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Interpreting Data, SM-Bank 017

Body mass index (BMI), in kilograms per square metre, was recorded for a sample of 32 men and displayed in the ordered stem plot below.
  

  1. Describe the shape of the distribution.  (1 mark)

  2. Determine the median BMI for this group of men.  (1 mark)

  3. People with a BMI of 25 or over are considered to be overweight.
  4. What percentage of these men would be considered to be overweight?  (1 mark)

Show Answers Only

a.    \(\text{Positively skewed}\)

b.    \(24.55\)

c.    \(37.5\%\)

Show Worked Solution

a.   \(\text{The tail is to the right, therefore positively skewed}\)
 

b.   \(32\ \text{data points}\)

\(\text{Median}\) \(=\dfrac{\text{(16th + 17th)}}{2}\)
  \(=\dfrac{ (24.5 + 24.6)}{2}\)
  \(= 24.55\)

 

c.    \(\text{Percentage}\) \(=\dfrac{12}{32}\times 100\)
    \(=37.5\%\)

Interpreting Data, SM-Bank 015 MC

The back-to-back ordered stem-and-leaf plot below shows the distribution of maximum temperatures (in °Celsius) of two towns, Beachside and Flattown, over 21 days in January.
 


 

For this distribution, which of the following is not true?

  1. The range of temperatures for Flattown is greater than the range of temperatures for Beachside.
  2. The median temperature for Beachside is lower than the median temperature for Flatttown.
  3. The distribution of temperatures for Beachside is positively skewed.
  4. The maximum temperatures for Flattown are generally lower than those of Beachside.
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Options A}\ \rightarrow\ \text{Flattown Range}=28,\ \ \text{Beachside Range}=23\ \checkmark\)

\(\text{Options B}\ \rightarrow\ \text{Flattown Median}=37,\ \ \text{Beachside Median}=23\ \checkmark\)

\(\text{Options C}\ \rightarrow \ \text{Beachside distribution has a tail to the right, so positively skewed}\ \checkmark\)

\(\text{Options D}\ \rightarrow \ \text{Flattown has 8 max temps that are}\geq\ \text{to those of Beachside.  ×}\)
 

\(\Rightarrow D\)

Interpreting Data, SM-Bank 006

Hannah is planning an Australian snowboarding trip this winter and is using the chart below to help decide when she should take her holidays and where she should go.
 

Hannah wishes to compare the 2 resorts using statistical information.

  1. Complete the statistical information in the table below.  (2 marks)
     
    \begin{array} {|l|c|c|}
    \hline
    \rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt} &\ \ \ \ \ \ \  \textbf{Resort 1}\ \ \ \ \ \ \  \rule[-1ex]{0pt}{0pt} &\ \ \ \ \ \ \  \textbf{Resort 2}\ \ \ \ \ \ \  \\
    \hline
    \rule{0pt}{2.5ex} \textbf{Range of snowfall (cm)} \rule[-1ex]{0pt}{0pt}&  & \\
    \hline
    \rule{0pt}{2.5ex} \textbf{Mean of snowfall (cm)} \rule[-1ex]{0pt}{0pt} &   &  \\
    \hline
    \rule{0pt}{2.5ex} \textbf{Median of snowfall (cm)} \rule[-1ex]{0pt}{0pt} & &  \\
    \hline
    \end{array}
  2. Using your results in the table above (a), which resort should Hannah choose to visit for her snowboarding holiday?
    Justify your answer with at least 1 reference to the table.  (1 mark)

  3. Complete the table below, and use it to decide in which month Hannah should book her snowboarding holiday?
    Justify your answer with at least 1 reference to the table and 1 to the graph.  (3 marks)

    \begin{array} {|l|c|c|}
    \hline
    \rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt} &\ \ \ \ \ \ \  \textbf{Mean Snowfall}\ \ \ \ \ \ \    \\
    \hline
    \rule{0pt}{2.5ex} \textbf{June} \rule[-1ex]{0pt}{0pt}& \\
    \hline
    \rule{0pt}{2.5ex} \textbf{July} \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} \textbf{August} \rule[-1ex]{0pt}{0pt} & \\
    \hline
    \rule{0pt}{2.5ex} \textbf{September} \rule[-1ex]{0pt}{0pt} & \\
    \hline
    \end{array}

Show Answers Only

a.

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt} &\ \ \ \ \ \ \  \textbf{Resort 1}\ \ \ \ \ \ \  \rule[-1ex]{0pt}{0pt} &\ \ \ \ \ \ \  \textbf{Resort 2}\ \ \ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex} \textbf{Range of snowfall (cm)} \rule[-1ex]{0pt}{0pt} &7 &10\\
\hline
\rule{0pt}{2.5ex} \textbf{Mean of snowfall (cm)} \rule[-1ex]{0pt}{0pt} & 10.5  & 10.5 \\
\hline
\rule{0pt}{2.5ex} \textbf{Median of snowfall (cm)} \rule[-1ex]{0pt}{0pt} & 14 & 13.5 \\
\hline
\end{array}

b.    \(\text{See worked solution}\)

c.  

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt} &\ \ \ \ \ \ \  \textbf{Mean Snowfall}\ \ \ \ \ \ \    \\
\hline
\rule{0pt}{2.5ex} \textbf{June} \rule[-1ex]{0pt}{0pt}& 6.5 \\
\hline
\rule{0pt}{2.5ex} \textbf{July} \rule[-1ex]{0pt}{0pt} & 13.5 \\
\hline
\rule{0pt}{2.5ex} \textbf{August} \rule[-1ex]{0pt}{0pt} & 16\\
\hline
\rule{0pt}{2.5ex} \textbf{September} \rule[-1ex]{0pt}{0pt} & 8\\
\hline
\end{array}

\(\text{See worked solution}\)

Show Worked Solution

a.

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt} &\ \ \ \ \ \ \  \textbf{Resort 1}\ \ \ \ \ \ \  \rule[-1ex]{0pt}{0pt} &\ \ \ \ \ \ \  \textbf{Resort 2}\ \ \ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex} \textbf{Range of snowfall (cm)} \rule[-1ex]{0pt}{0pt} & 14-7=7 & 17-7=10\\
\hline
\rule{0pt}{2.5ex} \textbf{Mean of snowfall (cm)} \rule[-1ex]{0pt}{0pt} & \dfrac{6+14+13+9}{4}=10.5  & \dfrac{7+11+17+7}{4}=10.5 \\
\hline
\rule{0pt}{2.5ex} \textbf{Median of snowfall (cm)} \rule[-1ex]{0pt}{0pt} & \dfrac{11+17}{2}=14 & \dfrac{14+13}{2}=13.5 \\
\hline
\end{array}

 

b.    \(\text{The mean snowfall for both resorts is the same.}\)

\(\text{The median snowfall for Resort 2 is higher than Resort 1.}\)

\(\text{The range of snowfall for Resort 2 is higher than Resort 1.}\)

\(\text{Based on these findings Hannah should choose Resort 2.}\)

c.   

\begin{array} {|l|c|c|}
\hline
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt} &\ \ \ \ \ \ \  \textbf{Mean Snowfall}\ \ \ \ \ \ \    \\
\hline
\rule{0pt}{2.5ex} \textbf{June} \rule[-1ex]{0pt}{0pt}& \dfrac{6+7}{2}=6.5 \\
\hline
\rule{0pt}{2.5ex} \textbf{July} \rule[-1ex]{0pt}{0pt} & \dfrac{14+13}{2}=13.5 \\
\hline
\rule{0pt}{2.5ex} \textbf{August} \rule[-1ex]{0pt}{0pt} & \dfrac{15+17}{2}=16\\
\hline
\rule{0pt}{2.5ex} \textbf{September} \rule[-1ex]{0pt}{0pt} & \dfrac{9+7}{2}=8\\
\hline
\end{array}

 
\(\text{Based on both the information in the graph and the table above,}\)

\(\text{Hannah should holiday in August.}\)

\(\text{The mean snowfall is highest in this month from the table}\)

\(\text{and, from the graph, Resort 2 has its highest snowfall in August which is 3 cm}\)

\(\text{than Resort 1’s highest in July.}\)

Interpreting Data, SM-Bank 005

While packaging cookies for the Easter show, Johnny recorded the number of broken cookies in each batch.

Broken cookies per batch

\(05\ ,\ 12\ ,\ 09\ ,\ 02\ ,\ 31\ ,\ 11\ ,\ 10\ ,\ 18\ ,\ 20\)

\(25\ ,\ 23\ ,\ 06\ ,\ 15\ ,\ 21\ ,\ 30\ ,\ 35\ ,\ 19\ ,\ 49\)

 

  1. Use the data to complete the stem-and-leaf plot below.  (1 mark)
     

  2. What is the range of broken cookies?  (1 mark)
  3. What is the median number of broken cookies?  (1 mark)
  4. Is the distribution symmetrical, positively skewed or negatively skewed?  (1 mark)
Show Answers Only

a.

b.    \(47\)

c.    \(18.5\)

d.    \(\text{Positively skewed}\)

Show Worked Solution

a.

b.    \(\text{Range} = 49-2=47\)

c.    \(\text{18 scores → Median}\)

\(=\dfrac{\text{9th + 10th}}{2}\)

\(=\dfrac{18+19}{2}=18.5\)

d.    \(\text{Positively skewed}\)