SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

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:

        \(\textit{volume} = 0.780 + 0.953 \times \textit{weight} \)

    1. Name the response variable in this equation.
    2. Complete the following sentence by filling in the blank space provided.
    3. 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.
  2. 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.
  3. 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.
  4. The number of megapixels needed to construct an accurate electronic image of an oyster is approximately normally distributed.
  5. 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.
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.    \(z\text{-score (4.6)}\ = 2\ \ \Rightarrow \bar x + 2 \times s_x = 4.6\ …\ (1)\)

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

\( (1)-(2) \)

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

\(\bar x = 4.4\) 

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`

CORE, FUR2-NHT 2019 VCAA 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.

     

    Write your answers in the boxes provided below.

     

    Round your answer to one decimal place.  (1 mark)
     

         mean = hours                   standard deviation = hours
     

  3. The average sleep time for a human is eight hours.

     

    What percentage of this sample of mammals has an average sleep time that is less than the average sleep time for a human.

     

    Round your answer to one decimal place.  (1 mark)

  4. The sample is increase in size by adding in the average sleep time of the little brown bat.

     

    Its average sleep time is 19.9 hours.

     

    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 2018 VCAA 1-2 MC

The dot plot below displays the difference in travel time between the morning peak and the evening peak travel times for the same journey on 25 days.
 


 

Part 1

The percentage of days when there was five minutes difference in travel time between the morning peak and the evening peak travel times is

  1.    0%
  2.    5%
  3.  20%
  4.  25%
  5.  28%
     

Part 2

The median difference in travel time is

  1.  3.0 minutes.
  2.  3.5 minutes.
  3.  4.0 minutes.
  4.  4.5 minutes.
  5.  5.0 minutes.
Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ A`

Show Worked Solution

`text(Part 1)`

`text(Percentage)` `= text(days with 5 m difference)/text(total days) xx 100`
  `= 5/25 xx 100`
  `= 20\ text(%)`

`=> C`

 

`text(Part 2)`

`text(Median)` `= 13text(th data point)`
  `= 3\ text(minutes)`

`=> A`

CORE, FUR2 2017 VCAA 1

The number of eggs counted in a sample of 12 clusters of moth eggs is recorded in the table below.
 

   

  1. From the information given, determine

    1. the range  (1 mark)
    2. the percentage of clusters in this sample that contain more than 170 eggs.  (1 mark)

In a large population of moths, the number of eggs per cluster is approximately normally distributed with a mean of 165 eggs and a standard deviation of 25 eggs.

  1. Using the 68–95–99.7% rule, determine

    1. the percentage of clusters expected to contain more than 140 eggs  (1 mark)
    2. the number of clusters expected to have less than 215 eggs in a sample of 1000 clusters.  (1 mark)
  2. The standardised number of eggs in one cluster is given by  `z = –2.4`

     

    Determine the actual number of eggs in this cluster.  (1 mark)

Show Answers Only
    1. `72`
    2. `25text(%)`
    1. `84text(%)`
    2. `975`
  3. `105\ text(eggs)`
Show Worked Solution
a.i.    `text(Range)` `= 197 – 125`
    `= 72`

 

a.ii.   `text(3 clusters > 170 eggs)`

`:.\ text(Percentage)` `= 3/12 xx 100text(%)`
  `= 25text(%)`

 

b.i.    `ztext{-score (140)}` `= (140 – 165)/25`
    `= −1`

`:.\ text(Percentage over 140)`

`= 68 + 16`

`= 84text(%)`

 

b.ii.    `ztext{-score (215)}` `= (215 – 165)/25`
    `= 2`

 

`:.\ text(Percentage less than 215)`

`= 97.5text(%) xx 1000`

`= 975`

 

c.    `text(Using)\ \ \ z` `= (x – barx)/s`
  `−2.4` `= (x – 165)/25`
  `x` `= (−2.4 xx 25) + 165`
    `= 105\ text(eggs)`

CORE, FUR1 2017 VCAA 4 MC

The histogram below shows the distribution of the log10 (area), with area in square kilometres, of 17 islands.
 

The median area of these islands, in square kilometres, is between

  1. 2 and 3
  2. 3 and 4
  3. 10 and 100
  4. 1000 and 10 000
  5. 10 000 and 100 000
Show Answers Only

`D`

Show Worked Solution

`text(17 data points)\ =>\ text(median is 9th)`

`3 < text(log)_10(text(area)) < 4`

`1000 < text(area) < 10\ 000`

`=> D`

CORE, FUR2 2016 VCAA 1

The dot plot below shows the distribution of daily rainfall, in millimetres, at a weather station for 30 days in September.
 

 

  1. Write down the
    1. range  (1 mark)
    2. median.  (1 mark)
  2. Circle the data point on the dot plot above that corresponds to the third quartile `(Q_3).`  (1 mark)
  3. Write down the
    1. the number of days on which no rainfall was recorded  (1 mark)
    2. the percentage of days on which the daily rainfall exceeded 12 mm.  (1 mark)
  4. Use the grid below to construct a histogram that displays the distribution of daily rainfall for the month of September. Use interval widths of two with the first interval starting at 0.  (2 marks) 

 

Show Answers Only
    1. `17.8\ text(mm)`
    2. `0`
  2. `text(See Worked Solutions)`

     

    1. `16\ text(days)`
    2. `10\text(%)`
  4. `text(See Worked Solutions)`
Show Worked Solution
a.i.    `text(Range)` `=\ text(High) – text(Low)`
    `= 17.8 – 0`
    `= 17.8\ text(mm)`
     

a.ii.   `text(30 data points)`

`text(Median)` `= text{15th + 16th}/2`
  `= 0`

 

b.   

 

c.i.   `16\ text(days)`

c.ii.    `text(Percentage)` `= 3/30 xx 100`
    `= 10\ text(%)`

 

d.   

CORE, FUR2 2009 VCAA 1

Table 1 shows the number of rainy days recorded in a high rainfall area for each month during 2008.
 

CORE, FUR2 2009 VCAA 11

 

The dot plot below displays the distribution of the number of rainy days for the 12 months of 2008.
 

CORE, FUR2 2009 VCAA 12
 

  1. Circle the dot on the dot plot that represents the number of rainy days in April 2008.  (1 mark)
  2. For the year 2008, determine

     

    1. the median number of rainy days per month  (1 mark)
    2. the percentage of months that have more than 10 rainy days. Write your answer correct to the nearest per cent.  (1 mark) 
Show Answers Only
  1.  
    CORE, FUR2 2009 VCAA 12 Answer
    1. `15.5`
    2. `text(92%)`
Show Worked Solution
a.    CORE, FUR2 2009 VCAA 12 Answer

 

b.i.    `text(Median)` `= text{(6th + 7th)}/2`
    `=(15+16)/2`
    `=15.5`

 

b.ii.   `text(Months with more than 10 rainy days)`

`=11/12 xx text(100%)`

`=91.66…`

`=92text(%)\ \ text{(nearest %)}`

CORE, FUR2 2012 VCAA 1

The dot plot below displays the maximum daily temperature (in °C) recorded at a weather station on each of the 30 days in November 2011. 
  

CORE, FUR2 2012 VCAA 1
 

  1. From this dot plot, determine

     

    1. the median maximum daily temperature, correct to the nearest degree  (1 mark)
    2. the percentage of days on which the maximum temperature was less than 16 °C.

       

      Write your answer, correct to one decimal place.  (1 mark)

Records show that the minimum daily temperature for November at this weather station is approximately normally distributed with a mean of 9.5 °C and a standard deviation of 2.25 °C.

  1. Determine the percentage of days in November that are expected to have a minimum daily temperature less than 14 °C at this weather station.

     

    Write your answer, correct to one decimal place.  (1 mark)

Show Answers Only
    1. `20 text(°C)`
    2. `23.3text(%)`
  2. `97.5text(%)`
Show Worked Solution

a.i.   `text(The median is the average of the 15th and 16th)`

`\ \ text{data points (30 data points in total).}`.

`:.\ text(Median = 20 °C)`
 

a.ii.   `text{Percentage of days less than 16 °C}`

`= 7/30 xx 100text(%)`

`=23.333…`

`=23.3text{%  (to 1 d.p.)}`
 

b.    `z text{-score (14)}` `=(x-barx)/s`
    `=(14-9.5)/2.25`
    `=2`

 

CORE, FUR2 2012 VCAA 1 Answer 
 

`text(Percentage of days with a minimum below 14 °C)`

`= 95text(%) + 2.5text(%)`

`= 97.5text(%)`

CORE, FUR2 2013 VCAA 2

The development index for each country is a whole number between 0 and 100.

The dot plot below displays the values of the development index for each of the 28 countries that has a high development index.
 

CORE, FUR2 2013 VCAA 21 
 

  1. Using the information in the dot plot, determine each of the following.
     

     

    Core, FUR2 2013 VCAA 2_2   (1 mark)
     

  2. Write down an appropriate calculation and use it to explain why the country with a development index of 70 is an outlier for this group of countries.  (2 marks)
Show Answers Only
  1. `text(Mode = 78,  Range = 9)`
  2. `text(See Worked Solutions)`
Show Worked Solution

a.   `text(Mode = 78)`

`text(Range = 79 − 70 = 9)`

 

b.   `text(An outlier occurs if a data point is below)`

`Q_1 − 1.5 xx IQR`

 

`Q_1 = 75, \ \ Q_3 = 78, and IQR = 78-75=3`

`:. Q_1 − 1.5 xx IQR` `= 75 − 1.5 xx 3`
  `= 70.5`

 

`:. 70\ text{is an outlier  (70 < 70.5)}`

CORE, FUR1 2006 VCAA 5-6 MC

The distribution of test marks obtained by a large group of students is displayed in the percentage frequency histogram below.
 

Part 1

The pass mark on the test was 30 marks.

The percentage of students who passed the test is

A.     `7text(%)`

B.   `22text(%)`

C.   `50text(%)`

D.   `78text(%)`

E.   `87text(%)`

 

Part 2

The median mark lies between

A.   `35 and 40`

B.   `40 and 45`

C.   `45 and 50`

D.   `50 and 55`

E.   `55 and 60`

Show Answers Only

`text (Part 1:)\ D`

`text (Part 2:)\ B`

Show Worked Solution

`text (Part 1)`

`text(Adding up the percentage bars above 30)`

`=7+11+14+16+18+12`

`=78text(%)`

`rArr D`

 

`text (Part 2)`

`text(Adding up the percentage bars from the left,)`

`text(the 50th and 51st percentile lie in the 40–45)`

`text(mark interval.)`

`rArr B`

CORE, FUR1 2007 VCAA 1-2 MC

The dot plot below shows the distribution of the number of bedrooms in each of 21 apartments advertised for sale in a new high-rise apartment block.
 

 

Part 1

The mode of this distribution is

A.   `1`

B.   `2`

C.   `3`

D.   `7`

E.   `8`

 

Part 2

The median of this distribution is

A.   `1`

B.   `2`

C.   `3`

D.   `4`

E.   `5`

Show Answers Only

`text (Part 1:)\ A`

`text (Part 2:)\ B`

Show Worked Solution

`text (Part 1)`

`rArr A`

 

`text (Part 2)`

 `text(The median of 21 data points is the 11th value.)`

`:.\ text(Median) = 2`

`rArr B`

CORE, FUR1 2008 VCAA 8-9 MC

The weights (in g) and lengths (in cm) of 12 fish were recorded and plotted in the scatterplot below. The least squares regression line that enables the weight of these fish to be predicted from their length has also been plotted.
 

Part 1

The least squares regression line predicts that the weight (in g) of a fish of length 30 cm would be closest to

A.   `240`

B.   `252`

C.   `262`

D.   `274`

E.   `310`

 
Part 2

The median weight (in g) of the 12 fish is closest to

A.   `346`

B.   `375`

C.   `440`

D.   `450`

E.   `475`

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text{The regression line crosses the 30cm length (on the}`

`xtext{-axis) at approx 262.}`

`=>C`

 

`text(Part 2)`

`text(12 weight data points – the median will be the average of)`

`text(the 6th and 7th.)`

`text(From the graph,)`

`text(6th highest weight = 430 g)`

`text(7th highest weight = 450 g)`

`:.\ text(Median)` `=(430+450)/2`
  `=440\ text(g)`

`=> C` 

CORE, FUR1 2011 VCAA 1-3 MC

The histogram below displays the distribution of the percentage of Internet users in 160 countries in 2007.
 

Part 1

The shape of the histogram is best described as

A.   approximately symmetric.

B.   bell shaped. 

C.   positively skewed.

D.   negatively skewed.

E.   bi-modal.

 

Part 2

The number of countries in which less than 10% of people are Internet users is closest to

A.   `10`

B.   `16`

C.   `22`

D.   `32`

E.   `54`

 

Part 3

From the histogram, the median percentage of Internet users is closest to

A.   `10text(%)`

B.   `15text(%)`

C.   `20text(%)`

D.   `30text(%)`

E.   `40text(%)`

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ E`

`text(Part 3:)\ C`

Show Worked Solution

`text(Part 1)`

`text(The shape of the histogram has a definite tail)`

`text(on the right side which means it is positively)`

`text(skewed.)`

`=>  C`

 

`text(Part 2)`

`text(The histogram shows that 32% of countries fall)`

`text(between 0–5%, and 22% fall between 5–10%.)`

`:.\ text(Users below 10%)`

`= 32 + 22`

`= 54 text(%)`

`=>  E`

 

`text(Part 3)`

♦ Mean mark 45%.

`text(Total countries = 160)`

`text(Adding bars from the left hand side, there are)`

`text{80 countries in the first 4 bars (i.e. half of 160).}`

`:.\ text(Median is closest to 20%)`

`=>  C`