5.3 Boise Cascade Corporation6.3 American Oil Company7.2 Truck Safety Inspection Geoff Scott Robert Minello Andrew Rivas
Boise Cascade Corporation • 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 • Discrete data: random variable values are determined by counting • We know lambda • What happens in one segment has no effect on the other segments • No value for q, cannot use Binomial Distribution
Lambda: λ = 12 Time: t = 1 hour • 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: • P(x≤6) (Each scale station can scale 6 trucks per hour) • 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%
P(1 scale station) P(0<x≤6) = 0.0458 • P(2 scale stations) P(6<x≤12) = 0.5303 • P(3 scale stations) P(12<x≤18) = 0.3866 • P(4 scale stations) P(18<x≤24) = 0.0367
Decision • 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 • Uses electronic sensing equipment • Developed a new enhancement for equipment • Enhancement requires 800 capacitors that must operate within within ±0.5 microns from the standard of 12 microns
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. • 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 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 • Idaho department of Law Enforcement started new truck inspection program. • Objective is to reduce the number of trucks with safety defects operating in Idaho. • Jane Lund wants to estimate the number of defective trucks currently operating. • Need to find a statistically sampling plan.
Sampling Plan • Random sampling will be used as the practical sampling plan. • 8 weigh stations will be used to obtain the sample according to the problem. • We are using 30 to represent the sample size (n) • According to the central limit theorem, the bigger the sample size, the better approximation to the normal distribution
Process • Calculate the sample proportions • Sample proportion - p = x/n • Calculate the mean of sample proportion • Sum of the sample proportion / 8 scale stations This value is the unbiased estimator for population proportion (π)
Process • When you calculate π you can then find the standard error • total number of trucks from all 8 weigh stations • (8)(30) =240 • 240 = n
Is it normally distributed • In order for the unbiased estimator to be considered normally distributed it must • Satisfy • nπ>5 • n(1-π)>5
Conclusion • 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. • Keep the same sample size, stations and method. • If her program was a success then π before > π after.