

a.i.   \(\text{Mean = 1.6}\)

a.ii.  \(\text{Std dev = 0.185}\)

b.   
               

c.   \(IQR (LLT) = 0.65-0.39=0.26\)

\(\text{Lower fence}\ =Q_1-1.5 \times IQR = 0.39-1.5 \times o.26=0\)

\(\text{Since 0.22 > 0, 0.22 is not an outlier.}\)

d.   \(HHT = 2.130 + (-1.054) \times LLT\)

Show Worked Solution

a.i.   \(\text{Mean = 1.6}\)

a.ii.  \(\text{Std dev = 0.185}\)
 

b.   \(\text{Order \(HHT\) data:}\)

\(1.42, 1.42, 1.42, [1.42], 1,44, 1.47, 1.55 | 1.55, 1.64, 1.65, [1.74],\)

\(1.83, 1.90, 1.92\)

\(\text{High = 1.92, Low = 1.42, \(Q_1=1.42, Q_3=1.74\), Median = 1.55}\)

 
c.   \(IQR (LLT) = 0.65-0.39=0.26\)

\(\text{Lower fence}\ =Q_1-1.5 \times IQR = 0.39-1.5 \times 0.26=0\)

\(\text{Since 0.22 > 0, 0.22 is not an outlier.}\)
 

d.   \(HHT\ \text{is the response \((y)\) variable.}\)

\(\text{By CAS:}\)

\(HHT = 2.130 + (-1.054) \times LLT\)

Data Analysis, GEN1 2024 NHT 3 MC

The dot plot below displays the number of errors made in a test, for a sample of 15 students.
 

The mean and standard deviation for the number of errors are closest to

  1. mean \(=2.60 \quad\) standard deviation \(=7.47\)
  2. mean \(=2.70 \quad\) standard deviation \(=7.47\)
  3. mean \(=8.00 \quad\) standard deviation \(=3.00\)
  4. mean \(=7.47 \quad\) standard deviation \(=2.60\)
  5. mean \(=7.47 \quad\) standard deviation \(=2.70\)
Show Answers Only

\(E\)

Show Worked Solution

\(\text{Using CAS (input dataset):}\)

\(3,3,4,5,6,7,8,8,8,9,10,10,10,10,11\)

\(\text{Mean = 7.47,}\ \ \sigma =2.70\)

\(\Rightarrow E\)

Data Analysis, GEN1 2024 VCAA 13-14 MC

A school runs an orientation program for new staff each January.

The time series plot below shows the number of new staff, new, for each year, year, from 2011 to 2022 (inclusive).
 

Part 1

The time series is smoothed using seven-median smoothing.

The smoothed value of new for the year 2016 is

  1. 10
  2. 11
  3. 12
  4. 13

 
Part 2

The number of new staff in 2023 is added to the total number of new staff from the previous 12 years.

For these 13 years, the mean number of new staff is 11 .

The number of new staff in 2023 is

  1. 11
  2. 16
  3. 17
  4. 19
Show Answers Only

Part 1: \(A\)

Part 2: \(B\)

Show Worked Solution

Part 1

\(\text{Ordered scores}\)

\begin{array} {|c|c|c|c|c|c|}
\hline 2018 & 2014 & 2013 & 2019 & 2015 & 2017 & 2016 \\
\hline 6 & 6 & 7 & 10 & 11 & 12 & 13  \\
\hline
\end{array}

\(\text{Smoothed value of }new\ \text{for the }year\ 2016\ \text{is 10.}\)

\(\Rightarrow A\)

 
Part 2

\(\text{mean}\) \(=\dfrac{\Sigma\text{scores}}{\text{Number of scores}}\)
\(11\) \(=\dfrac{x+14+11+7+6+11+13+12+6+10+15+12+10}{13}\)
\(x+127\) \(=143\)
\(x\) \(=16\)

  
\(\Rightarrow B\)

Data Analysis, GEN1 2022 VCAA 4 MC

The age, in years, of a sample of 14 possums is displayed in the dot plot below.
 

The mean and the standard deviation of age for this sample of possums are closest to

  1. mean = 4.25    standard deviation = 2.6
  2. mean = 4.8      standard deviation = 2.4
  3. mean = 4.8      standard deviation = 2.5
  4. mean = 4.9      standard deviation = 2.4
  5. mean = 4.9      standard deviation = 2.5
Show Answers Only

\(E\)

Show Worked Solution

\(\text{By calculator:}\) 

\(\text{Sample standard deviation} = 2.525…\)

\(\text{Mean} = 4.928…\)

\(\Rightarrow E\)

Data Analysis, GEN2 2023 VCAA 1

Data was collected to investigate the use of electronic images to automate the sizing of oysters for sale. The variables in this study were:

    • ID: identity number of the oyster
    • weight: weight of the oyster in grams (g)
    • volume: volume of the oyster in cubic centimetres (cm³)
    • image size: oyster size determined from its electronic image (in megapixels)
    • size: oyster size when offered for sale: small, medium or large

The data collected for a sample of 15 oysters is displayed in the table.
 

  1. Write down the number of categorical variables in the table.   (1 mark)

  2. Determine, in grams:
    1. the mean weight of all the oysters in this sample.   (1 mark)

    2. the median weight of the large oysters in this sample.   (1 mark)
  3. When a least squares line is used to model the association between oyster weight and volume, the equation is: 

    1. \(\textit{volume} = 0.780 + 0.953 \times \textit{weight} \)
    1. Name the response variable in this equation.   (1 mark)

    2. Complete the following sentence by filling in the blank space provided.   (1 mark)

      This equation predicts that, on average, each 10 g increase in the weight of an oyster is associated with a ________________ cm³ increase in its volume.
  1. A least squares line can also be used to model the association between an oyster's volume, in cm³, and its electronic image size, in megapixels. In this model, image size is the explanatory variable.
  2. Using data from the table, determine the equation of this least squares line. Use the template below to write your answer. Round the values of the intercept and slope to four significant figures.   (2 marks)

  3. The number of megapixels needed to construct an accurate electronic image of an oyster is approximately normally distributed.
  4. Measurements made on recently harvested oysters showed that:
    • 97.5% of the electronic images contain less than 4.6 megapixels
    • 84% of the electronic images contain more than 4.3 megapixels.
  1. Use the 68-95-99.7% rule to determine, in megapixels, the mean and standard deviation of this normal distribution.   (2 marks)

Show Answers Only

a.    \(\text{Categorical variables = 2 (ID and size)}\)

b.i.  \(\text{Mean weight}\ = \dfrac{\text{sum of oyster weights}}{15} = \dfrac{171.3}{15} = 11.42 \)

b.ii.  \(\text{Median}\ = 11.4 \)

c.i.   \(\text{Volume}\)

c.ii.  \(\text{Increase}\ = 0.953 \times 10 = 9.53\ \text{cm}^{3} \)

d.    \(\text{Volume}\ = 0.002857 + 2.571 \times \text{image size} \)

e.    \(s_x = 0.1 \)

\(\bar x = 4.4\) 

Show Worked Solution

a.    \(\text{Categorical variables = 2 (ID and size)}\)
 

b.i.  \(\text{Mean weight}\ = \dfrac{\text{sum of oyster weights}}{15} = \dfrac{171.3}{15} = 11.42 \)
 

b.ii.  \(\text{15 data points}\ \ \Rightarrow \ \ \text{Median = 8th data point (in order)}\)

 \(\text{Median}\ = 11.4 \)
 

c.i.   \(\text{Volume}\)
 

c.ii.  \(\text{Increase}\ = 0.953 \times 10 = 9.53\ \text{cm}^{3} \)
 

d.    \(\text{Input the image size column values}\ (x)\ \text{and volume} \)

\(\text{values}\ (y)\ \text{into the calculator:}\)

\(\textit{Volume}\ = 0.002857 + 2.571 \times \textit{image size} \)
 

e.    \(z\text{-score (4.6)}\ = 2\ \ \Rightarrow \bar x + 2 \times s_x = 4.6\ …\ (1)\)

\(z\text{-score (4.3)}\ = -1\ \ \Rightarrow \bar x-s_x = 4.3\ …\ (2)\)

\( (1)-(2) \)

\(3 s_x = 0.3 \ \ \Rightarrow\ \ s_x = 0.1 \)

\(\bar x = 4.4\) 

Data Analysis, GEN1 2023 VCAA 3 MC

Gemma’s favourite online word puzzle allows her 12 attempts to guess a mystery word. Her number of attempts for the last five days is displayed in the table below.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Day} \rule[-1ex]{0pt}{0pt} & \textbf{Number of attempts} \\
\hline
\rule{0pt}{2.5ex} 1 \rule[-1ex]{0pt}{0pt} & 8 \\
\hline
\rule{0pt}{2.5ex} 2 \rule[-1ex]{0pt}{0pt} & 11 \\
\hline
\rule{0pt}{2.5ex} 3 \rule[-1ex]{0pt}{0pt} & 5 \\
\hline
\rule{0pt}{2.5ex} 4 \rule[-1ex]{0pt}{0pt} & 6 \\
\hline
\rule{0pt}{2.5ex} 5 \rule[-1ex]{0pt}{0pt} & 9 \\
\hline
\end{array}

On day six, how many attempts can she make so that the mean number of attempts for these six days is exactly eight?

  1. 5
  2. 6
  3. 7
  4. 8
  5. 9
Show Answers Only

\(E\)

Show Worked Solution

\(\text{Let}\ x=\ \text{number of attempts on day 6}\)

\(8\) \(=\dfrac{x+8+11+5+6+9}{6}\)  
\(8 \times 6\) \(=x+39\)  
\(x\) \(=48-39\)  
  \(=9\)  

 
\(\Rightarrow E\)

CORE, FUR2 2021 VCAA 1

In the sport of heptathlon, athletes compete in seven events.

These events are the 100 m hurdles, high jump, shot-put, javelin, 200 m run, 800 m run and long jump.

Fifteen female athletes competed to qualify for the heptathlon at the Olympic Games.

Their results for three of the heptathlon events – high jump, shot-put and javelin – are shown in Table 1

  1. Write down the number of numerical variables in Table 1.   (1 mark)

  2. Complete Table 2 below by calculating the mean height jumped for the high jump, in metres, by the 15 athletes. Write your answer in the space provided in the table.   (1 mark)

  3. In shot-put, athletes throw a heavy spherical ball (a shot) as far as they can. Athlete number six, Jamilia, threw the shot 14.50 m.
  4. Calculate Jamilia's standardised score (`z`). Round your answer to one decimal place.   (1 mark)

  5. In the qualifying competition, the heights jumped in the high jump are expected to be approximately normally distributed.
  6. Chara's jump in this competition would give her a standardised score of  `z = –1.0`
  7. Use the 68–95–99.7% rule to calculate the percentage of athletes who would be expected to jump higher than Chara in the qualifying competition.   (1 mark)

  8. The boxplot below was constructed to show the distribution of high jump heights for all 15 athletes in the qualifying competition.

 

  1. Explain why the boxplot has no whisker at its upper end.   (1 mark)

  2. For the javelin qualifying competition (refer to Table 1), another boxplot is used to display the distribution of athlete's results.
  3. An athlete whose result is displayed as an outlier at the upper end of the plot is considered to be a potential medal winner in the event.
  4. What is the minimum distance that an athlete needs to throw the javelin to be considered a potential medal winner?   (2 marks)

Show Answers Only

  1. `3`
  2. `1.81`
  3. `0.5 \ text{(to d.p.)}`
  4. `84text(%)`
  5. `text{See Worked Solutions}`
  6. `46.89 \ text{m}`

Show Worked Solution

a.    `3 \ text{High jump, shot-put and javelin}`

 `text{Athlete number is not a numerical variable}`
  

b.     `text{High jump mean}`

`= (1.76 + 1.79 + 1.83 + 1.82 + 1.87 + 1.73 + 1.68 + 1.82 +`

`1.83 + 1.87 + 1.87 + 1.80 + 1.83 + 1.87 + 1.78) ÷ 15`

`= 1.81`
 

c.   `z text{-score} (14.50)` `= {14.50-13.74}/{1.43}`
    `= 0.531 …`
    `= 0.5 \ text{(to 1 d.p.)}`

 
d.  `P (z text{-score} > -1 ) = 84text(%)`
 

e.  `text{If the} \ Q_3 \ text{value is also the highest value in the data set,}`

`text{there is no whisker at the upper end of a boxplot.}`
 

f.  `text{Javelin (ascending):}`

`38.12, 39.22, 40.62, 40.88, 41.22, 41.32, 42.33, 42.41, `

`42.51, 42.65, 42.75, 42.88, 45.64, 45.68, 46.53`

`Q_1 = 40.88 \ \ , \ Q_3 = 42.88 \ \ , \ \ IQR = 42.88-40.88 = 2`

`text{Upper Fence}` `= Q_3 + 1.5  xx IQR`
  `= 42.88 + 1.5 xx 2`
  `= 45.88`

 
`:. \ text{Minimum distance = 45.89 m  (longer than upper fence value)}`

CORE, FUR1 2020 VCAA 1-3 MC

The times between successive nerve impulses (time), in milliseconds, were recorded.

Table 1 shows the mean and the five-number summary calculated using 800 recorded data values.
 


 

Part 1

The difference, in milliseconds, between the mean time and the median time is

  1.  10
  2.  70
  3.  150
  4.  220
  5.  230

 
Part 2

Of these 800 times, the number of times that are longer than 300 milliseconds is closest to

  1. 20
  2. 25
  3. 75
  4. 200
  5. 400

 
Part 3

The shape of the distribution of these 800 times is best described as

  1. approximately symmetric.
  2. positively skewed.
  3. positively skewed with one or more outliers.
  4. negatively skewed.
  5. negatively skewed with one or more outliers.
Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ D`

`text(Part 3:)\ C`

Show Worked Solution

Part 1

`text(Difference)` `= 220 -150`
  `= 70`

`=> B`
 

Part 2

`Q_3 = 300`

`:.\ text(Impulses longer than 300 milliseconds)`

`= 25text(%) xx 800`

`= 200`

`=> D`
 

Part 3

`text(Distribution has a long tail to the right)`

♦ Mean mark 50%.

`:.\ text(Positively skewed)`

`text(Upper fence)` `= Q_3 + 1.5 xx IQR`
  `= 300 + 1.5 (300 – 70)`
  `= 645`

 
`=> C`

Data Analysis, GEN2 2019 NHT 1

The table below displays the average sleep time, in hours, for a sample of 19 types of mammals.
 

  1. Which of the two variables, type of mammal or average sleep time, is a nominal variable?   (1 mark)

  2. Determine the mean and standard deviation of the variable average sleep time for this sample of mammals.
  3. Round your answer to one decimal place.   (1 mark)

  4. The average sleep time for a human is eight hours.
  5. What percentage of this sample of mammals has an average sleep time that is less than the average sleep time for a human.
  6. Round your answer to one decimal place.   (1 mark)

  7. The sample is increase in size by adding in the average sleep time of the little brown bat.
  8. Its average sleep time is 19.9 hours.
  9. By how many many hours will the range for average sleep time increase when the average sleep time for the little brown bat is added to the sample?   (1 mark)

Show Answers Only
  1. `text(type of mammal)`
  2. `text(mean)= 9.2 \ text(hours)`

     

    `sigma = 4.2 \ text(hours)`

  3. `31.6text(%)`
  4. `5.4 \ text(hours)`
Show Worked Solution

a.    `text(type of mammal is nominal)`

 
b.    `text(mean)= 9.2 \ text(hours) \ \ text{(by CAS)}`

`sigma = 4.2 \ text(hours) \ \ text{(by CAS)}`
 

c.    `text(Percentage)` `= (6)/(19) xx 100`
  `= 0.3157 …`
  `= 31.6text(%)`

 

d.    `text(Range increase)` `= 19.9-14.5`
  `= 5.4 \ text(hours)`

CORE, FUR1 2015 VCAA 3 MC

The dot plot below displays the difference between female and male life expectancy, in years, for a sample of 20 countries.
 

CORE, FUR1 2015 VCAA 3 MC
 

The mean (`barx`) and standard deviation (`s`) for this data are

A.   `text(mean)\ = 2.32` `\ \ \ \ \ text(standard deviation)\ = 5.25`
B.   `text(mean)\ = 2.38` `\ \ \ \ \ text(standard deviation)\ = 5.25`
C.   `text(mean)\ = 5.0` `\ \ \ \ \ text(standard deviation)\ = 2.0`
D.   `text(mean)\ = 5.25` `\ \ \ \ \ text(standard deviation)\ = 2.32`
E.   `text(mean)\ = 5.25` `\ \ \ \ \ text(standard deviation)\ = 2.38`
Show Answers Only

`E`

Show Worked Solution

`text(By calculator.)`

`=> E`

CORE, FUR1 2011 VCAA 6-8 MC

When blood pressure is measured, both the systolic (or maximum) pressure and the diastolic (or minimum) pressure are recorded.

Table 1 displays the blood pressure readings, in mmHg, that result from fifteen successive measurements of the same person's blood pressure.
 

core 2011 VCAA 6-8

Part 1

Correct to one decimal place, the mean and standard deviation of this person's systolic blood pressure measurements are respectively

A.   `124.9 and 4.4`

B.   `125.0 and 5.8`

C.   `125.0 and 6.0`

D.   `125.9 and 5.8`

E.   `125.9 and 6.0`

 

Part 2

Using systolic blood pressure (systolic) as the response variable, and diastolic blood pressure (diastolic) as the explanatory variable, a least squares regression line is fitted to the data in Table 1.

The equation of the least squares regression line is closest to

A.   `text(systolic) = 70.3 + 0.790 xx text(diastolic)`

B.   `text(diastolic) = 70.3 + 0.790 xx text(systolic)`

C.   `text(systolic) = 29.3 + 0.330 xx text(diastolic)`

D.   `text(diastolic) = 0.330 + 29.3 xx text(systolic)`

E.   `text(systolic) = 0.790 + 70.3 xx text(diastolic)`

 

Part 3

From the fifteen blood pressure measurements for this person, it can be concluded that the percentage of the variation in systolic blood pressure that is explained by the variation in diastolic blood pressure is closest to

A.   `25.8text(%)`

B.   `50.8text(%)`

C.   `55.4text(%)`

D.   `71.9text(%)`

E.   `79.0text(%)`

Show Answers Only

`text(Part 1:)\ E`

`text(Part 2:)\ A`

`text(Part 3:)\ A`

Show Worked Solution

`text(Part 1)`

`text{By calculator (using sample standard deviation)}`

`text{the results are: }`

`text(Mean = 125.9, std dev = 6.0)`

`=>E`

 

`text(Part 2)`

`text{By calculator (making sure diastolic values are}`

`text{the explanatory or}\ x text{-variable), the regression line}`

`text(can be expressed as follows,)`

`text(systolic) = 70.3 + 0.790 xx text(diastolic)`

`=>A`

 

`text(Part 3)`

`text{By calculator, the regression line (above) should}`

`text(have found) \ \ r^2 = 0.258,\ text(which means that)`

`text(25.8% of the variation in systolic pressure can be)`

`text(explained by variation in diastolic pressure.)`

`=>A`

CORE, FUR1 2008 VCAA 5 MC

A sample of 14 people were asked to indicate the time (in hours) they had spent watching television on the previous night. The results are displayed in the dot plot below.
 

    2008 5
 

Correct to one decimal place, the mean and standard deviation of these times are respectively

A.   `bar x=2.0\ \ \ \ \ s=1.5`

B.   `bar x=2.1\ \ \ \ \ s=1.5`  

C.   `bar x=2.1\ \ \ \ \ s=1.6`

D.   `bar x=2.6\ \ \ \ \ s=1.2`

E.   `bar x=2.6\ \ \ \ \ s=1.3` 

Show Answers Only

`C`

Show Worked Solution

`text(Data points are:)`

`0,0,0,1,1,2,2,2,2,3,3,4,4,5`

`text(By calculator (using sample standard deviation))`

`bar x=2.1,\ \ s=1.6`

`=>  C`

CORE, FUR1 2014 VCAA 3-5 MC

The following table shows the data collected from a sample of seven drivers who entered a supermarket car park. The variables in the table are:

distance – the distance that each driver travelled to the supermarket from their home
 

    • sex – the sex of the driver (female, male)
    • number of children – the number of children in the car
    • type of car – the type of car (sedan, wagon, other)
    • postcode – the postcode of the driver’s home.
       

Part 1

The mean,  `barx`, and the standard deviation, `s_x`, of the variable, distance, are closest to

A.  `barx = 2.5\ \ \ \ \ \ \s_x = 3.3`

B.  `barx = 2.8\ \ \ \ \ \ \s_x = 1.7`

C.  `barx = 2.8\ \ \ \ \ \ \s_x = 1.8`

D.  `barx = 2.9\ \ \ \ \ \ \s_x = 1.7`

E.  `barx = 3.3\ \ \ \ \ \ \s_x = 2.5`

 

Part 2

The number of categorical variables in this data set is

A.  `0`

B.  `1`

C.  `2`

D.  `3`

E.  `4`

 

Part 3

The number of female drivers with three children in the car is

A.  `0`

B.  `1`

C.  `2`

D.  `3`

E.  `4`

 

Show Answers Only

`text(Part 1:) \ C`

`text(Part 2:)\ D`

`text(Part 3:)\ B`

Show Worked Solution

`text(Part 1)`

`text(By calculator)`

`text(Distance)\ \ barx` `=2.8`
`s_x` `≈1.822`

 
`=>C`

 

`text(Part 2)`

♦♦ Mean mark 29%.
MARKER’S NOTE: Postcodes here are categorical variables. Ask yourself “Does it make sense to calculate the mean of this variable?” If the answer is “No”, the variable is categorical.

`text(Categorical variables are sex, type)`

`text(of car, and post code.)`

`=>D`

 

`text(Part 3)`

`text(1 female driver has 3 children.)`

`=>B`

CORE, FUR1 2012 VCAA 3 MC

The total weight of nine oranges is 1.53 kg.

Using this information, the mean weight of an orange would be calculated to be closest to

A.   115 g

B.   138 g

C.   153 g

D.   162 g

E.   170 g

Show Answers Only

`E`

Show Worked Solution
`text(Mean Weight)` `= text (Total weight)/ text (# Oranges)`
  `= 1.53/9`
  `= 0.17\ text(kg)`
  `= 170\ text(g)`

 
`rArr E`