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Statistics, STD1 S1 2023 HSC 11

A company employs 50 people.

The annual income of the employees is shown in the grouped frequency distribution table.

\begin{array} {|c|c|c|c|}
\hline
\textit{Annual income} & \textit{Class centre} & \textit{Number of} & fx \\ \text{(\$)} & (x) & \textit{employees}\ (f) &  \\
\hline
\rule{0pt}{2.5ex} \text{40 000 – 49 999} \rule[-1ex]{0pt}{0pt} & 45\ 000 & 12 & 540\ 000 \\
\hline
\rule{0pt}{2.5ex} \text{50 000 – 59 999} \rule[-1ex]{0pt}{0pt} & 55\ 000 & 13 & 715\ 000 \\
\hline\rule{0pt}{2.5ex} \text{60 000 – 69 999} \rule[-1ex]{0pt}{0pt} & 65\ 000 & 15 & A \\
\hline\rule{0pt}{2.5ex} \text{70 000 – 79 999} \rule[-1ex]{0pt}{0pt} & 75\ 000 & 7 & 525\ 000 \\
\hline\rule{0pt}{2.5ex} \text{80 000 – 89 999} \rule[-1ex]{0pt}{0pt} & 85\ 000 & 3 & 255\ 000 \\
\hline
\hline\rule{0pt}{2.5ex}  \rule[-1ex]{0pt}{0pt} &  & \textit{Total}\ = 50 & \textit{Total = B} \\
\hline
\end{array}  

  1. What are the values of \(A\) and \(B\)?  (2 marks)

  2. Find the mean for this distribution.  (1 mark)

Show Answers Only

a.    \(A=$975\ 000\), \(B=$3\ 010\ 000\)

b.    \($60\ 200\)

Show Worked Solution

a.    \(A=65\ 000\times 15 = $975\ 000\)

\(B=540\ 000+715\ 000+975\ 000+525\ 000+255\ 000=$3\ 010\ 000\)

  

b.    \(\text{Mean}=\dfrac{\text{Total }fx}{\text{Total }f}=\dfrac{3\ 010\ 000}{50}=$60\ 200\)

♦♦♦ Mean mark (b) 12%.

Statistics, STD2 S1 2016 HSC 21 MC

A grouped data frequency table is shown.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Class Interval} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \textit{Frequency}\ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex} \text{1 – 5} \rule[-1ex]{0pt}{0pt} & 3 \\
\hline
\rule{0pt}{2.5ex} \text{6 – 10} \rule[-1ex]{0pt}{0pt} & 6 \\
\hline
\rule{0pt}{2.5ex} \text{11 – 15} \rule[-1ex]{0pt}{0pt} & 8 \\
\hline
\rule{0pt}{2.5ex} \text{16 – 20} \rule[-1ex]{0pt}{0pt} & 9 \\
\hline
\end{array}

What is the mean for this set of data?

  1.    6.5
  2.    10.5
  3.    11.9
  4.    12.4
Show Answers Only

`=> D`

Show Worked Solution

`text(Using the centre of each class interval:)`

♦ Mean mark 43%.
`text(Mean)` `= (3 xx 3 + 8 xx 6 + 13 xx 8 + 18 xx 9)/(3 + 6 + 8 + 9)`
  `= 12.42…`

`=> D`

Statistics, STD2 S1 2006 HSC 23c

Vicki wants to investigate the number of hours spent on homework by students at her high school.

  1. Briefly describe a valid method of randomly selecting 200 students for a sample.  (1 mark)

  2. Vicki chooses her sample and asks each student how many hours (to the nearest hour) they usually spend on homework during one week.

     

    The responses are shown in the frequency table.
     
         2UG-2006-23c

    What is the mean amount of time spent on homework?  (2 marks)

Show Answers Only
  1. `text(A valid method would be using a stratified sample.)`

     

    `text(The number of students sampled in each year is)`

     

    `text(proportional to the size of each year.)`

  2. `text(7.275 hours)`
Show Worked Solution

i.   `text(A valid method would be using a stratified sample.)`

`text(The number of students sampled in each year is)`

`text(proportional to the size of each year.)`

MARKER’S COMMENT: This “routine” exercise of finding a mean from grouped data was incorrectly answered by most students! The best responses copied the table and inserted a class-centre column (see solution).

 

ii.    2UG-2006-23c Answer

 

`text(Mean)` `= text(Sum of Scores) / text(Total scores)`
  `= 1455/200`
  `= 7.275\ text(hours)`

Statistics, STD2 S1 2014 HSC 14 MC

Twenty Year 12 students were surveyed. These students were asked how many hours of sport they play per week, to the nearest hour.

The results are shown in the frequency table. 

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Hours per week} \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \textit{Frequency}\ \ \ \ \  \\
\hline
\rule{0pt}{2.5ex} \text{0 – 2} \rule[-1ex]{0pt}{0pt} & 5 \\
\hline
\rule{0pt}{2.5ex} \text{3 – 5} \rule[-1ex]{0pt}{0pt} & 10 \\
\hline
\rule{0pt}{2.5ex} \text{6 – 8} \rule[-1ex]{0pt}{0pt} & 3 \\
\hline
\rule{0pt}{2.5ex} \text{9 – 11} \rule[-1ex]{0pt}{0pt} & 2 \\
\hline
\end{array}

 What is the mean number of hours of sport played by the students per week?

  1.    3.3
  2.    4.3
  3.    5.0
  4.    5.3
Show Answers Only

`B`

Show Worked Solution

`text(Using the class centres)`

`text(Total hours)` `= (1 xx 5) + (4 xx 10) + (7 xx 3) + (10 xx 2)`
  `= 5 + 40 + 21 + 20`
  `= 86`
♦ Mean mark 45%
COMMENT: The mean is calculated using “class centres” in grouped data.
`text(Mean hours)` `= 86/20 = 4.3`

`=>  B`