##### Document Text Contents

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Syed Nasir Danial ([email protected]) MUN ID 201176310

Page 1 of 3

Safety & Risk Engineering

ENGI 9115

Solution to Assignment # 1

Course Supervisor: Dr. Faisal Khan

Name: Syed Nasir Danial Email:[email protected]

Dated: May 10, 2012

Problem 2: A worker is told her chances of being killed by a particular process are 1 in every 500 years.

Should the worker be satisfied or alarmed? What is the FAR (assuming normal working hours) and the

deaths per person per year?

Solution:

The sample space is long enough to cover the period of 500 years.

The sample space comprises workers and the probability distribution is such that for a worker, say, Wi

which is sampled at the ith time unit:

Prob(Wi will die) = 1.0

Since the probability is at maximum which means that the worker must die in 500 years, therefore, the

number of fatalities in 500 years is 1.

Although, it is difficult to say if the person is really alarmed without knowing the actual probability

distribution but assuming a uniform distribution, and because the maximum duration of lifetime a worker

can work is 50 years, if we divide the 500yrs interval into 10 pieces, the probability will not be changed

because of uniformity in the distribution. But in case, the distribution is not uniform, say it is skewed

towards left or right, and possesses some correlation structure then changing the sample size will have a

definite impact on the probability of selection.

Anyways, under the assumption of a uniform distribution, which I think is often natural to a layman, the

worker will be alarmed.

Now, the FAR is calculated as:

.

100

The deaths per person per year is called Fatality Rate and is calculated as:

8 50 6 ; assuming 6 days per week, and 50 weeks in a year

= 0.24

Page 2

Syed Nasir Danial ([email protected]) MUN ID 201176310

Page 2 of 3

Problem 5: A car leaves New York City and travels 2800-miles distance to Los Angles at an average

speed of 50 mile per hour. An alternative travel plan is to fly on a commercial airline for 4 and 1/2 hr.

Which travel is safest, based on FAR in general, and for this trip, fatality rate (this trip)?

Solution

Based on FAR, in general1, travelling by car is safer because the corresponding FAR value is 57 which is

around a quarter (only 23.75%) for the value of FAR for travelling by air.

However, if we keep the parameters of the given case, we can calculate the number of fatalities both for

car and for aeroplane as given here as under:

By definition, FAR is defined as:

.

For this trip by car

The total passenger hours = = 2000000

57

.

, ,

=> no. of fatalities by car = fatality rate by car= 1.14.

For this trip by air

The speed by air is 2800/4.5 = 622.22 miles / hr.

The total passenger hours =

.

= 160714.3

240

.

.

no. of fatalities by air =fatality rate by air = 0.38

Comparing the two values, we find that:

no. of fatalities by air < no. of fatalities by car

=> Travelling by air is safer for this trip because the fatality rate is less.

1

see Loss Prevention by Frank P. Lees (1980)

Syed Nasir Danial ([email protected]) MUN ID 201176310

Page 1 of 3

Safety & Risk Engineering

ENGI 9115

Solution to Assignment # 1

Course Supervisor: Dr. Faisal Khan

Name: Syed Nasir Danial Email:[email protected]

Dated: May 10, 2012

Problem 2: A worker is told her chances of being killed by a particular process are 1 in every 500 years.

Should the worker be satisfied or alarmed? What is the FAR (assuming normal working hours) and the

deaths per person per year?

Solution:

The sample space is long enough to cover the period of 500 years.

The sample space comprises workers and the probability distribution is such that for a worker, say, Wi

which is sampled at the ith time unit:

Prob(Wi will die) = 1.0

Since the probability is at maximum which means that the worker must die in 500 years, therefore, the

number of fatalities in 500 years is 1.

Although, it is difficult to say if the person is really alarmed without knowing the actual probability

distribution but assuming a uniform distribution, and because the maximum duration of lifetime a worker

can work is 50 years, if we divide the 500yrs interval into 10 pieces, the probability will not be changed

because of uniformity in the distribution. But in case, the distribution is not uniform, say it is skewed

towards left or right, and possesses some correlation structure then changing the sample size will have a

definite impact on the probability of selection.

Anyways, under the assumption of a uniform distribution, which I think is often natural to a layman, the

worker will be alarmed.

Now, the FAR is calculated as:

.

100

The deaths per person per year is called Fatality Rate and is calculated as:

8 50 6 ; assuming 6 days per week, and 50 weeks in a year

= 0.24

Page 2

Syed Nasir Danial ([email protected]) MUN ID 201176310

Page 2 of 3

Problem 5: A car leaves New York City and travels 2800-miles distance to Los Angles at an average

speed of 50 mile per hour. An alternative travel plan is to fly on a commercial airline for 4 and 1/2 hr.

Which travel is safest, based on FAR in general, and for this trip, fatality rate (this trip)?

Solution

Based on FAR, in general1, travelling by car is safer because the corresponding FAR value is 57 which is

around a quarter (only 23.75%) for the value of FAR for travelling by air.

However, if we keep the parameters of the given case, we can calculate the number of fatalities both for

car and for aeroplane as given here as under:

By definition, FAR is defined as:

.

For this trip by car

The total passenger hours = = 2000000

57

.

, ,

=> no. of fatalities by car = fatality rate by car= 1.14.

For this trip by air

The speed by air is 2800/4.5 = 622.22 miles / hr.

The total passenger hours =

.

= 160714.3

240

.

.

no. of fatalities by air =fatality rate by air = 0.38

Comparing the two values, we find that:

no. of fatalities by air < no. of fatalities by car

=> Travelling by air is safer for this trip because the fatality rate is less.

1

see Loss Prevention by Frank P. Lees (1980)