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Statistics, STD2 S1 2016 VCE-G 2*

A weather station records daily maximum temperatures

The five-number summary for the distribution of maximum temperatures for the month of February is displayed in the table below. 

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt} & \textbf{Temperature (°C)} \\
\hline
\rule{0pt}{2.5ex} \text{Minimum} \rule[-1ex]{0pt}{0pt} & 16 \\
\hline
\rule{0pt}{2.5ex} Q_1 \rule[-1ex]{0pt}{0pt} & 21 \\
\hline
\rule{0pt}{2.5ex} \text{Median} \rule[-1ex]{0pt}{0pt} & 25 \\
\hline
\rule{0pt}{2.5ex} Q_3 \rule[-1ex]{0pt}{0pt} & 31 \\
\hline
\rule{0pt}{2.5ex} \text{Maximum} \rule[-1ex]{0pt}{0pt} & 39 \\
\hline
\end{array}

  1. Use the five-number summary above to construct a boxplot on the grid below.   (1 mark)
      


  1. Show, using calculations, that there are no outliers in the dataset.   (2 marks)

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Show Worked Solution
a.   

 

b.   `IQR = Q_3-Q_1=31-21=10`

`text(Lower fence)` `= Q_1-1.5 xx IQR`
  `= 21-1.5 xx 10`
  `= 6`

 

`text(Upper fence)` `= Q_3 + 1.5 xx IQR`
  `= 31 + 1.5 xx 10`
  `= 46`

 

`text{Since 6 < 16 (minimum) and 46 > 39 (maximum)}`

`=>\ \text{There are no outliers.}`

Statistics, STD2 S1 2019 VCE-G 2*

The five-number summary below was determined from the sleep time, in hours, of a sample of 59 types of mammals.

\begin{array} {|l|c|}
\hline
\rule{0pt}{2.5ex} \ \ \ \textbf{Statistic} \rule[-1ex]{0pt}{0pt} & \textbf{Sleep time (hours)} \\
\hline
\rule{0pt}{2.5ex} \text{minimum} \rule[-1ex]{0pt}{0pt} & \text{2.5} \\
\hline
\rule{0pt}{2.5ex} \text{first quartile} \rule[-1ex]{0pt}{0pt} & \text{8.0} \\
\hline
\rule{0pt}{2.5ex} \text{median} \rule[-1ex]{0pt}{0pt} & \text{10.5} \\
\hline
\rule{0pt}{2.5ex} \text{third quartile} \rule[-1ex]{0pt}{0pt} & \text{13.5} \\
\hline
\rule{0pt}{2.5ex} \text{maximum} \rule[-1ex]{0pt}{0pt} & \text{20.0} \\
\hline
\end{array}

  1. Show with calculations, that a boxplot constructed from this five-number summary will not include outliers.   (2 marks)
  2. Construct the boxplot below.   (1 mark)
     
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  1. `text(Proof (See Worked Solution))`
  2.  
Show Worked Solution

a.    `IQR = Q_3-Q_1 = 13.5-8.0 = 5.5`

`text(Lower fence)` `= Q_1-1.5 xx IQR`
  `= 8-1.5 xx 5.5`
  `= -0.25`

 

`text(Upper fence)` `= Q_3 + 1.5 xx IQR`
  `= 13.5 + 1.5 xx 5.5`
  `= 21.75`

 
`text(S) text(ince) \ -0.25 < 2.5 \ text{(minimum value) and} \ 21.75 > 20.0 \ text{(maximum value)}`

`=> \ text(no outliers)`
 

b. 

Statistics, STD2 S1 2019 HSC 39

Two netball teams, Team A and Team B, each played 15 games in a tournament. For each team, the number of goals scored in each game was recorded.

The frequency table shows the data for Team A.
 


 

The data for Team B was analysed to create the box-plot shown.
 

 
 

Compare the distributions of the number of goals scored by the two teams. Support your answer with the construction of a box-plot for the data for Team A.  (5 marks)

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`text(See Worked Solution)`

Show Worked Solution

`text(Team A: High = 28, Low = 19,)\ Q_1 = 23, Q_3 = 27,\ text{Median = 26}`
 

`text(Team A’s distribution is negatively skewed while)`

♦♦ Mean mark 28%.

`text(Team B’s distribution is slightly positively skewed.)`

`text(The standard deviation of Team A’s distribution is)`

`text(smaller than Team B, as both its IQR and range is)`

`text(smaller.)`

`text(Team B is a more successful team at scoring goals)`

`text(as each value in its 5-point summary is higher than)`

`text(Team A’s equivalent value.)`

Statistics, STD2 S1 2006 HSC 24c

The heights of the 60 members of a choir were recorded. These results were grouped and then displayed as a cumulative frequency histogram and polygon.

The shortest person in the choir is 140 cm and the tallest is 190 cm.
 

2UG-2006-24c1

Draw an accurate box-and-whisker plot to represent the data.  (3 marks)

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`text{See Worked Solutions}`

Show Worked Solution

`text(Low) = 140`

`text(High) = 190`

`text(Median) = 150\ \ \ \ text{(# People = 30)}`

`Q_1 = 145\ \ \ \ text{(# People = 15)}`

`Q_3 = 170\ \ \ \ text{(# People = 45)}`

`text(Box and Whisker)`

HSC Data 13

Statistics, STD2 S1 2014 HSC 29c

Terry and Kim each sat twenty class tests. Terry’s results on the tests are displayed in the box-and-whisker plot shown in part (i).
 

  1. Kim’s  5-number summary for the tests is  67,  69,  71,  73,  75.

     

    Draw a box-and-whisker plot to display Kim’s results below that of Terry’s results.   (1 mark)
     
         

  2. What percentage of Terry’s results were below 69?     (1 mark)

  3. Terry claims that his results were better than Kim’s. Is he correct?

     

    Justify your answer by referring to the summary statistics and the skewness of the distributions.    (4 marks)

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  1. `text(See Worked Solutions)`
  2. `text(50%)`
  3. `text(See Worked Solutions)`
Show Worked Solution
i.    

 

♦ Mean mark 39%

ii.  `text(50%)`

 

iii.  `text(Terry’s results are more positively skewed than)`

♦♦ Mean mark 29%
COMMENT: Examiners look favourably on using language of location in answers, particularly the areas they have specifically pointed students towards (skewness in this example).

`text(Kim’s and also have a higher limit high.)`

`text(However, Kim’s results are more consistent,)`

`text(showing a tighter IQR. They also have a)`

`text(significantly higher median than Terry’s and)`

`text(are evenly skewed.)`

`:.\ text(Kim’s results were better.)`