Feasibility of Activity Recognition using a Single Chest Mounted Accelerometer

Question 1
You work for a consulting engineering firm who has been contracted by the
Australian Border Force to evaluate two commercial solutions which are being
considered for the next generation border control system. This system utilises
iris recognition to verify individuals as they enter the country through
international airports.
The system employs verification only: a person presents their passport; the
system then checks to see if the identity of the person matches the identity of
the person named on the passport. This match is conducted by comparing an iris
image of the person taken at the border, with the iris image stored on their e-
passport.
In the border control system, the verification test has two outcomes:
1. V+: the person is verified (i.e., their iris image taken at the border
matches that stored on their e-passport); or
2. V-: the person is not verified.
In reality, the person could be:
1. P+: a genuine traveller, who has their own unique and valid passport; or
2. P-: an imposter, who is not genuine and has a fake or stolen passport (for
example).
Each commercial solution has strengths and weaknesses. The specifications
released by the commercial providers are given in the table below. It is also
known from historical data that the probability of a person being an imposter is
0.12%.
Commercial Solution Solution 1 –
Eyematch
Solution 2 –
Bullseye
The system correctly verifies a
genuine traveller. 98.26% 98.35%
The system incorrectly verifies an
imposter. 5.12% 5.18%
1a Calculate conditional probabilities
Your firm is to compute the conditional probability (for both solutions) that there
is an imposter, given the verification is positive; i.e. p(P-|V+). Show your
calculations using Bayes’ rule.Hence, first the conditional probability of taking solution 1-Eyematch is
p(P-|V+) = p(P-∩V+)/(p(V+) = (p(V+|P-)*p(P-))/(p(V+|P-)*p(P-) + p(V+|P+)*p(P+)

= (0.0512*0.0012)/(0.0512*0.0012 + 0.9826*(1-0.0012)) = 0.00626%
Now, second conditional probability of taking solution 2-Bullseye is
p(P-|V+) = p(P-∩V+)/(p(V+) = (p(V+|P-)*p(P-))/(p(V+|P-)*p(P-) + p(V+|P+)*p(P+)
= (0.0518*0.0012)/(0.0518*0.0012 + 0.9835*(1-0.0012)) = 0.00633%
1b Provide a recommendation
Provide your recommendations to the Australian Border Force on the best choice
of the solution if they want to minimise this conditional probability.
You should draw on the unit content relating to dealing with uncertainty to
answer this question.Hence, for the solution 1 the conditional probability is lower than the conditional
probability of a random person being an imposter given that the verification is
positive. So, the Australian Border Force should employ solution 1 as in this case
the conditional probability is minimum.

Question 2
Part A
 Automotive Excellence produces a range of components needed to produce
automobile engines.
 Production of different component categories varies from month to month.
 The intent of this graph is to summarise the variation in numbers of
component categories from month to month in 2017.
Cylinder Blocks
Pistons
Cylinder Heads
Crank Shafts
Connecting Rods
Oil Sumps
Camshafts
Valves
Injectors
Push Rods
Manifolds
Gaskets
0% 10% 20% 30% 40% 50% 60% 70% 80% 90%100%
Automotive Excellence
Production Amounts in 2017
Minumim
Maximum
Median
2A.1 What’s wrong with this graphic?
List and briefly explain each of the problems that you detect in the design of this
graphic drawing on the principles presented on visualisation and from your wider
reading.In this particular graphic, the difference between the minimum, median and
maximum number of components produced in months is not clearly shown but a
rough estimate by visualization can be observed. Also, the scale is represented
in percentage so the actual values of min, max and median can not be
determined perfectly.
2A.2 How would you redesign this graphic?
Develop an alternative graphic that addresses the problems you detected

Cylinder Blocks
Pistons
Cylinder Heads
Crank Shafts
Connecting Rods
Oil Sumps
Camshafts
Valves
Injectors
Push Rods
Manifolds
Gaskets
0
200
400
600
800
1000
1200
Number of Components of Produced
Minimum Maximum Median
2A.3 What do you now observe in your redesigned graphic?
What are the salient problems or features of the data that you now observe?Now, in the above clustered column chart the minimum, maximum and median
values of each component can be exactly determined from the graph and the
difference between them can be calculated from the graph.

Part B
 Automotive Excellence has gaskets which come in a range of sizes for
different engine blocks and for different components.
 Through a sophisticated product tracking system, Automotive Excellence
monitors the working life of these gaskets and have collected time-to-
failure data on hundreds of gaskets.
 The Automotive Excellence product quality team have summarised this
data by calculating the mean failure times of a sample of each size. The
sizes are reported by their overall length in mm.
 The intent of this graph is to help understand whether there is a
relationship between gasket size and failure times.
0 2 4 6 8 10 12
0
500
1000
1500
2000
2500
3000
Mean Time to Failure
Mean time to failure
2B.1 What’s wrong with this graphic?
List and briefly explain each of the problems that you detect in the design of this
graphic drawing on the principles presented on visualisation and from your wider
reading.In the above graphic the gasket size (in mm) is not taken in the x axis and hence
this graphic is not able to represent the mean time of failure as a function of
gasket size in mm. Also, there is no axis labels in the graph.
2B.2 How would you redesign this graphic?
Develop an alternative graphic that addresses the problems you detected

0 100 200 300 400 500 600
0
500
1000
1500
2000
2500
3000
f(x) = 1.78 x + 1252.26
R² = 0.38
Mean time to failure
Size in mm
Mean time
2B.3 What do you now observe in your redesigned graphic?
What are the salient problems or features of the data that you now observe?Now, in the above scatterplot the mean time is represented with respect to the
size as given by the linear regression line. The regression model is y = 1.7822x
+1252.3, where y = mean time x = size in mm. The model represents only
38.11% percentage of explained variation in mean time by the gasket size
variation. Hence, the model is a good model for predicting the Mean-time, it is
possible that the mean time depends on other external variables.