Transcript Engineering Statistics

Engineering Statistics
Chapter 2
Special Variables
2B Poisson Distribution
Rare Events
• In many statistical problems, an event which
seldom happens is considered. The rareness is of
course relative. Even though people who die of
cancer because of smoking in Malaysia is in tens
of thousands annually, compared to the
population, it is still a rare event.
• In such cases, it is found that a distribution using
exponential function will model the probabilities
of the event based on the mean number of
occurrences.
Poisson Distribution
If the mean number occurrences X of a rare event for a
certain period is , then X is said to follow the
Poisson distribution if
P(X = r) = e-r/r!
We represent a Poisson distribution with a mean  by
P().
Since  is the average of a number of event, its value
can be a non-integer. In addition, there is no top
limit for the value of X. This means that, even
though  may be 5, 3, 1, even 0.2, the values for X
can range from 0 to .
First example
1.
On the average, 2.5 cars are found parking illegally on a
busy road during a traffic check. Assume this number
follows the Poisson distribution. Find the probability
(i) No illegally-parked car is found during a round by a
police;
(ii) Up to 3 cars are found illegally parked on the road
during a check.
Solution: Let I represent the number of illegally parked car
during a check on the road. Then we assume I~P(2.5)
(i) P(I=0) = e-2.52.50/0! = 0.0821
(ii) P(I3) = e-2.5 + e-2.52.5/1! + e-2.52.52/2! + e-2.52.53/3! =
0.08208+0.20521+0.25621+0.21376 =0.7573 (4 d.p).
Using Tables
• Just as for binomial distributions, statistical tables
for Poisson distributions are constructed to assist
in calculations. As Poisson distributions have only
one parameter each, the tables are simpler.
• The UTM tables for Poisson distributions are also
cumulative from below, showing P(Xr) for each
r. Your experience in using the binomial
distribution tables should make it easier to read the
Poisson distribution tables.
Example 2
•
The marriage bureau records an average of 5.5
marriages each week. Assume this number follows the
Poisson distribution. What is probability
(i) 5 marriages are registered this week?
(ii) 5 or more marriages are recorded this week?
Solution: Let M represent the number of marriages for a
week. Then M~P(5.5)
(i) P(M=5) = e-5.55.55/5! = 0.1714.
(ii) P(M5) = 1 – P(M4) = 1 – 0.3575 (table)
= 0.6425.
Multiple Intervals
• If the mean number of occurrences of an event is 
for a certain period, then it makes sense that the
mean number for double the period would be 2,
for triple the period 3 and so on.
• For example, if on the average, 3.2 voters vote in a
minute, then for 2 minutes, the average should be
6.4, and so on.
• Hence if X~P() for a period of 5 minutes, then
Y~P(2) if Y represent the number of events for
double the period.
Example 3
•
On the average, there are 1.2 traffic lights on 1 km of a
city road. Assume this number follows the Poisson
distribution. What is the probability we find
(i) 5 traffic lights or less on a stretch of road 5 km long?
(ii) More than 3 traffic lights on a road 3 km long?
Solution: In this case, the ‘period’ is actually a distance. The
theory remains the same as for time interval.
(i) Let L5 represent the number of traffic lights on a 5 km
long road. Then L5~P(6).  P(L55) = 0.445
(ii) Let L3 represent the number of traffic lights on a 3 km
long road. Then L3~P(3.6).  P(L3>3) = 1 – P(L33) =
1 – 0.5152 = 0.4848.
Example 4 – sub-intervals
•
It is estimated that the number of critical cases received
at the emergency ward averages 4.8 per hour. Assume
this number follows the Poisson distribution. What is the
probability
(i) No critical case is received in 10 minutes?
(ii) At least 3 critical cases come in during a half-hour
interval?
Solution: Just as multiple intervals yield multiples for
average, the reverse is true too. Thus
(i) Let C1/6 represent the number of critical cases in 10
minutes, then C1/6 ~P(0.8). So P(C1/6=0)=e-0.8 = 0.4493.
(ii) Let C½ represent number of critical cases in ½ hour, then
C½ ~P(2.4). So P(C½3) = 1 – P(C½  2) = 0..
Combining Poisson Distributions
• Unlike the binomial distributions, given X1~P(1) and
X2~P(2), we can sum the two variables: X1+ X2~P(1+2).
For example if the weekly mean number of accidents on
route A is 2.2 and that on route B (separate from and
independent of A) is 1.5 then the combined mean number is
3.7 and
A~P(2.2) and B~P(1.5)  A+B~P(3.7).
Unfortunately, there is still no short-cut for finding
probabilities of the type P(A>B) or P(A>B+2) and similar
events. For such problems, only computer programs
specially designed to solve them will reduce the tedium of
works needed.
Example 5
• The number of passengers reporting sick during a domestic
flight averages 0.8; for international flight, 1.2. Two
planes, one on domestic route and the other international,
just landed at an airport. What is the probability 4
passengers are sick?
Solution: To solve this problem without combining the two
variables, you need to find P(D=0), … P(D=4) and P(I=0),
… P(I=4), then combine them to give the answer as
P(D=0)P(I=4) + P(D=1)P(I=3) + P(D=2)P(I=2) +
P(D=3)P(I=1) + P(D=4)P(I=0).
Combining, we have D+I~P(2). So P(D+I=4) = e-224/4! =
0.0902. You might like to try the first method and compare
your result with this.
Interpolation again
• Just as for binomial distributions, the Poisson table
gives probabilities for only a small set of  . There
will be problems when  is not exactly one of
those given in the tables. As usual, you may
decide to carry out separate calculations to obtain
accurate answers.
• Alternately, interpolations will provide sufficiently
good answers when the problem demands fast
answers.
Example 6 – interpolation example
•
The mean number of goats getting choked by
palm leaves in a integrated oil palm plantation is
1.8 per day. During a week, what is the
probability
(i) At least 10 goats are choked on palm leaves?
(ii) 8 to 15 goats are choked on palm leaves?
Solution: Let G represent the number of choking
goats for a week (7 days). Then G~P(12.6). We
set G1~P(12) and G2~P(13)
Example 6 - solution
P(G110) = 1 – P(G19) = 1 – 0.2424 = 0.7578
P(G210) = 1 – P(G29) = 1 – 0.1658 = 0.8342
Hence P(G10) = 0.7578 + (0.8342 – 0.7578)  0.6
= 0.8036.
(ii) P(8G115) = P(G115) – P(G17)
= 0.8444 – 0.0895 = 0.7549
P(8G215) = P(G215) – P(G27)
= 0.7636 – 0.0540 = 0.7096
Hence P(8G15) = 0.7549+ (0.7096 – 0.7549)0.6
= 0.7277.
Compound Problems
• As shown in the part on binomial
distribution, the result of an analysis may be
used as the basis for another problem.
• The next few examples combine the
Poisson distribution with binomial
distribution.
Example 7
(i) On the average, a police station receives
5.2 reports of lost ID cards per day. What
is the probability the number of loss
reports exceeds 8, the critical number for
a red day?
(ii) The station tracks the records for 10 days.
What is the probability it has more than 3
red days during the period?
Example 7 (Solution)
(i) Let L represent the number of lost cards.
L~P(5.2).
P(L>9) = 1–P(L9) =
(ii) Let R represent the number of red days in
10 days. R~Bin(10, k)
P(R>3) = 1–P(R3) =
Example 8
•
It is estimated that on the average, there are 2.2
floods a year in a district in Trengganu.
(i) What is the probability a district suffers more
than 5 floods for a certain year?
(ii) A district getting 5 floods or more in a year will
be allocated special subsidies in the next
financial years. There are 12 districts in
Trengganu. What is the probability at least 2
districts will get such assistance for the next
year?
Example 8 (Solution)
(i) F=number of floods in a year. F~P(2.2).
P(F5) = 1–P(F4) = r
(ii) S=districts getting subsidies. S~Bin(12, r)
P(S2)