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5.3 Boise Cascade Corporation

6.3 American Oil Company7.2 Truck Safety Inspection

Geoff ScottRobert MinelloAndrew 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 = 12Time: 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

Distribution

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 = 12σ = 1According 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 enhancement61.7% 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 stationsThis 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.