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Statistics, STD1 2026 HSC 22

The stem-and-leaf plot shows the age of 24 volunteers at a charity event.
 

\(\begin{array}{c|l}
\text{Stem} & \quad \quad \text{Leaf} \\
\hline \rule{0pt}{2.5ex}1 & \ 3\ \ \ \ 5 \\
2 & \ 0\ \ \ \ 2\ \ \ \ 5 \\
3 & \ 0\ \ \ \ 2\ \ \ \ 4\ \ \ \ 6 \\
4 & \ 1\ \ \ \ 3\ \ \ \ 3\ \ \ \ 7\ \ \ \ 8\ \ \ \ 9 \\
5 & \ 0\ \ \ \ 2\ \ \ \ 2\ \ \ \ \textbf{X} \\
6 & \ 1\ \ \ \ 4\ \ \ \ 6\ \  \\
7 & \ 0\ \ \ \ 2\ \  \\
\end{array} \qquad \qquad \boxed{\begin{array}{l} \text{KEY:} \\ 3\,|\,2=32 \end{array}}\)
 

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

    --- 2 WORK AREA LINES (style=lined) ---

  2. It is known that the upper quartile of the dataset is 55.
  3. Find the value of \(\textbf{X}\).   (2 marks)

    --- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

a.    \(\text{Approximately symmetric}\)

b.    \(\textbf{X}=8\)

Show Worked Solution

a.    \(\text{Approximately symmetric}\)
 

b.    \(\text{24 data values}\ \Rightarrow\ Q_3\ \text{is the average of the 18th and 19th values}\)

\(\text{18th value}=52, \ \ \text{19th value}=50+\textbf{X}\)

\(\dfrac{52+(50+\textbf{X})}{2}\) \(=55\)
\(102+\textbf{X}\) \(=110\)
\(\textbf{X}\) \(=8\)

Filed Under: Displaying Data - Other Charts Tagged With: Band 5, Band 6, smc-6531-10-Stem-and-Leaf

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