Measures of Centre and Spread

Statistics, STD2 EQ-Bank 5 MC

Which graph represents a negatively skewed distribution?

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$D$

Show Worked Solution

$\text{Negatively skewed graphs have a long tail to the left.}$

$\Rightarrow D$

Statistics, STD2 EQ-Bank 21 MC

Twenty-five people were surveyed about the amount of money they spent on groceries per week, to the nearest dollar.

The results are shown in the frequency table.

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \textit{Amount Spent} ($) \rule[-1ex]{0pt}{0pt} & \ \ \ \ \ \textit{Frequency}\ \ \ \ \ \\
\hline
\rule{0pt}{2.5ex} 50-99 \rule[-1ex]{0pt}{0pt} & 4 \\
\hline
\rule{0pt}{2.5ex} 100-149 \rule[-1ex]{0pt}{0pt} & 9 \\
\hline
\rule{0pt}{2.5ex} 150-199 \rule[-1ex]{0pt}{0pt} & 8 \\
\hline
\rule{0pt}{2.5ex} 200-249 \rule[-1ex]{0pt}{0pt} & 4 \\
\hline
\end{array}

What is the mean amount spent on groceries by these people per week?

$143.50
$148.50
$153.50
$158.50

Show Answers Only

$B$

Show Worked Solution

$\text{Using the class centres}$

$\text{Class centres: } 74.5, 124.5, 174.5, 224.5$

$\text{Total hours}$	$=(74.5 \times 4) + (124.5 \times 9) + (174.5 \times 8) + (224.5 \times 4)$
	$=298 + 1120.5 + 1396 + 898=3712.5$

$\text{Mean amount}=\dfrac{3712.5}{25} =$148.50$

$\Rightarrow B$

Statistics, STD2 EQ-Bank 19 MC

Thirty students were surveyed about the number of hours they spent on homework 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} 0-4 \rule[-1ex]{0pt}{0pt} & 8 \\
\hline
\rule{0pt}{2.5ex} 5-9 \rule[-1ex]{0pt}{0pt} & 10 \\
\hline
\rule{0pt}{2.5ex} 10-14 \rule[-1ex]{0pt}{0pt} & 7 \\
\hline
\rule{0pt}{2.5ex} 15-19 \rule[-1ex]{0pt}{0pt} & 5 \\
\hline
\end{array}

What is the mean number of hours of homework completed by the students per week?

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$B$

Show Worked Solution

$\text{Using the class centres}$

$\text{Class centres: } 2, 7, 12, 17$

$\text{Total hours}$	$=(2\times 8) + (7\times 10) + (12\times 7) + (17\times 5)$
	$=16 + 70 + 84 + 85=255$

$\text{Mean hours}=\dfrac{255}{30} =8.5$

$\Rightarrow B$

Statistics, STD2 EQ-Bank 9 MC

A dataset has an interquartile range (IQR) of 18.

The upper quartile (Q₃) is 45.

What is the maximum value that would NOT be classified as an outlier?

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$D$

Show Worked Solution

$1.5 \times \text{IQR} = 1.5 \times 18 = 27$

$\text{Upper boundary} = Q_3 + 27 = 45 + 27 = 72$

$\text{Values up to and including 72 are not outliers}$

$\text{Values above 72 are outliers}$

$\therefore\ \text{Maximum value that is not an outlier} = 72$

$\Rightarrow D$

Statistics, STD2 EQ-Bank 8 MC

The following data shows the ages of people attending a fitness class:

16, 24, 26, 28, 30, 32, 34, 36, 52, 62

How many outliers are in this dataset?

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$C$

Show Worked Solution

$\text{Arrange in order (already ordered)}\ n = 10 $

$\text{Median }(Q_2) = \dfrac{30 + 32}{2} = 31 $

$Q_1 = 26$

$Q_3 =36$

$\text{IQR} = 36-26 = 10 $

$1.5 \times \text{IQR} = 1.5 \times 10 = 15$

$\text{Lower boundary} = 26-15 = 11 $

$\text{Upper boundary} = 36+15 = 51$

$\text{Values 52 and 62 are both above 51} \therefore\ 2 \text{ outliers}$

$\Rightarrow C$

Statistics, STD2 EQ-Bank 3 MC

A dataset has a lower quartile (Q₁) of 25 and an upper quartile (Q₃₎ of 35.

Which of the following values would be classified as an outlier?

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$A$

Show Worked Solution

$\text{IQR}=Q_3-Q_1$

$=35-25=10$

$\text{Lower boundary}=25-1.5\times 10=10$

$\text{Upper boundary}=35+1.5\times 10=50$

$\text{Check each value:} $

$2 < 10 \text{ (outlier)} $

$10 = 10 \text{ (on boundary, not an outlier)}$

$48 < 50 \text{ (not an outlier)} $

$50 = 50 \text{ (on boundary, not an outlier)}$

$\therefore\ 2\ \text{is the only outlier}$

$\Rightarrow A$

Statistics, STD2 EQ-Bank 1 MC

A dataset has the following values arranged in order:

12, 15, 18, 21, 24, 27, 30, 33, 36

What is the interquartile range (IQR) of this dataset?

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$B$

Show Worked Solution

$\text{Number of data values: }n=9$

$\text{Median }(Q_2): =24$

$\text{Lower half of data: }12, 15, 18, 21$

$Q_1=\dfrac{15+18}{2}\ =16.5$

$\text{Upper half of data: } 27, 30, 33, 36$

$Q_3=\dfrac{30+33}{2}\ =31.5$

$\text{IQR}=Q_3-Q_1$

$=31.5-16.5=15$

$\Rightarrow C$

Statistics, STD2 EQ-Bank 12 MC

A teacher recorded the following test scores (out of 50) for 8 students:

32, 38, 41, 44, 46, 48, 50, 50

What is the standard deviation of this set of data, correct to one decimal place?

5.2 marks
5.6 marks
5.9 marks
6.7 marks

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$C$

Show Worked Solution

$\text{Using a calculator to find the population standard deviation:}$

$\sigma = 5.914\dots= 5.9\ \text{marks (to 1 d.p.)}$

$\Rightarrow C$

Statistics, STD2 EQ-Bank 11 MC

A basketball player recorded the following number of points scored in 9 games:

18, 22, 25, 28, 31, 34, 38, 41, 45

What is the standard deviation of this set of data, correct to one decimal place?

8.5 points
9.1 points
10.2 points
11.5 points

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$A$

Show Worked Solution

$\text{Using a calculator to find the population standard deviation:}$

$\sigma = 8.485…= 8.5\ \text{points (to 1 d.p.)}$

$\Rightarrow A$

Statistics, STD2 EQ-Bank 10 MC

A science class recorded the following heights (in cm) of sunflower plants after 8 weeks:

142, 156, 163, 171, 178, 185, 192, 201

What is the standard deviation of this set of data, correct to one decimal place?

17.8 cm
18.3 cm
20.5 cm
22.1 cm

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$B$

Show Worked Solution

$\text{Using a calculator to find the population standard deviation:}$

$\sigma = 18.255…= 18.3\ \text{cm (to 1 d.p.)}$

$\Rightarrow B$

Statistics, STD2 EQ-Bank 2 MC

In a small business, the seven employees earn the following wages per week:

$\$300, \ \$490, \ \$520, \ \$590, \ \$660, \ \$680, \ \$970$

What is the population standard deviation of the wages, correct to 2 decimal places?

$187.25
$191.04
$201.58
$215.73

Show Answers Only

$B$

Show Worked Solution

$\text{Using a calculator to find the population standard deviation:}$

$\sigma = 191.044…= \$191.04\ \text{(to 2 d.p.)}$

$\Rightarrow B$

Statistics, STD2 S1 2023 HSC 15 MC

Ashan's mathematics class needs to complete six tests, each worth 100 marks.

After completing the first five tests, Ashan calculated that he would need a mark of 90 in the final test in order to have a mean mark of 80 for the six tests.

What was Ashan's mean mark after completing the first five tests?

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`A`

Show Worked Solution

`text{Total marks after 6 tests}\ = 80xx6=480`

`text{Actual marks after 5 tests}\ = 480-90=390`

`text{Average after 5 tests}\ = 390/5=78`

`=>A`

♦ Mean mark 40%.

Statistics, STD2 S1 2022 HSC 5 MC

Consider the following dataset.

`{:[13,16,17,17,21,24]:}`

Which row of the table shows how the median and mean are affected when a score of 5 is added to the dataset?

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`D`

Show Worked Solution

`text{Mean decreases}-text{Mean}_(text{old})=18,\ text(Mean)_(text{new})=16.14`

`text{Median remains 17.}`

`=>D`

Statistics, STD2 S1 2021 HSC 17

The five-number summary of a dataset is given.

Lowest score = 1

Lowest quartile (`Q_1`) = 4

Median (`Q_2`) = 7

Upper quartile (`Q_3`) = 10

Highest score = 20

Is 20 an outlier? Justify your answer with calculations. (2 marks)

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`text(20 is an outlier – See Worked Solution)`

Show Worked Solution

`IQR = Q_3-Q_1 = 10-4=6`

COMMENT: Stating 20 > 19 is necessary to justify this answer.

`text(Upper Fence)`	`= Q_3 + 1.5 xx IQR`
	`=10 + 1.5 xx 6=19`

`text{S}text{ince 20 > 19, 20 is an outlier.}`

Statistics, STD1 S1 2020 HSC 24

The ages in years, of ten people at the local cinema last Saturday afternoon are shown.

$38 \ \ 25 \ \ 38 \ \ 46 \ \ 55 \ \ 68 \ \ 72 \ \ 55 \ \ 36 \ \ 38$

The mean of this dataset is 47.1 years.
How many of the ten people were aged between the mean age and the median age? (2 marks)

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On Wednesday, ten people all aged 70 went to this same cinema.
Would the standard deviation of the age dataset from Wednesday be larger than, smaller than or equal to the standard deviation of the age dataset given in part (a)? Briefly explain your answer without performing any calculations. (2 marks)

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a. $1$

b. $\text{Standard deviation is a measure of how much the}$

$\text{ages of individuals differ from the mean age of the group.}$

$\therefore\ \text{Standard deviation of Wednesday’s group would be}$

$\text{less as the mean is 70 and everyone’s age is 70.}$

Show Worked Solution

a. $\text{Reorder ages in ascending order:}$

$25, 36, 38, 38, 38, 46, 55, 55, 68, 72$

$\text{Median} = \dfrac{\text{5th + 6th}}{2} = \dfrac{38 + 46}{2} = 42$

$\therefore\ \text{People with age between}\ 42-47.1 = 1\ \rightarrow\ (46)$

♦ Mean mark (a) 39%.

b. $\text{Standard deviation is a measure of spread, so how much the}$

$\text{ages of individuals differ from the mean age of the group.}$

$\therefore\ \text{Standard deviation of Wednesday’s group would be}$

$\text{less as the mean is 70 and everyone’s age is 70.}$

♦♦♦ Mean mark (b) 20%.

Statistics, STD2 S2 2020 HSC 28

Consider the following dataset.

`1 5 9 10 15`

Suppose a new value, `x`, is added to this dataset, giving the following.

`1 5 9 10 15 x`

It is known that `x` is greater than 15. It is also known that the difference between the means of the two datasets is equal to ten times the difference between the medians of the two datasets.

Calculate the value of `x`. (4 marks)

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`38`

Show Worked Solution

`text{Dataset 1 : Median} = 9`

Mean mark 53%.

`text{Dataset 2 : Median} = frac{9 + 10}{2} = 9.5`

`:.\ text{Difference between means}`

`= 10 xx (9.5-9)= 5`

`overset_ x_1 = frac{1 + 5 + 9 + 10 + 15}{5} = 8`

`therefore overset_ x_2 = 8 + 5 = 13`

`text{Sum of all data points in Dataset 2}`

`= 6 xx 13= 78`

`78`	`= 1 + 5 + 9 + 10 + 15 + x`
`therefore \ x`	`= 78-40= 38`

Statistics, STD2 S1 2019 HSC 19

The heights, in centimetres, of 10 players on a basketball team are shown.

170, 180, 185, 188, 192, 193, 193, 194, 196, 202

Is the height of the shortest player on the team considered an outlier? Justify your answer with calculations. (3 marks)

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`text(See Worked Solutions)`

Show Worked Solution

`Q_1 = 185, \ Q_3 = 194`

Mean mark 51%.
COMMENT: The last statement must be made to achieve full marks here!

`IQR = 194 – 185 = 9`

`text(Shortest player = 170)`

`text(Outlier height:)`

`Q_1 – 1.5 xx IQR `	`= 185 – 1.5 xx 9`
	`= 171.5`

`:.\ text(S)text(ince 170 < 171.5, 170 is an outlier.)`

Statistics, STD2 S1 2018 HSC 26e

A cumulative frequency table for a data set is shown.

\begin{array} {|c|c|}
\hline
\ \ \ \ \ \ \ \textit{Score}\ \ \ \ \ \ \ & \ \ \ \ \ \textit{Cumulative}\ \ \ \ \ \\ & \textit{frequency} \\
\hline
\rule{0pt}{2.5ex} \text{1} \rule[-1ex]{0pt}{0pt} & 5 \\
\hline
\rule{0pt}{2.5ex} \text{2} \rule[-1ex]{0pt}{0pt} & 9 \\
\hline
\rule{0pt}{2.5ex} \text{3} \rule[-1ex]{0pt}{0pt} & 16 \\
\hline
\rule{0pt}{2.5ex} \text{4} \rule[-1ex]{0pt}{0pt} & 20 \\
\hline
\rule{0pt}{2.5ex} \text{5} \rule[-1ex]{0pt}{0pt} & 34 \\
\hline
\rule{0pt}{2.5ex} \text{6} \rule[-1ex]{0pt}{0pt} & 42 \\
\hline
\end{array}

What is the interquartile range of this data set? (2 marks)

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`2`

Show Worked Solution

`text(42 data points ⇒ median) = text(21st + 22nd)/2`

♦♦ Mean mark 27%.

`text(Q)_1= 11text(th data point) = 3`

`text(Q)_3= 32text(nd data point) = 5`

`:.\ text(IQR)= 5-3=2`

Statistics, STD2 S1 2018 HSC 11 MC

A set of data is summarised in this frequency distribution table.

Which of the following is true about the data?

Mode = 7, median = 5.5
Mode = 7, median = 6
Mode = 9, median = 5.5
Mode = 9, median = 6

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`B`

Show Worked Solution

`text{Mode = 7 (highest frequency of 9)}`

`text(Median = average of 15th and 16th data points.)`

`:.\ text(Median = 6)`

`=> B`

Statistics, STD2 S1 2018 HSC 1 MC

A set of scores has the following five-number summary.

lower extreme = 2
lower quartile = 5
median = 6
upper quartile = 8
upper extreme = 9

What is the range?

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`D`

Show Worked Solution

`text(Range)`	`=\ text(upper extreme − lower extreme)`
	`= 9-2= 7`

`=> D`

Statistics, STD2 S1 2017 HSC 30a

A set of data has a lower quartile (`Q_L`) of 10 and an upper quartile (`Q_U`) of 16.

What is the maximum possible range for this set of data if there are no outliers? (2 marks)

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`24`

Show Worked Solution

`IQR = 16-10 = 6`

♦♦ Mean mark 34%.

`text(If no outliers,)`

`text(Upper limit)`	`= Q_U + 1.5 xx IQR`
	`= 16 + 1.5 xx 6= 25`

`text(Lower limit)`	`= Q_L-1.5 xx IQR`
	`= 10-1.5 xx 6= 1`

`:.\ text(Maximum range)= 25-1=24`

Statistics, STD2 S1 2017 HSC 27a

Jamal surveyed eight households in his street. He asked them how many kilolitres (kL) of water they used in the last year. Here are the results.

`220, 105, 101, 450, 37, 338, 151, 205`

Calculate the mean of this set of data. (1 mark)

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What is the standard deviation of this set of data, correct to one decimal place? (1 mark)

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a. `200.875`

b. `127.4\ \ text{(1 d.p.)}`

Show Worked Solution

a.	`text(Mean)`	`= (220 + 105 + 101 + 450 + 37 + 338 + 151 + 205) ÷ 8`
		`= 200.875`

♦ Mean mark part (b) 47%.
IMPORTANT: The population standard deviation is required here.

b.	`text(Std Dev)`	`= 127.357…\ \ text{(by calc)}`
		`= 127.4\ \ text{(1 d.p.)}`

Statistics, STD2 S1 2016 HSC 27b

A small population consists of three students of heights 153 cm, 168 cm and 174 cm. Samples of varying sizes can be taken from this population.

What is the mean of the mean heights of all the possible samples? Justify your answer. (2 marks)

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`165\ text(cm)`

Show Worked Solution

`text(If sample is 1 person,)`

♦ Mean mark 40%.

`text{Possible mean(s): 153, 168 or 174.}`

`text(If sample is 2 people,)`

`text{Possible mean(s):}quad(153 + 168)/2`	`= 160.5`
`(153 + 174)/2`	`= 163.5`
`(168 + 174)/2`	`= 171`

`text(If sample is 3 people,)`

`text(Mean:)quad(153 + 168 + 174)/3 = 165`

`:.\ text(Mean of all mean heights)`

`= (153 + 168 + 174 + 160.5 + 163.5 + 171 + 165)/7`

`= 165\ text(cm)`

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?

6.5
10.5
11.9
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 2016 HSC 19 MC

A soccer referee wrote down the number of goals scored in 9 different games during the season.

`2, \ 3, \ 3, \ 3, \ 5, \ 5, \ 8, \ 9, \ ...`

The last number has been omitted. The range of the data is 10.

What is the five-number summary for this data set?

`2, 3, 5, 8.5, 12`
`2, 3, 5, 8.5, 10`
`2, 3, 5, 8, 12`
`2, 3, 5, 8, 10`

Show Answers Only

`A`

Show Worked Solution

`text{Since range is 10} \ => \ text{Last data point = 12}`

`text{Q}_1 = 3`

`text{Q}_3 = (8 + 9)/2 = 8.5`

`text(Median = 5)`

`=> A`

♦ Mean mark 46%.

Statistics, STD2 S1 2015 HSC 27d

In a small business, the seven employees earn the following wages per week:

$\$300, \ \$490, \ \$520, \ \$590, \ \$660, \ \$680, \ \$970$

Is the wage of $970 an outlier for this set of data? Justify your answer with calculations. (3 marks)

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Each employee receives a $20 pay increase.

What effect will this have on the standard deviation? (1 mark)

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a. $\text{See Worked Solutions.} $

b. $\text{The standard deviation will remain the same.}$

Show Worked Solution

a. $300, 490, 520, 590, 660, 680, 970$

$\text{Median}= 590$

$Q_1= 490,\ Q_3= 680$

$IQR= 680-490 = 190$

$\text{Outlier if \$970 is greater than:} $

$Q_3 + 1.5 \times IQR = 680 + 1.5 \times 190 = \$965 $

$\therefore\ \text{The wage \$970 per week is an outlier.}$

♦ Mean mark (a) 39%.

b. $\text{All values increase by \$20, but so too does the mean.} $

$\text{Therefore the spread about the new mean will not change} $

$\text{and therefore the standard deviation will remain the same.} $

Statistics, STD2 S1 2005 HSC 27d

Nine students were selected at random from a school, and their ages were recorded.

\begin{array} {|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Ages} \rule[-1ex]{0pt}{0pt} \\
\hline
\rule{0pt}{2.5ex} \ \ \ \text{12 11 16} \ \ \ \rule[-1ex]{0pt}{0pt} \\ \rule{0pt}{2.5ex} \text{14 16 15} \rule[-1ex]{0pt}{0pt} \\ \rule{0pt}{2.5ex} \text{14 15 14} \rule[-1ex]{0pt}{0pt} \\
\hline
\end{array}

What is the sample standard deviation, correct to two decimal places? (2 marks)

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Briefly explain what is meant by the term standard deviation. (1 mark)

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a. `text{1.69 (to 2 d.p.)}`

b. `text(Standard deviation is a measure of how much)`

`text(members of a data group differ from the mean)`

`text(value of the group.)`

Show Worked Solution

a. `text(Sample standard deviation)`

`= 1.6914…\ text{(by calculator)}`

`= 1.69\ \ \ text{(to 2 d.p.)}`

b. `text(Standard deviation is a measure of how much)`

`text(members of a data group differ from the mean)`

`text(value of the group.)`

Statistics, STD2 S1 2006 HSC 23c

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

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

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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.

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

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a. `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.)`

b. `text(7.275 hours)`

Show Worked Solution

a. `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).

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

Statistics, STD2 S1 2006 HSC 12 MC

The mean of a set of 5 scores is 62.

What is the new mean of the set of scores after a score of 14 is added?

Show Answers Only

`B`

Show Worked Solution

`text(Mean of 5 scores) = 62`

`:.\ text(Total of 5 scores) = 62 xx 5 = 310`

`text(Add a score of 14)`

`text(Total of 6 scores) = 310 + 14 = 324`

`:.\ text(New mean)= 324/6=54`

`=> B`

Statistics, STD2 S1 2005 HSC 1 MC

What is the mean of the set of scores?

`3, \ 4, \ 5, \ 6, \ 6, \ 8, \ 8, \ 8, \ 15`

Show Answers Only

`B`

Show Worked Solution

`text(Mean)`	`= ((3 + 4 + 5 +6 + 6 + 8 + 8 + 8 + 15))/9`
	`= 63/9= 7`

`=> B`

Statistics, STD2 S1 2004 HSC 6-7 MC

Use the set of scores 1, 3, 3, 3, 4, 5, 7, 7, 12 to answer Questions 6 and 7.

Question 6

What is the range of the set of scores?

Question 7

What are the median and the mode of the set of scores?

Median 3, mode 5
Median 3, mode 3
Median 4, mode 5
Median 4, mode 3

Show Answers Only

`text(Question 6:)\ C`

`text(Question 7:)\ D`

Show Worked Solution

`text(Question 6)`

`text(Range)= text(High)-text(Low)= 12-1=11`

`=> C`

`text(Question 7)`

`text(9 scores)`

`:.\ text(Median)= (9 + 1) / 2=5 text(th score)= 4`

`text(Mode) = 3`

`=> D`

Statistics, STD2 S1 2007 HSC 24a

Consider the following set of scores:

`3, \ 5, \ 5, \ 6, \ 8, \ 8, \ 9, \ 10, \ 10, \ 50.`

Calculate the mean of the set of scores. (1 mark)

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What is the effect on the mean and on the median of removing the outlier? (2 marks)

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a. `11.4`

b. `text{If the outlier (50) is removed, the mean}`

`text(would become lower.)`

`text(Median will NOT change.)`

Show Worked Solution

a. `text(Total of scores)`

`= 3 + 5 + 5 + 6 + 8 + 8 + 9 + 10 + 10 +50`

`= 114`

`:.\ text(Mean) = 114/10 = 11.4`

b. `text(Mean)`

`text{If the outlier (50) is removed, the mean}`

`text(would become lower.)`

`text(Median)`

`text(The current median (10 data points))`

`= text(5th + 6th)/2 = (8 + 8)/2 = 8`

`text(The new median (9 data points))`

`=\ text(5th value)= 8`

`:.\ text(Median will NOT change.)`

Statistics, STD2 S1 2007 HSC 21 MC

This set of data is arranged in order from smallest to largest.

`5, \ 6, \ 11, \ x, \ 13, \ 18, \ 25`

The range is six less than twice the value of `x`.

Which one of the following is true?

The median is 12 and the interquartile range is 7.
The median is 12 and the interquartile range is 12.
The median is 13 and the interquartile range is 7.
The median is 13 and the interquartile range is 12.

Show Answers Only

`D`

Show Worked Solution

`5, 6, 11, x, 13, 18, 25`

`text(Range)`	`= 2x-6`
`25-5`	`= 2x-6`
`2x`	`= 26`
`x`	`= 13`
`:.\ text(Median)`	`= 13`

`Q_1 = 6\ \ \ \ \ Q_3 = 18`

`:.\ text(IQR) = 12`

`=> D`

Statistics, STD2 S1 2008 HSC 23f

Christina has completed three Mathematics tests. Her mean mark is 72%.

What mark (out of 100) does she have to get in her next test to increase her mean mark to 73%? (2 marks)

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

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`76`

Show Worked Solution

`text(Total marks in 3 tests)= 3 xx 72= 216`

`text(We need 4-test mean) = 73`

`text{i.e. Total Marks (4 tests)}-:4= 73`

`text(Total Marks)\ text{(4 tests)}= 292`

`:.\ text(4th test score)= 292-216= 76`

Statistics, STD2 S1 2008 HSC 13 MC

The height of each student in a class was measured and it was found that the mean height was 160 cm.

Two students were absent. When their heights were included in the data for the class, the mean height did not change.

Which of the following heights are possible for the two absent students?

155 cm and 162 cm
152 cm and 167 cm
149 cm and 171 cm
143 cm and 178 cm

Show Answers Only

`C`

Show Worked Solution

`text(S) text(ince the mean doesn’t change)`

`=>\ text(2 absent students must have a)`

`text(mean height of 160 cm.)`

`text(Considering each option given,)`

`(149 + 171) -: 2 = 160`

`=> C`

Statistics, STD2 S1 2008 HSC 8 MC

What is the median of the following set of scores?

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`C`

Show Worked Solution

`text(Median)=(n+1)/2=(33+1)/2=\ text (17th score)`

`:.\ text(Median is 14)`

`=> C`

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} 0-2 \rule[-1ex]{0pt}{0pt} & 5 \\
\hline
\rule{0pt}{2.5ex} 3-5 \rule[-1ex]{0pt}{0pt} & 10 \\
\hline
\rule{0pt}{2.5ex} 6-8 \rule[-1ex]{0pt}{0pt} & 3 \\
\hline
\rule{0pt}{2.5ex} 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?

Show Answers Only

$B$

Show Worked Solution

$\text{Using the class centres}$

$\text{Total hours}$	$=(1\times 5) + (4\times 10) + (7\times 3) + (10\times 2)$
	$= 5 + 40 + 21 + 20= 86$

♦ Mean mark 45%
COMMENT: The mean is calculated using “class centres” in grouped data.

$\text{Mean hours}=\dfrac{86}{20} =4.3$

$\Rightarrow B$

Statistics, STD2 S1 2011 HSC 17 MC

The heights of the players in a basketball team were recorded as 1.8 m, 1.83 m, 1.84 m, 1.86 m and 1.92 m. When a sixth player joined the team, the average height of the players increased by 1 centimetre.

What was the height of the sixth player?

1.85 m
1.86 m
1.91 m
1.93 m

Show Answers Only

`C`

Show Worked Solution

`text(Old Mean)`	`=(1.8+1.83+1.84+1.86+1.92)-:5`
	`=9.25/5=1.85\ \ text(m)`

`text{S}text{ince the new mean = 1.86m (given)}`

`text(New Mean)`	`=text(Height of all 6 players) -: 6`
`:.1.86`	`=(9.25+h)/6\ \ \ \ (h\ text{= height of new player})`
`h`	`=(6xx1.86)-9.25=1.91\ \ text(m)`

`=> C`

Statistics, STD2 S1 2011 HSC 14 MC

A data set of nine scores has a median of 7.

The scores 6, 6, 12 and 17 are added to this data set.

What is the median of the data set now?

Show Answers Only

`B`

Show Worked Solution

`text(S)text(ince an even amount of scores are added below and)`

`text(above the existing median, it will not change.)`

`=>B`

Statistics, STD2 S5 2013 HSC 29b

Ali’s class sits two Geography tests. The results of her class on the first Geography test are shown.

`58,\ \ 74,\ \ 65,\ \ 66,\ \ 73,\ \ 71,\ \ 72,\ \ 74,\ \ 62,\ \ 70`

The mean was 68.5 for the first test.

Calculate the standard deviation for the first test. Give your answer correct to one decimal place. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
On the second Geography test, the mean for the class was 74.4 and the standard deviation was 12.4.

Ali scored 62 on the first test. Calculate the mark that she needed to obtain in the second test to ensure that her performance relative to the class was maintained. (3 marks)

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

Show Answers Only

a. `5.2\ \ \ text{(to 1 d.p.)}`

b. `text(Ali needs to score 58.9)`

Show Worked Solution

♦ Mean mark (a) 39%
COMMENT: Make sure you are confident with this function on your calculator!

a. `sigma=5.2201…=5.2\ text{(to 1 d.p.)}`

b. `z =(x-mu)/sigma`

`z text{-score (1st test)}`	`= (62-68.5)/5.2`
	`=-1.25`

`text(2nd test has)\ z text(-score of)\ \-1.25 :`

♦♦ Mean mark (b) 21%.
MARKER’S COMMENT: When “performance relative to the class is maintained”, `z text(-scores)` are the same in each test.

`-1.25`	`= (x-74.4)/12.4`
`x-74.4`	`=-15.5`
`x`	`=58.9`

`:.\ text(Ali needs to score 58.9)`

Statistics, STD2 S1 2013 HSC 26b

Write down a set of six data values that has a range of 12, a mode of 12 and a minimum value of 12. (2 marks)

Show Answers Only

`12, 12, 12, 16, 18, 24`

Show Worked Solution

`12, 12, 12, 16, 18, 24`

`text(NB. There are many correct solutions.)`

Statistics, STD2 S1 2009 HSC 3 MC

The eye colours of a sample of children were recorded.

When analysing this data, which of the following could be found?

Mean
Median
Mode
Range

Show Answers Only

`C`

Show Worked Solution

`text(Eye colour is categorical data)`

`:.\ text(Only the mode can be found)`

`=> C`

Statistics, STD2 S1 2009 HSC 21 MC

The mean of a set of ten scores is 14. Another two scores are included and the new mean is 16.

What is the mean of the two additional scores?

Show Answers Only

`D`

Show Worked Solution

♦♦♦ Mean mark 28%.

`text(If ) bar x\ text(of 10 scores = 14)`

`=>text(Sum of 10 scores)= 10 xx 14 = 140`

`text(With 2 additional scores,)\ \ bar x = 16 `

`=>text(Sum of 12 scores)= 12 xx 16 = 192`

`:.\ text(Value of 2 extra scores)= 192-140=52`

`:.\ text(Mean of 2 extra scores)= 52/2 = 26`

`=> D`

Statistics, STD2 S1 2013 HSC 14 MC

The July sales prices for properties in a suburb were:

$552 000, $595 000, $607 000, $607 000, $682 000, and $685 000.

On 1 August, another property in the same suburb was sold for over one million dollars.

If the property had been sold in July, what effect would it have had on the mean and median sale prices for July?

Both the mean and median would have changed.
Neither the mean nor the median would have changed.
The mean would have changed and the median would have stayed the same.
The mean would have stayed the same and the median would have changed.

Show Answers Only

`C`

Show Worked Solution

`text(Mean increases because new house is sold above)`

`text(the existing average.)`

`text(Initial median)= (607\ 000+607\ 000)/2=607\ 000`

`text(New median)=607\ 000\ \ \ text{(4th value in a list of 7)}`

`=>\ C`

\(\text{Total hours}\)	\(=(2\times 8) + (7\times 10) + (12\times 7) + (17\times 5)\)
	\(=16 + 70 + 84 + 85=255\)

\(\text{Total hours}\)	\(=(1\times 5) + (4\times 10) + (7\times 3) + (10\times 2)\)
	\(= 5 + 40 + 21 + 20= 86\)