smc-600-10-Single z-score

Data Analysis, GEN1 2025 VCAA 5 MC

The heights of females aged between 16 and 18 years within a population are normally distributed.

Analysis of the heights of this group of females showed that:

2.5% of the heights were greater than 178.9 cm
16% of the heights were less than 157.6 cm.

Using the 68-95-99.7% rule, the mean and standard deviation of the heights of these females are respectively

150.5 and 21.3
154.9 and 7.1
164.7 and 7.1
171.8 and 14.2

Show Answers Only

$C$

Show Worked Solution

$\bar{x}+2z$	$=178.9\ \ …\ (1)$
$\bar{x}-z$	$=157.6\ …\ (2)$

$\text{Subtract}\ (1)-(2):$

$3z=21.3\ \ \Rightarrow\ \ z=7.1$

$\text{Substitute}\ \ z=7.1\ \ \text{into (2):}$

$\bar{x}-7.1=157.6\ \ \Rightarrow\ \ \bar{x}=164.7$

$\Rightarrow C$

Data Analysis, GEN1 2024 NHT 5 MC

The heights of a group of Year 9 students were measured and the standard deviation was found to be 12.25 cm .

One student with a height of 174.6 cm had a standardised score of $z=0.45$

The mean height of this group of students, in centimetres, was closest to

161.9
169.1
180.1
186.4
187.3

Show Answers Only

$B$

Show Worked Solution

$\text{Using}\ \ z=\dfrac{x-\overline{x}}{s_x}: $

$0.45$	$=\dfrac{174.6-\overline{x}}{12.25} $
$\overline{x}=$	$174.6-(0.45 \times 12.25) = 169.0875$

$\Rightarrow B$

Data Analysis, GEN2 2024 VCAA 1

Table 1 lists the Olympic year, $\textit{year}$, and the gold medal-winning height for the men's high jump, $\textit{Mgold}$, in metres, for each Olympic Games held from 1928 to 2020. No Olympic Games were held in 1940 or 1944, and the 2020 Olympic Games were held in 2021.

Table 1

\begin{array}{|c|c|}
\hline \quad \textit{year} \quad & \textit{Mgold}\,\text{(m)} \\
\hline 1928 & 1.94 \\
\hline 1932 & 1.97 \\
\hline 1936 & 2.03 \\
\hline 1948 & 1.98 \\
\hline 1952 & 2.04 \\
\hline 1956 & 2.12 \\
\hline 1960 & 2.16 \\
\hline 1964 & 2.18 \\
\hline 1968 & 2.24 \\
\hline 1972 & 2.23 \\
\hline 1976 & 2.25 \\
\hline 1980 & 2.36 \\
\hline 1984 & 2.35 \\
\hline 1988 & 2.38 \\
\hline 1992 & 2.34 \\
\hline 1996 & 2.39 \\
\hline 2000 & 2.35 \\
\hline 2004 & 2.36 \\
\hline 2008 & 2.36 \\
\hline 2012 & 2.33 \\
\hline 2016 & 2.38 \\
\hline 2020 & 2.37 \\
\hline
\end{array}

For the data in Table 1, determine:
i. the maximum $\textit{Mgold}$ in metres (1 mark)

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ii. the percentage of $\textit{Mgold}$ values greater than 2.25 m. (1 mark)

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The mean of these $\textit{Mgold}$ values is 2.23 m, and the standard deviation is 0.15 m.
Calculate the standardised $z$-score for the 2000 $\textit{Mgold}$ of 2.35 m. (1 mark)

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Construct a boxplot for the $\textit{Mgold}$ data in Table 1 on the grid below. (2 marks)

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A least squares line can also be used to model the association between $\textit{Mgold}$ and $\textit{year}$.
Using the data from Table 1, determine the equation of the least squares line for this data set.
Use the template below to write your answer.
Round the values of the intercept and slope to three significant figures. (2 marks)

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The coefficient of determination is 0.857
Interpret the coefficient of determination in terms of $\textit{Mgold}$ and $\textit{year}$. (1 mark)

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Show Answers Only

a.i. $2.39$

a.ii. $50\%$

b. $0.8$

e. $\text{A coefficient of determination of 85.7% shows the variation in}$

$\text{the}\ Mgold\ \text{that is explained by the variation in the }year.$

Show Worked Solution

a.i. $2.39$

a.ii. $\dfrac{11}{22}=50\%$

b.	$z$	$=\dfrac{x-\overline x}{s_x}$
		$=\dfrac{2.35-2.23}{0.15}$
		$=0.8$

c. $Q_2=\dfrac{2.33+2.25}{2}=2.29$

$Q_1=2.12, \ Q_3=2.36$

$\text{Min}\ =1.94, \ \text{Max}\ =2.39$

d. $\text{Using CAS:}$

Mean mark (d) 52%.
Mean mark (e) 52%.

e. $\text{A coefficient of determination of 85.7% shows the variation in}$

$\text{the}\ Mgold\ \text{that is explained by the variation in the }year.$

Data Analysis, GEN1 2024 VCAA 7 MC

Fiona plays nine holes of golf each week, and records her score.

Her mean score for all rounds in 2024 is 55.7

In one round, when she recorded a score of 48, her standardised score was $z=-1.75$

The standard deviation for score in 2024 is

1.1
2.3
4.4
6.95

Show Answers Only

$C$

Show Worked Solution

$z$	$=\dfrac{x-\overline{x}}{s_x}$
$-1.75$	$=\dfrac{48-55.7}{s_x}$
$s_x$	$=\dfrac{48-55.7}{-1.75}$
	$=4.4$

$\Rightarrow C$

Data Analysis, GEN1 2023 VCAA 5 MC

The heights of a group of Year 8 students have a mean of $163.56 cm and a standard deviation of $8.14 cm.

One student's height has a standardised $z$-score of –0.85 .

This student's height, in centimetres, is closest to

155.4
156.6
162.7
170.5
171.7

Show Answers Only

$B$

Show Worked Solution

$\bar x = 163.56, \ s_x = 8.14$

$\text{Find}\ x\ \text{given}\ \ z=-0.85:$

$z$	$= \dfrac{x-\bar x}{s_x}$
$-0.85$	$=\dfrac{x-163.56}{8.14}$
$x-163.56$	$=8.14 \times -0.85$
$x$	$=163.56-6.919$
	$=156.641$

$\Rightarrow B$

Data Analysis, GEN1 2023 VCAA 4 MC

The time spent by visitors in a museum is approximately normally distributed with a mean of 82 minutes and a standard deviation of 11 minutes.

2380 visitors are expected to visit the museum today.

Using the 68–95–99.7% rule, the number of these visitors who are expected to spend between 60 and 104 minutes in the museum is

1128
1618
2256
2261
2373

Show Answers Only

$D$

Show Worked Solution

$\bar x = 82, \ s_x = 11$

$z\text{-score (60)} = \dfrac{x-\bar x}{s_x} = \dfrac{60-82}{11} = -2 $

$z\text{-score (104)} = \dfrac{104-82}{11} = 2 $

$\text{95% of data points lie between}\ z=\pm 2 $

$\text{Number of visitors}\ = 95\% \times 2380 = 2261$

$\Rightarrow D$

CORE, FUR2 2021 VCAA 1

In the sport of heptathlon, athletes compete in seven events.

These events are the 100 m hurdles, high jump, shot-put, javelin, 200 m run, 800 m run and long jump.

Fifteen female athletes competed to qualify for the heptathlon at the Olympic Games.

Their results for three of the heptathlon events – high jump, shot-put and javelin – are shown in Table 1

Write down the number of numerical variables in Table 1. (1 mark)

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Complete Table 2 below by calculating the mean height jumped for the high jump, in metres, by the 15 athletes. Write your answer in the space provided in the table. (1 mark)

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In shot-put, athletes throw a heavy spherical ball (a shot) as far as they can. Athlete number six, Jamilia, threw the shot 14.50 m.
Calculate Jamilia's standardised score (`z`). Round your answer to one decimal place. (1 mark)

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In the qualifying competition, the heights jumped in the high jump are expected to be approximately normally distributed.
Chara's jump in this competition would give her a standardised score of `z = –1.0`
Use the 68–95–99.7% rule to calculate the percentage of athletes who would be expected to jump higher than Chara in the qualifying competition. (1 mark)

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The boxplot below was constructed to show the distribution of high jump heights for all 15 athletes in the qualifying competition.

Explain why the boxplot has no whisker at its upper end. (1 mark)

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For the javelin qualifying competition (refer to Table 1), another boxplot is used to display the distribution of athlete's results.
An athlete whose result is displayed as an outlier at the upper end of the plot is considered to be a potential medal winner in the event.
What is the minimum distance that an athlete needs to throw the javelin to be considered a potential medal winner? (2 marks)

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Show Answers Only

`3`
`1.81`
`0.5 \ text{(to d.p.)}`
`84text(%)`
`text{See Worked Solutions}`
`46.89 \ text{m}`

Show Worked Solution

a. `3 \ text{High jump, shot-put and javelin}`

`text{Athlete number is not a numerical variable}`

b. `text{High jump mean}`

`= (1.76 + 1.79 + 1.83 + 1.82 + 1.87 + 1.73 + 1.68 + 1.82 +`

`1.83 + 1.87 + 1.87 + 1.80 + 1.83 + 1.87 + 1.78) ÷ 15`

`= 1.81`

c.	`z text{-score} (14.50)`	`= {14.50-13.74}/{1.43}`
		`= 0.531 …`
		`= 0.5 \ text{(to 1 d.p.)}`

d. `P (z text{-score} > -1 ) = 84text(%)`

e. `text{If the} \ Q_3 \ text{value is also the highest value in the data set,}`

`text{there is no whisker at the upper end of a boxplot.}`

f. `text{Javelin (ascending):}`

`38.12, 39.22, 40.62, 40.88, 41.22, 41.32, 42.33, 42.41, `

`42.51, 42.65, 42.75, 42.88, 45.64, 45.68, 46.53`

`Q_1 = 40.88 \ \ , \ Q_3 = 42.88 \ \ , \ \ IQR = 42.88-40.88 = 2`

`text{Upper Fence}`	`= Q_3 + 1.5 xx IQR`
	`= 42.88 + 1.5 xx 2`
	`= 45.88`

`:. \ text{Minimum distance = 45.89 m (longer than upper fence value)}`

CORE, FUR1 2021 VCAA 7-8 MC

800 participants auditioned for a stage musical. Each participant was required to complete a series of ability tests for which they received an overall score.

The overall scores were approximately normally distributed with a mean score of 69.5 points and a standard deviation of 6.5 points.

Part 1

Only the participants who scored at least 76.0 points in the audition were considered successful,

Using the 68-95-99.7% rule, how many of the participants were considered unsuccessful?

Part 2

To be offered a leading role in the stage musical, a participant must achieve a standardised score of at least 1.80

Three participants' names and their overall scores are given in the table below.

Which one of the following statements is true?

Only Amy was offered a leading role.
Only Cherie was offered a leading role.
Only Brian was not offered a leading role
Both Brian and Cherie were offered leading roles.
All three participants were offered leading roles.

Show Answers Only

`text(Part 1:)\ D`

`text(Part 2:)\ C`

Show Worked Solution

`text{Part 1}`

`mu = 69.5 \ , \ sigma= 6.5`

`ztext{-score} \ (76) = {76 – 69.5}/6.5 = 1`

`:.\ text{Unsuccessful}`	`=84text(%) xx 800`
	`= 672`

`=> D`

`text{Part 2}`

`ztext{-score}`	`= {x – mu}/sigma`
`1.8`	`= {x – 69.5}/6.5`
`x`	`= 1.8 xx 6.5 + 69.5`
	`= 81.2`

`:. \ text{Amy and Cherie were both offered leading roles (only}`

`text{Brian wasn’t).}`

`=> C`

CORE, FUR1 2020 VCAA 8 MC

The wing length of a species of bird is approximately normally distributed with a mean of 61 mm and a standard deviation of 2 mm.

Using the 68–95–99.7% rule, for a random sample of 10 000 of these birds, the number of these birds with a wing length of less than 57 mm is closest to

Show Answers Only

`D`

Show Worked Solution

`z text(-score)\ (57) = (x – bar x)/s = (57 – 61)/2 = -2`

`:.\ text(Number less than 57)`

`= 2.5text(%) xx 10\ 000`

`= 250`

`=> D`

Data Analysis, GEN1 2019 NHT 5-7 MC

The birth weights of a large population of babies are approximately normally distributed with a mean of 3300 g and a standard deviation of 550 g.

Part 1

A baby selected at random from this population has a standardised weight of `z = – 0.75`

Which one of the following calculations will result in the actual birth weight of this baby?

`text(actual birth weight)\ = 550 - 0.75 × 3300`
`text(actual birth weight)\ = 550 + 0.75 × 3300`
`text(actual birth weight)\ = 3300 - 0.75 × 550`
`text(actual birth weight)\ = 3300 + 0.75/550`
`text(actual birth weight)\ = 3300 - 0.75/550`

Part 2

Using the 68–95–99.7% rule, the percentage of babies with a birth weight of less than 1650 g is closest to

0.14%
0.15%
0.17%
0.3%
2.5%

Part 3

A sample of 600 babies was drawn at random from this population.

Using the 68–95–99.7% rule, the number of these babies with a birth weight between 2200 g and 3850 g is closest to

Show Answers Only

`text(Part 1:)\ \ C`

`text(Part 2:)\ \ B`

`text(Part 3:)\ \ E`

Show Worked Solution

`text(Part 1)`

`text(Actual weight)`	`= text(mean) + z xx text(std dev)`
	`= 3300 – 0.75 xx 550`

`=> C`

`text(Part 2)`

`z-text(score)`	`= (x – barx)/5`
	`= (1650 – 3300)/550`
	`= −3`

`:. P(x < 1650)`	`= P(z < −3)`
	`= 0.3/2`
	`= 0.15\ text(%)`

`=>B`

`text(Part 3)`

`ztext(-score)\ (2200) = (2200 – 3300)/550 = −2`

`ztext(-score)\ (3850) = (3850 – 3300)/550 = 1`

`text(Percentage)`	`= (47.5 + 34)text(%) xx 600`
	`= 81.5text(%) xx 600`
	`= 48text(%)`

`=>\ E`

CORE, FUR2 2019 VCAA 2

The parallel boxplots below show the maximum daily temperature and minimum daily temperature, in degrees Celsius, for 30 days in November 2017.

Use the information in the boxplots to complete the following sentences.
For November 2017
i. the interquartile range for the minimum daily temperature was _____ °C (1 mark)

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ii. the median value for maximum daily temperature was _____ °C higher than the median value for minimum daily temperature (1 mark)

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iii. the number of days on which the maximum daily temperature was less than the median value for minimum daily temperature was _____ (1 mark)

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The temperature difference between the minimum daily temperature and the maximum daily temperature in November 2017 at this location is approximately normally distributed with a mean of 9.4 °C and a standard deviation of 3.2 °C.
Determine the number of days in November 2017 for which this temperature difference is expected to be greater than 9.4 °C. (1 mark)

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Show Answers Only

1. `5text(°C)`
2. `10text(°C)`
3. `1\ text(day)`
`15\ text(days)`

Show Worked Solution

a.i. `text(IQR)\ = 17-12= 5text(°C)`

a.ii.	`text{Median (maximum temperature)}`	`= 25`
	`text{Median (minimum temperature)}`	`= 15`

`:.\ text(Maximum is 10°C higher)`

a.iii. `text{Median (minimum temperature)} = 15text(°C)`

`text(1 day) => text(maximum temperature is below)\ 15text(°C)`

b.	`text(Number of days)`	`= 0.50 xx 30`
		`= 15\ text(days)`

CORE, FUR1 2019 VCAA 7 MC

The volume of a cup of soup served by a machine is normally distributed with a mean of 240 mL and a standard deviation of 5 mL.

A fast-food store used this machine to serve 160 cups of soup.

The number of these cups of soup that are expected to contain less than 230 mL of soup is closest to

Show Answers Only

`B`

Show Worked Solution

`z text(-score)\ (230)`	`= (230 – 240)/5`
	`= -2`

`:.\ text(Number of cups) < 230\ text(mL)`

`= 2.5/100 xx 160`

`= 4`

`=> B`

CORE, FUR1 2018 VCAA 3-5 MC

The pulse rates of a population of Year 12 students are approximately normally distributed with a mean of 69 beats per minute and a standard deviation of 4 beats per minute.

Part 1

A student selected at random from this population has a standardised pulse rate of `z = –2.5`

This student’s actual pulse rate is

59 beats per minute.
63 beats per minute.
65 beats per minute.
73 beats per minute.
79 beats per minute.

Part 2

Another student selected at random from this population has a standardised pulse rate of `z=–1.`

The percentage of students in this population with a pulse rate greater than this student is closest to

2.5%
5%
16%
68%
84%

Part 3

A sample of 200 students was selected at random from this population.

The number of these students with a pulse rate of less than 61 beats per minute or greater than 73 beats per minute is closest to

Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ E`

`text(Part 3:)\ B`

Show Worked Solution

`text(Part 1)`

`ztext(-score)`	`= (x – barx)/s`
`−2.5`	`= (x – 69)/4`
`x – 69`	`= −10`
`:. x`	`= 59`

`=> A`

`text(Part 2)`

`ztext(-score) = −1`

`text(% above)`	`= 34 + 50`
	`= 84 text(%)`

`=> E`

`text(Part 3)`

`z-text(score)\ (61)`	`= (61 – 69)/4 = −2`
`z-text(score)\ (73)`	`= (73 – 69)/4 = 1`

`text(Percentage)`	`= 2.5 + 16`
	`= 18.5text(%)`

`:.\ text(Number of students)`	`= 18.5 text(%) xx 200`
	`= 37`

`=> B`

CORE, FUR2 2017 VCAA 1

The number of eggs counted in a sample of 12 clusters of moth eggs is recorded in the table below.

From the information given, determine
i. the range (1 mark)

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ii. the percentage of clusters in this sample that contain more than 170 eggs. (1 mark)

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In a large population of moths, the number of eggs per cluster is approximately normally distributed with a mean of 165 eggs and a standard deviation of 25 eggs.

Using the 68–95–99.7% rule, determine
i. the percentage of clusters expected to contain more than 140 eggs. (1 mark)

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ii. the number of clusters expected to have less than 215 eggs in a sample of 1000 clusters. (1 mark)

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The standardised number of eggs in one cluster is given by `z = –2.4`
Determine the actual number of eggs in this cluster. (1 mark)

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Show Answers Only

a. `72`

b.i. `25text(%)`

b.ii. `-1`

c. `105\ text(eggs)`

Show Worked Solution

a.i. `text(Range)\ = 197-125= 72`

a.ii. `text(3 clusters > 170 eggs)`

`:.\ text(Percentage)`	`= 3/12 xx 100text(%)`
	`= 25text(%)`

b.i.	`ztext{-score (140)}`	`= (140-165)/25`
		`= −1`

`:.\ text(Percentage over 140)`

`= 68 + 16`

`= 84text(%)`

b.ii.	`ztext{-score (215)}`	`= (215-165)/25`
		`= 2`

`:.\ text(Percentage less than 215)`

`= 97.5text(%) xx 1000`

`= 975`

c.	`text(Using)\ \ \ z`	`= (x-barx)/s`
	`−2.4`	`= (x-165)/25`
	`x`	`= (−2.4 xx 25) + 165`
		`= 105\ text(eggs)`

CORE, FUR1 2017 VCAA 8-10 MC

The scatterplot below shows the wrist circumference and ankle circumference, both in centimetres, of 13 people. A least squares line has been fitted to the scatterplot with ankle circumference as the explanatory variable.

Part 1

The equation of the least squares line is closest to

ankle = 10.2 + 0.342 × wrist
wrist = 10.2 + 0.342 × ankle
ankle = 17.4 + 0.342 × wrist
wrist = 17.4 + 0.342 × ankle
wrist = 17.4 + 0.731 × ankle

Part 2

When the least squares line on the scatterplot is used to predict the wrist circumference of the person with an ankle circumference of 24 cm, the residual will be closest to

`–0.7`
`–0.4`
`–0.1`
`0.4`
`0.7`

Part 3

The residuals for this least squares line have a mean of 0.02 cm and a standard deviation of 0.4 cm.

The value of the residual for one of the data points is found to be – 0.3 cm.

The standardised value of this residual is

`–0.8`
`–0.7`
`–0.3`
`0.7`
`0.8`

Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ A`

`text(Part 3:)\ A`

Show Worked Solution

`text(Part 1)`

`ytext(-intercept = 10.2)`

`(text(note)\ xtext(-axis in graph begins at 21 cm))`

`:. text(wrist) = 10.2 + 0.342 xx text(ankle)`

`=> B`

`text(Part 2)`

`text(Predicted wrist)`	`= 10.2 + 0.342 xx 24`
	`= 18.4`

`text(Residual)`	`=\ text(actual − predicted)`
	`~~ 17.7 – 18.4`
	`~~ −0.7`

`=> A`

`text(Part 3)`

`barx = 0.02,qquad s_x = 0.4`

`text(If)\ \ x = −0.3,`

`z`	`= (x – barx)/s_x`
	`= (−0.3 – 0.02)/0.4`
	`= −0.8`

`=> A`

CORE, FUR1 2016 VCAA 4-5 MC

The weights of male players in a basketball competition are approximately normally distributed with a mean of 78.6 kg and a standard deviation of 9.3 kg.

Part 1

There are 456 male players in the competition.

The expected number of male players in the competition with weights above 60 kg is closest to

`3`
`11`
`23`
`433`
`445`

Part 2

Brett and Sanjeeva both play in the basketball competition.

When the weights of all players in the competition are considered, Brett has a standardised weight of `z` = – 0.96 and Sanjeeva has a standardised weight of `z` = – 0.26

Which one of the following statements is not true?

Brett and Sanjeeva are both below the mean weight for players in the basketball competition.
Sanjeeva weighs more than Brett.
If Sanjeeva increases his weight by 2 kg, he would be above the mean weight for players in the basketball competition.
Brett weighs more than 68 kg.
More than 50% of the players in the basketball competition weigh more than Sanjeeva

Show Answers Only

`text(Part 1:)\ E`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(Find)\ ztext(-score of 60 kg)`

`ztext(-score)`	`= (x – barx)/s`
	`= (60 – 78.6)/9.3`
	`= −2`

`text(Players with weight above 60 kg)`

`= 97.5text(%) xx 456`

`= 445`

`=> E`

`text(Part 2)`

`text(Calculate the weight of each player:)`

`text(Brett)`

`−0.96`	`= (x – 78.6)/9.3`
`x`	`= (9.3 xx −0.96) + 78.6`
	`~~ 69.7\ text(kg)`

`text(Sanjeev)`

`−0.26`	`= (x – 78.6)/9.3`
`x`	`= (9.3 xx −0.26) + 78.6`
	`~~ 76.2\ text(kg)`

`text(Consider)\ C,`

`76.182 + 2 = 78.182 < 78.6`

`:. C\ text(is not true.)`

`=> C`

CORE, FUR2 2006 VCAA 1

Table 1 shows the heights (in cm) of three groups of randomly chosen boys aged 18 months, 27 months and 36 months respectively.

Complete Table 2 by calculating the standard deviation of the heights of the 18-month-old boys.

Write your answer correct to one decimal place. (1 mark)

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A 27-month-old boy has a height of 83.1 cm.

Calculate his standardised height (`z` score) relative to this sample of 27-month-old boys.
Write your answer correct to one decimal place. (1 mark)

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The heights of the 36-month-old boys are normally distributed.

A 36-month-old boy has a standardised height of 2.

Approximately what percentage of 36-month-old boys will be shorter than this child? (1 mark)

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Using the data from Table 1, boxplots have been constructed to display the distributions of heights of 36-month-old and 27-month-old boys as shown below.

Complete the display by constructing and drawing a boxplot that shows the distribution of heights for the 18-month-old boys. (2 marks)

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Use the appropriate boxplot to determine the median height (in centimetres) of the 27-month-old boys. (1 mark)

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The three parallel boxplots suggest that height and age (18 months, 27 months, 36 months) are positively related.

Explain why, giving reference to an appropriate statistic. (1 mark)

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Show Answers Only

`3.8`
`−1.4`
`2.5 text(%)`
`89.5`
`text(Median height increases as age increases.)`

Show Worked Solution

a. `text(By calculator,)`

`text(standard deviation) = 3.8`

b.	`z`	`= (x – barx)/s`
		`= (83.1 – 89.3)/(4.5)`
		`=-1.377…`
		`= -1.4\ \ text{(1 d.p.)}`

♦♦ MARKER’S COMMENT: Attention required here as this standard question was “very poorly answered”.

c. `text{2.5% (see graph below)}`

d. `text(Range = 76 – 89.8,)\ Q_1 = 80,\ Q_3 = 85.8,\ text(Median = 83,)`

e. `89.5`

MARKER’S COMMENT: A boxplot statistic was required, so mean values were not relevant.

f. `text(The median height increases with age.)`

CORE, FUR2 2012 VCAA 1

The dot plot below displays the maximum daily temperature (in °C) recorded at a weather station on each of the 30 days in November 2011.

From this dot plot, determine
i. the median maximum daily temperature, correct to the nearest degree (1 mark)

--- 2 WORK AREA LINES (style=lined) ---
ii. the percentage of days on which the maximum temperature was less than 16 °C.
Write your answer, correct to one decimal place. (1 mark)

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Records show that the minimum daily temperature for November at this weather station is approximately normally distributed with a mean of 9.5 °C and a standard deviation of 2.25 °C.

Determine the percentage of days in November that are expected to have a minimum daily temperature less than 14 °C at this weather station.
Write your answer, correct to one decimal place. (1 mark)

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

Show Answers Only

1. `20 text(°C)`
2. `23.3text(%)`
`97.5text(%)`

Show Worked Solution

a.i. `text(The median is the average of the 15th and 16th)`

`\ \ text{data points (30 data points in total).}`.

`:.\ text(Median = 20 °C)`

a.ii. `text{Percentage of days less than 16 °C}`

`= 7/30 xx 100text(%)`

`=23.333…`

`=23.3text{% (to 1 d.p.)}`

b.	`z text{-score (14)}`	`=(x-barx)/s`
		`=(14-9.5)/2.25`
		`=2`

`text(Percentage of days with a minimum below 14 °C)`

`= 95text(%) + 2.5text(%)`

`= 97.5text(%)`

CORE, FUR2 2013 VCAA 3

The development index and the average pay rate for workers, in dollars per hour, for a selection of 25 countries are displayed in the scatterplot below.

The table below contains the values of some statistics that have been calculated for this data.

Determine the standardised value of the development index (`z` score) for a country with a development index of 91.
Write your answer, correct to one decimal place. (1 mark)

--- 3 WORK AREA LINES (style=lined) ---
Use the information in the table to show that the equation of the least squares regression line for a country’s development index, `y`, in terms of its average pay rate, `x`, is given by
`y = 81.3 + 0.272x` (2 marks)

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The country with an average pay rate of $14.30 per hour has a development index of 83.
Determine the residual value when the least squares regression line given in part (b) is used to predict this country’s development index.
Write your answer, correct to one decimal place. (2 marks)

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

Show Answers Only

`1.8`
`text(See Worked Solutions)`
`−2.2`

Show Worked Solution

MARKER’S COMMENT: Many students made a careless error by using the Average pay rate (`x`-variable).

a.	`z`	`= (y-bar y)/s`
		`= (91-85.6)/2.99`
		`=1.806…`
		`= 1.8\ \ text{(to 1 d.p.)}`

♦ Parts (b) and (c) had a combined mean mark of 48%.

b.	`b`	`= r xx s_y/s_x`
		`= 0.488 xx 2.99/5.37`
		`=0.2717…`

`a`	`= bary-b barx`
	`= 85.6-0.272 xx 15.7`
	`= 81.329…`

`:.\ text(The least squares line is)`

`y = 81.3 + 0.272x`

c.	`text(Predicted)`	`= 81.3 + 0.272 xx 14.30`
		`= 85.1896…`

`:.\ text(Residual)`	`= 83-85.1896…`
	`=-2.1896…`
	`= – 2.2`

CORE, FUR2 2014 VCAA 4

The scatterplot below shows the population density, in people per square kilometre, and the area, in square kilometres, of 38 inner suburbs of the same city.

For this scatterplot, `r^2 = 0.141`

Describe the association between the variables population density and area for these suburbs in terms of strength, direction and form. (1 mark)

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The mean and standard deviation of the variables population density and area for these 38 inner suburbs are shown in the table below.
i. One of these suburbs has a population density of 3082 people per square kilometre.
Determine the standard `z`-score of this suburb’s population density.
Write your answer, correct to one decimal place. (1 mark)

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

Assume the areas of these inner suburbs are approximately normally distributed.

ii. How many of these 38 suburbs are expected to have an area that is two standard deviations or more above the mean?
Write your answer, correct to the nearest whole number. (1 mark)

--- 1 WORK AREA LINES (style=lined) ---
iii. How many of these 38 inner suburbs actually have an area that is two standard deviations or more above the mean? (1 mark)

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

Show Answers Only

`text(The association is weak, negative and linear.)`
1. `− 0.8`
2. `1`
3. `2`

Show Worked Solution

♦♦ Mean mark 30%.
MARKER’S COMMENT: The most common error was to describe the association as positive.

a.	`r`	`= −sqrt(0.141)`
		`= −0.3755`

`:.\ text(The association is weak, negative and linear.)`

b.i	`z`	`=(x-bar x)/s`
		`= (3082-4370)/1560`
		`= -0.82…`
		`= -0.8\ \ text{(to 1 d.p.)}`

b.ii. `2.5text(%) xx 38 =0.95=1\ text{(nearest whole)}`

b.iii. `text{Area size (2 std deviation above mean)}`

♦♦ Very few students answered this part correctly.

`= 3.4 + 2 xx 1.6`

`=6.6\ text(km²)`

`:.\ text{From the graph, 2 inner suburbs have an area}`

`text(2 standard deviations above the mean.)`

CORE, FUR1 2015 VCAA 4-5 MC

The foot lengths of a sample of 2400 women were approximately normally distributed with a mean of 23.8 cm and a standard deviation of 1.2 cm.

Part 1

The expected number of these women with foot lengths less than 21.4 cm is closest to

A. `60`

B. `120`

C. ` 384`

D. `2280`

E. `2340`

Part 2

The standardised foot length of one of these women is `z` = – 1.3

Her actual foot length, in centimetres, is closest to

A. `22.2`

B. `22.7`

C. `25.3`

D. `25.6`

E. `31.2`

Show Answers Only

`text(Part 1:)\ A`

`text(Part 2:)\ A`

Show Worked Solution

`text(Part 1)`

`z text{-score (21.4)}`	`=(x – barx)/s`
	`=(21.4-23.8)/1.2`
	`= – 2`

`:.\ text(Number of women with foot length under 21.4 cm)`

`=2.5text(%) xx 2400 = 60`

`=> A`

`text(Part 2)`

`text(If)\ \ z = – 1.3`

`– 1.3`	`=(x-23.8)/1.2`
`x-23.8`	`= – 1.3 xx 1.2`
`:. x`	`=23.8 – 1.56`
	`=22.24`

`=> A`

CORE, FUR1 2006 VCAA 4 MC

The head circumference (in cm) of a population of infant boys is normally distributed with a mean of 49.5 cm and a standard deviation of 1.5 cm.

Four hundred of these boys are selected at random and each boy’s head circumference is measured.

The number of these boys with a head circumference of less than 48.0 cm is closest to

A. `3`

B. `10`

C. `64`

D. `272`

E. `336`

Show Answers Only

`C`

Show Worked Solution

`bar x=49.5,\ \ \ s=1.5`

`z text{-score (49.5)`	`=(x-bar x)/s`
	`=(48.0-49.5)/1.5`
	`=– 1`

`:.\ text(Number of boys with a head under 48.0 cm)`

`=16text(%) xx 400`

`=64`

`=> C`

CORE, FUR1 2007 VCAA 3 MC

A student obtains a mark of 56 on a test for which the mean mark is 67 and the standard deviation is 10.2.

The student’s standardised mark (standard `z`-score) is closest to

A. `– 1.08`

B. `– 1.01`

C. `1.01`

D. `1.08`

E. `49.4`

Show Answers Only

`A`

Show Worked Solution

`z text(-score)`	`=(x-bar x)/s`
	`=(56-67)/10.2`
	`= – 1.0784…`

`rArr A`

CORE, FUR1 2011 VCAA 9-10 MC

The length of a type of ant is approximately normally distributed with a mean of 4.8 mm and a standard deviation of 1.2 mm.

Part 1

From this information it can be concluded that around 95% of the lengths of these ants should lie between

A. `text(2.4 mm and 6.0 mm)`

B. `text(2.4 mm and 7.2 mm)`

C. `text(3.6 mm and 6.0 mm)`

D. `text(3.6 mm and 7.2 mm)`

E. `text(4.8 mm and 7.2 mm)`

Part 2

A standardised ant length of `z = text(−0.5)` corresponds to an actual ant length of

A. `text(2.4 mm)`

B. `text(3.6 mm)`

C. `text(4.2 mm)`

D. `text(5.4 mm)`

E. `text(7.0 mm)`

Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(95% of scores lie between ±2 std dev)`

`bar x = 4.8, \ \ \ s = 1.2`

`text(Lower limit)`	`= bar x – 2text(s)`
	`= 4.8 – 2(1.2)`
	`= 2.4\ text(mm)`
`text(Upper limit)`	`= bar x + 2text(s)`
	`= 4.8 + 2(1.2)`
	`= 7.2\ text(mm)`

`=>B`

`text(Part 2)`

`z`	`= \ \ (x – bar x)/s`
`-0.5`	`= \ \ (x – 4.8)/1.2`
`-0.6`	`= \ \ x – 4.8`
`x`	`= \ \ 4.2\ text(mm)`

`=>C`

CORE, FUR1 2009 VCAA 7 MC

The level of oil use in certain countries is approximately normally distributed with a mean of 42.2 units and a standard deviation of 10.2 units.

The percentage of these countries in which the level of oil use is greater than 32 units is closest to

A. 5%

B. 16%

C. 34%

D. 84%

E. 97.5%

Show Answers Only

`D`

Show Worked Solution

`bar x`

`=42.2`

`s`

`=10.2`

`z text(-score (32))`	`=(x – barx)/s`
	`=(32-42.2)/10.2`
	`=–1`

`text(68% lie between)\ z text(-score of –1 and 1)`

`=>\ text(34% lie between z-score –1 and 0)`

`text(50% lie above z-score of 0)`

`∴\ text(% above 32 units)`	`=34 + 50`
	`= 82text(%)`

`rArr D`

CORE, FUR1 2008 VCAA 6-7 MC

The pulse rates of a large group of 18-year-old students are approximately normally distributed with a mean of 75 beats/minute and a standard deviation of 11 beats/minute.

Part 1

The percentage of 18-year-old students with pulse rates less than 75 beats/minute is closest to

A. 32%

B. 50%

C. 68%

D. 84%

E. 97.5%

Part 2

The percentage of 18-year-old students with pulse rates less than 53 beats/minute or greater than 86 beats/minute is closest to

A. 2.5%

B. 5%

C. 16%

D. 18.5%

E. 21%

Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

`barx=75,\ \ \ s=11`

`text(In a normal distribution, mean = median.)`

`:.\ text(50% of group are below mean of 75)`

`=>B`

`text(Part 2)`

♦ Mean mark 44%.
MARKERS’ COMMENT: Two applications of the 68-95-99.7% rule are required. A good strategy is to first draw a normal curve and shade the required areas.

`barx=75,\ \ \ s=11`

`z text{-score (53)}`	`=(x-barx) /s`
	`=(53-75)/11`
	`= – 2`

`z text{-score (86)}`	`= (86-75)/11`
	`=1`

`text(From the diagram, the % of students that have a)`

`z text(-score below –2 or above 1)`

`=2.5+16`

`=18.5 text(%)`

`=>D`

CORE, FUR1 2010 VCAA 5-6 MC

The lengths of the left feet of a large sample of Year 12 students were measured and recorded. These foot lengths are approximately normally distributed with a mean of 24.2 cm and a standard deviation of 4.2 cm.

Part 1

A Year 12 student has a foot length of 23 cm.
The student’s standardised foot length (standard `z` score) is closest to

A. –1.2

B. –0.9

C. –0.3

D. 0.3

E. 1.2

Part 2

The percentage of students with foot lengths between 20.0 and 24.2 cm is closest to

A. 16%

B. 32%

C. 34%

D. 52%

E. 68%

Show Answers Only

`text(Part 1:)\ C`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`bar(x) = 24.2,`	`s=4.2`
`z text{-score (23)}`	`=(x – bar(x))/s`
	`= (23 – 24.2)/4.2`
	`= -0.285…`

`=> C`

`text(Part 2)`

`z text{-score (20)}`	`=(20- 24.2)/4.2`
	`= -1`
`z text{-score (24.2)}`	`= 0`

`text(68% have a)\ z text(-score between –1 and 1)`

`:.\ text(34% have a)\ z text(-score between –1 and 0)`

`=> C`

CORE, FUR1 2013 VCAA 5-6 MC

The time, in hours, that each student spent sleeping on a school night was recorded for 1550 secondary-school students. The distribution of these times was found to be approximately normal with a mean of 7.4 hours and a standard deviation of 0.7 hours.

Part 1

The time that 95% of these students spent sleeping on a school night could be

A. less than 6.0 hours.

B. between 6.0 and 8.8 hours.

C. between 6.7 and 8.8 hours.

D. less than 6.0 hours or greater than 8.8 hours.

E. less than 6.7 hours or greater than 9.5 hours.

Part 2

The number of these students who spent more than 8.1 hours sleeping on a school night was closest to

A. 16

B. 248

C. 1302

D. 1510

E. 1545

Show Answers Only

`text(Part 1:)\ B`

`text(Part 2:)\ B`

Show Worked Solution

`text(Part 1)`

`-2 < z text(-score) < 2,\ \ text(contains 95% of students.)`

`barx=7.4, \ \ \ s=0.7`

`z text(-score of +2)`	`= 7.4+(2×0.7)`
	`= 8.8\ text(hours)`

`z text(-score of –2)`	`= 7.4−(2×0.7)`
	`= 6.0\ text(hours)`

`:. 95 text(% students sleep between 6 and 8.8 hours.`

`=>B`

`text(Part 2)`

`text (Find)\ z text(-score of 8.1 hours)`

`ztext(-score)`	`= (x-barx)/s`
	`= (8.1-7.4)/0.7`
	`= 1`

`text(68% students have)\ \ –1 < z text(-score) < 1`

`:. 16 text(% have)\ z text(-score) > 1`

`:.\ text(# Students)`	`= 16text(%) xx1550`
	`= 248`

`=>B`

\(\bar{x}+2z\)	\(=178.9\ \ …\ (1)\)
\(\bar{x}-z\)	\(=157.6\ …\ (2)\)

\(0.45\)	\(=\dfrac{174.6-\overline{x}}{12.25} \)
\(\overline{x}=\)	\(174.6-(0.45 \times 12.25) = 169.0875\)

b.	\(z\)	\(=\dfrac{x-\overline x}{s_x}\)
		\(=\dfrac{2.35-2.23}{0.15}\)
		\(=0.8\)

\(z\)	\(=\dfrac{x-\overline{x}}{s_x}\)
\(-1.75\)	\(=\dfrac{48-55.7}{s_x}\)
\(s_x\)	\(=\dfrac{48-55.7}{-1.75}\)
	\(=4.4\)

\(z\)	\(= \dfrac{x-\bar x}{s_x}\)
\(-0.85\)	\(=\dfrac{x-163.56}{8.14}\)
\(x-163.56\)	\(=8.14 \times -0.85\)
\(x\)	\(=163.56-6.919\)
	\(=156.641\)