5.3 Boise Cascade Corporation
6.3 American Oil Company
7.2 Truck Safety Inspection
Geoff Scott
Robert Minello
Andrew Rivas
S
Boise Cascade Corporation
S How many scaling stations should be open during a given
hour to:
-Minimize the wait time for trucks
-Minimize excess idle time for workers
Poisson Distribution
S Discrete data: random variable values are determined by
counting
S We know lambda
S What happens in one segment has no effect on the other
segments
S No value for q, cannot use Binomial Distribution
Lambda: λ = 12
Time: t = 1 hour
S Manager observed that on average, 12 trucks (λ=12) arrived
between 7:00 AM and 8:00 AM (t=1).
λt = 12/1 = 12
Probability of only needing 1 scale
station:
S P(x≤6)
(Each scale station can scale 6 trucks per hour)
S P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)+P(x=6)
P(x) = (λt^x) (e^-λt) / x!
Where “e” represents error. e=2.718
P(3) = (12^3) (e^-12) / 3!
P(3) = 0.0018
*The probability of 3 trucks arriving is 0.18%
S P(1 scale station)
P(0<x≤6) = 0.0458
S P(2 scale stations)
P(6<x≤12) = 0.5303
S P(3 scale stations)
P(12<x≤18) = 0.3866
S P(4 scale stations)
P(18<x≤24) = 0.0367
Distribution
Decision
S 3 scale stations
-Why hire workers if there is only 3.73%
chance that they will have to work.
-3.73% chance that trucks will have to wait.
American Oil Company
S Uses electronic sensing equipment
S Developed a new enhancement for equipment
S Enhancement requires 800 capacitors that must operate
within within ±0.5 microns from the standard of 12 microns
S The supplier can only provide capacitors that operate
according to a normal distribution with a mean of 12
microns and a standard deviation of 1.
S How many capacitors should American Oil purchase in
order to fulfill the enhancement with a 98% chance that the
order will contain 800 usable capacitors?
Suppliers can provide:
µ = 12
σ=1
According to Normal Distribution
Enhancement requires:
11.5 ≤ µ ≤ 12.5
Acceptable Capacitors from
the supplier that fit the Enhancement
Finding the Area
Z = x-µ
σ
Z = 11.5-12 = -0.5
1
Z = 12.5-12 = 0.5
1
Look up the Z-values in the standard distribution table.
*Z0.5 = 0.1915
*Z-0.5 = 0.1915
0.1915 + 0.1915 = 0.383
38.3% of the suppliers capacitors will fit the enhancement
61.7% will not fit and be rejected
0.383 (x) = 800
X = 2088.77, or 2089 capacitors
0.98 (2089) = 2047.22, or 2048 capacitors
Decision
S American Oil Company should order 2048 capacitors in order to
fulfill their equipment enhancement with a 98% chance that they
will get 800 capacitors that fall into their ±0.5 micron from the
standard of 12 micron needs.
Truck Safety Inspection
S Idaho department of Law Enforcement started new truck
inspection program.
S Objective is to reduce the number of trucks with safety
defects operating in Idaho.
S Jane Lund wants to estimate the number of defective trucks
currently operating.
S Need to find a statistically sampling plan.
Sampling Plan
S Random sampling will be used as the practical sampling plan.
S 8 weigh stations will be used to obtain the sample according to the
problem.
S We are using 30 to represent the sample size (n)
S According to the central limit theorem, the bigger the sample size,
the better approximation to the normal distribution
Process
S Calculate the sample proportions
S Sample proportion - p = x/n
S Calculate the mean of sample proportion
S Sum of the sample proportion / 8 scale stations
This value is the unbiased estimator for population proportion (π)
Process
S When you calculate π you can then find the standard error
S total number of trucks from all 8 weigh stations
S (8)(30) =240
S 240 = n
Is it normally distributed
S In order for the unbiased estimator to be considered
normally distributed it must
S Satisfy
S
nπ>5
S
n(1-π)>5
Conclusion
S If Jane Lund wants to see the change her new program had
caused, she will repeat the entire process and compare to the
first time she did it.
S Keep the same sample size, stations and method.
S If her program was a success then π before > π after.