smc-266-60-MEAN Smoothing

Data Analysis, GEN1 2024 VCAA 15 MC

The table below shows the total number of cans of soft drink sold each month at a suburban cafe in 2023.

\begin{array}{|l|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Month } \rule[-1ex]{0pt}{0pt}& 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
\rule{0pt}{2.5ex} \textbf{Cans sold } \rule[-1ex]{0pt}{0pt}& 316 & 321 & 365 & 306 & 254 & 308 & 354 & 357 & 381 & 355 & 365 & 324 \\
\hline
\end{array}

The six-mean smoothed value of the number of cans sold, with centring, for month 5 is closest to

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$C$

Show Worked Solution

$\text{Six-mean smoothed:}$

$\text{Months:}\ 2-7$	$=\dfrac{321+365+306+254+308+354}{6}=318$
$\text{Months:}\ 3-8$	$=\dfrac{365+306+254+308+354+357}{6}=324$

$\therefore\ \text{Centred smoothed mean}$	$=\dfrac{318+324}{2}$
	$=321$

$C$

Data Analysis, GEN1 2022 VCAA 15 MC

The daily number of cups of coffee sold by a food truck over a three-week period is shown in the table below.

The six-mean smoothed number of cups of coffee, with centring, sold on Thursday in Week 2 is closest to

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$B$

Show Worked Solution

$\text{Six-mean smoothed average (Thu, week 2)}$

$=[(98+104+145+163+134+128)\ ÷\ 6 + $

$(104+145+163+134+128+206)\ ÷\ 6]\ ÷\ 2 $

$=137.66…$

$\Rightarrow B$

Data Analysis, GEN1 2023 VCAA 15 MC

The number of visitors to a public library each day for 10 consecutive days was recorded.

These results are shown in the table below.

\begin{array} {|l|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \textbf{Day number} \rule[-1ex]{0pt}{0pt} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
\rule{0pt}{2.5ex} \textbf{Number of visitors} \rule[-1ex]{0pt}{0pt} & 337 & 317 & 313 & 335 & 322 & 335 & 322 & 338 & 302 & 349 \\
\hline
\end{array}

The eight-mean smoothed number of visitors with centring for day number 6 is

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$C$

Show Worked Solution

$\text{8 Mean average (days 2 – 9):}$

$(317+313+335+322+335+322+338+302)\ ÷\ 8 = 323 $

$\text{8 Mean average (days 3 – 10):}$

$(313+335+322+335+322+338+302+349)\ ÷\ 8 = 327 $

$\text{Centred value}\ = \dfrac{323+327}{2} = 325$

$\Rightarrow C$

CORE, FUR1 2021 VCAA 14 MC

A garden centre sells garden soil.

The table below shows the daily quantity of garden soil sold, in cubic metres, over a one-week period.

The quantity of garden soil sold on Wednesday, Thursday and Friday is not shown.

The five-mean smoothed quantity of garden soil sold on Thursday is 206 m³.

The three-mean smoothed quantity of garden soil sold on Thursday, in cubic metres, is

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`B`

Show Worked Solution

`text{Let} \ x = \ text{amount sold Wednesday – Friday}`

♦ Mean mark 48%.

`{x + 186 + 346}/5`	`= 206`
`x`	`= (206 xx 5) – 186 – 346`
	`= 498 \ text{m}^3`

`:. \ text{Three-mean smoothed quantity (Thursday)}`

`=498/3`

`=166\ text(m³)`

`=> B`

CORE, FUR1 2020 VCAA 19-20 MC

The time series plot below displays the number of airline passengers, in thousands, each month during the period January to December 1960.

Part 1

During 1960, the median number of monthly airline passengers was closest to

461 000
465 000
471 000
573 000
621 000

Part 2

During the period January to May 1960, the total number of airline passengers was 2 160 000.

The five-mean smoothed number of passengers for March 1960 is

419 000
424 000
430 000
432 000
434 000

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`text(Part 1:)\ A`

`text(Part 2:)\ D`

Show Worked Solution

Part 1

♦ Mean mark part (1) 45%.

`text(12 data points)`

`text(Median)`	`= {text(6th + 7th data point)}/2`
	`~~ (460\ 000 + 460\ 000)/2`
	`~~ 460\ 000`

`=> A`

Part 2

`text(Five-mean smoothed number)`

`= (2\ 160\ 000)/5`

`= 432\ 000`

`=> D`

CORE, FUR1 2020 VCAA 17-18 MC

Table 4 below shows the monthly rainfall for 2019, in millimetres, recorded at a weather station, and the associated long-term seasonal indices for each month of the year.

Part 1

The deseasonalised rainfall for May 2019 is closest to

71.3 mm
75.8 mm
86.1 mm
88.1 mm
113.0 mm

Part 2

The six-mean smoothed monthly rainfall with centring for August 2019 is closest to

67.8 mm
75.9 mm
81.3 mm
83.4 mm
86.4 mm

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`text(Part 1:)\ B`

`text(Part 2:)\ C`

Show Worked Solution

Part 1

`text(Deseasonalised rainfall for May)`

`= 92.6/1.222`

`= 75.8\ text(mm)`

`=> B`

Part 2

`text{Six-mean smoothed average (Aug)}`

`=[(92.6 + 77.2 + 80 + 86.8 + 93.8 + 55.2) ÷6 +`

`(77.2 + 80 + 86.8 + 93.8 + 55.2 + 97.3) ÷ 6] ÷ 2`

`~~ (80.93 + 81.72) ÷ 2`

`~~ 81.3\ text(mm)`

`=> C`

CORE, FUR1 2019 VCAA 15-16 MC

The time series plot below shows the monthly rainfall at a weather station, in millimetres, for each month in 2017.

Part 1

The median monthly rainfall for 2017 was closest to

53 mm
82 mm
96 mm
103 mm
111 mm

Part 2

If seven-mean smoothing is used to smooth this time series plot, the number of smoothed data points would be

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`text(Part 1:)\ D`

`text(Part 2:)\ C`

Show Worked Solution

`text(Part 1)`

`text(12 data points)`

`text(Median)`	`= text{(6th + 7th)}/2`
	`~~ (97 + 109)/2`
	`~~ 103\ text(mm)`

`=> D`

`text(Part 2)`

`text(January – July)\ =>\ text(First data point)`

`text(February – August)\ =>\ text(S)text(econd data point)`

`qquad vdots`

`text(June – December)\ =>\ text(Sixth data point)`

`=> C`

CORE, FUR1 2018 VCAA 15 MC

The table below shows the monthly profit, in dollars, of a new coffee shop for the first nine months of 2018.

Using four-mean smoothing with centring, the smoothed profit for May is closest to

$2502
$3294
$3503
$3804
$4651

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`B`

Show Worked Solution

`text(4 point mean:)`

`text(Mar-Jun)`	`= (2402 + 2456 + 4651 + 3456)/4 = 3241.25`
`text(Apr-Jul)`	`= (2456 + 4651 + 3456 + 2823)/4 = 3346.5`

`:.\ text(Centred Mean)`	`= (3241.25 + 3346.5)/2`
	`= 3293.875`

`=> B`

CORE, FUR1 2017 VCAA 13-15 MC

The wind speed at a city location is measured throughout the day.

The time series plot below shows the daily maximum wind speed, in kilometres per hour, over a three-week period.

Part 1

The time series is best described as having

seasonality only.
irregular fluctuations only.
seasonality with irregular fluctuations.
a decreasing trend with irregular fluctuations.
an increasing trend with irregular fluctuations.

Part 2

The seven-median smoothed maximum wind speed, in kilometres per hour, for day 4 is closest to

`22`
`26`
`27`
`30`
`32`

Part 3

The table below shows the daily maximum wind speed, in kilometres per hour, for the days in week 2.

A four-point moving mean with centring is used to smooth the time series data above.

The smoothed maximum wind speed, in kilometres per hour, for day 11 is closest to

`22`
`24`
`26`
`28`
`30`

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`text(Part 1:)\ B`

`text(Part 2:)\ D`

`text(Part 3:)\ D`

Show Worked Solution

`text(Part 1)`

`text(The time series plot shows no obvious trend and)`

`text(is over too short a period to show seasonality.)`

`=> B`

`text(Part 2)`

`text(Consider the 7 values where day 4 is the middle)`

`text(data point.)`

`text(By inspection of the graph, the 4th highest point = 30.)`

`=> D`

`text(Part 3)`

`text(Mean for Day 9 – 12)`

`= (22 + 19 + 22 + 43)/4 = 26.5`

`text(Mean for Day 10 – 13)`

`= (19 + 22 + 43 + 37)/4 = 30.25`

`:. 4text(-point moving mean with centring)`

`= (26.5 + 30.25)/2`

`= 28.375`

`=> D`

CORE, FUR2 2016 VCAA 4

The time series plot below shows the minimum rainfall recorded at the weather station each month plotted against the month number (1 = January, 2 = February, and so on).

Rainfall is recorded in millimetres.

The data was collected over a period of one year.

Five-median smoothing has been used to smooth the time series plot above.

The first four smoothed points are shown as crosses (×).

Complete the five-median smoothing by marking smoothed values with crosses (×) on the time series plot above. (2 marks)

The maximum daily rainfall each month was also recorded at the weather station.

The table below shows the maximum daily rainfall each month for a period of one year.

The data in the table has been used to plot maximum daily rainfall against month number in the time series plot below.

Two-mean smoothing with centring has been used to smooth the time series plot above.

The smoothed values are marked with crosses (×).

Using the data given in the table, show that the two-mean smoothed rainfall centred on October is 157.25 mm. (2 marks)

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`text(See Worked Solutions)`
`text(Proof)\ text{(See Worked Solutions)}`

Show Worked Solution

♦ Mean mark of both Parts (a) and (b) was 49%.
MARKER’S COMMENT: Use the accurate table data when available. Reading values from the graph will cause inaccuracies.

b.	`text(Mean)\ _text(Sep-Oct)`	`= (124 + 140)/2`
		`= 132\ text(mm)`
	`text(Mean)\ _text(Oct-Nov)`	`= (140 + 225)/2`
		`= 182.5\ text(mm)`

`:.\ text{Two mean (smoothed) for October}`

`= (132 +182.5)/2`

`= 157.25\ text(mm … as required)`

CORE, FUR1 2006 VCAA 10 MC

The table below displays the total monthly rainfall (in mm) in a reservoir catchment area over a one-year period.

Using three mean moving average smoothing, the smoothed value for the total rainfall in April is closest to

A. `65`

B. `66`

C. `70`

D. `75`

E. `88`

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`C`

Show Worked Solution

`text(Three mean average for April)`

`=(35+99+75)/3`

`=69.66…`

`rArr C`

CORE, FUR1 2011 VCAA 13 MC

The table below shows the number of broadband users in Australia for each of the years from 2004 to 2008.

A two-point moving mean, with centring, is used to smooth the time series.

The smoothed value for the number of broadband users in Australia in 2006 is

A. `2 \ 958 \ 000`

B. `3 \ 379 \ 600`

C. `3 \ 455 \ 500`

D. `3 \ 661 \ 500`

E. `3 \ 900 \ 000`

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`D`

Show Worked Solution

`text(Two point mean for)`

`text(2005/06)`	`= (2 \ 016 \ 000 + 3 \ 900 \ 000) / 2`
	`= 2 \ 958 \ 000`

`text(2006/07)`	`= (3 \ 900 \ 000 + 4 \ 830 \ 000) / 2`
	`= 4 \ 365 \ 000`

`:.\ text(Centered mean)`	`= (2 \ 958 \ 000 + 4 \ 365 \ 000) / 2`
	`= 3 \ 661 \ 500`

`=> D`

CORE, FUR1 2008 VCAA 11-13 MC

The time series plot below shows the number of users each month of an online help service over a twelve-month period.

Part 1

The time series plot has

A. no trend.

B. no variability.

C. seasonality only.

D. an increasing trend with seasonality.

E. an increasing trend only.

Part 2

The data values used to construct the time series plot are given below.

2008 12

A four-point moving mean with centring is used to smooth timeline series.
The smoothed value of the number of users in month number 5 is closest to

A. `357`

B. `359`

C. `360`

D. `365`

E. `373`

Part 3

A least squares regression line is ﬁtted to the time series plot.
The equation of this least squares regression line is

number of users = 346 + 2.77 × month number

Let month number 1 = January 2007, month number 2 = February 2007, and so on.

Using the above information, the regression line predicts that the number of users in December 2009 will be closest to

A. `379`

B. `412`

C. `443`

D. `446`

E. `448`

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`text(Part 1:)\ E`

`text(Part 2:)\ C`

`text(Part 3:)\ D`

Show Worked Solution

`text(Part 1)`

♦ Mean mark 39%.
MARKERS’ COMMENT: 50% of students incorrectly read the three large random fluctuations in monthly sales as seasonality, which can’t be determined over only 12 months.

`text(The time series is clearly trending upwards with)`

`text(higher lows and higher highs occurring.)`

`text(The large fluctuations are random and should)`

`text(not be confused with seasonality.)`

`=>E`

`text(Part 2)`

`text(Mean for months 3-6)`

`=(354+356+373+353)/4`

`=359`

`text(Mean for months 4-7)`

`=(356+373+353+364)/4`

`=361.5`

`:.\ text(Four point moving mean with centring)`

`=(359+361.5)/2`

`=360.25`

`=> C`

`text(Part 3)`

`text(December 2009 will be month number 36.)`

`:.\ text(Number of users)`	`= 346+2.77xx36`
	`= 445.72`

`=> D`

CORE, FUR1 2013 VCAA 12-13 MC

The time series plot below displays the number of guests staying at a holiday resort during summer, autumn, winter and spring for the years 2007 to 2012 inclusive.

Part 1

Which one of the following best describes the pattern in the time series?

A. random variation only

B. decreasing trend with seasonality

C. seasonality only

D. increasing trend only

E. increasing trend with seasonality

Part 2

The table below shows the data from the times series plot for the years 2007 and 2008.

Using four-mean smoothing with centring, the smoothed number of guests for winter 2007 is closest to

A. `85`

B. `107`

C. `183`

D. `192`

E. `200`

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`text(Part 1:)\ C`

`text(Part 2:)\ D`

Show Worked Solution

`text(Part 1)`

`text(The pattern in the time series is seasonal only,)`

`text(with peaks appearing in Summer. There is no)`

`text(apparent year-to-year trend.)`

`=> C`

`text(Part 2)`

`text{Mean of guests (Season 1-4)}`

`=(390+126+85+130)/4`

`=182.75`

`text{Mean of guests (Season 2-5)}`

`=(126+85+85+130+460)/4`

`=200.25`

`:.\ text(Four-mean smoothing with centring)`

`=(182.75+200.25)/2`

`=191.5`

`=> D`

\(\text{Months:}\ 2-7\)	\(=\dfrac{321+365+306+254+308+354}{6}=318\)
\(\text{Months:}\ 3-8\)	\(=\dfrac{365+306+254+308+354+357}{6}=324\)

\(\therefore\ \text{Centred smoothed mean}\)	\(=\dfrac{318+324}{2}\)
	\(=321\)