Statistical Decision Making for Business Scenarios: Cases 5.3, 6.3, 7.2
Geoff Scott, Robert Minello, Andrew Rivas (Group 5)
In Case 5.3, we look at Boise Cascade Corporation. This company scales
lumber for its final destination. This lumber arrives by the truckload to scaling
stations that can scale up to 6 trucks per any given hour. The manager would like to
determine the number of scaling stations that should be open during any given hour
to minimize the wait times for the arriving trucks and to minimize the excess idle
time of the workers. To come to this number, the best course of action is to
determine how the data will be distributed. You are looking for probabilities of a
certain number of trucks arriving during any given hour, so you can use a Poisson
Probability Distribution.
The Poisson Probability Distribution can be used because the data is discrete.
The value for x is determined by counting and is always going to be a whole number.
The segments are also equal, meaning that the chance that a single truck arrives
during the given hour has no effect on the chance of another arriving. You can also
calculate Lambda for this problem, an essential element to finding the Poisson
Probability Distribution. This can be determined by the manager’s observation that
12 trucks arrived between 7:00 AM and 8:00 AM. Lambda is valued at 12 trucks
while the time variable is one hour. Although the data is discrete, you cannot use a
binomial distribution because there is no value for q, or the probability that a truck
does not arrive during a given hour.
At this point, we have 2 essential components of the Poisson Probability
Distribution formula. We know that lambda/time is 12 trucks per hour or 12, and
the “e” value can be equated to 2.718. This value is the error and it will always
equate to this value regardless of the values for x and lambda. You merely input
your “x” values to find the probability of the occurrence in the distribution. For
example, the probability that 3 trucks will arrive during a given hour can be
determined by inputting 3 into your x value in the formula. This would equate to
.0018, or 0.18% chance of this occurrence. The proportion of the distributed
associated with a single scale station can be determine by calculating the individual
frequencies of x that are included in the scale station. For one scale station, this can
be represented by the probability that x is less than or equal to 6. This number
calculates to 0.0458 or 4.58%. At 2 stations this is 0.5303, at 3 stations it is 0.3866,
and at a 4th scale station 0.0367. However, if you are to open a second scale station,
you must also fill the first, so by opening a second, you contain the probabilities of
both the first and second scale stations. This rule applies to each additional scale
station added. The percentage that is not included, calculated by the distance
between percentage included and 1.0, represents the probability that a truck will
arrive and will have to wait for an available station.
The decision is to open 3 scale stations during any given hour. At 3 scale
stations, you include 96.27% of your data, leaving you with less than a 4% chance
that a truck will have to wait. Compared to 2 scale stations that cover 57.61% of the
data, this is the optimal choice. At 4 scale stations, however, there is a 99.93%
chance that no trucks will have to wait. It would be unwise to operate at this level
due to the fact that you’d be hiring workers for a 3.73% chance that a truck will even
need to utilize this extra station. Consider you pay equal salaries to workers
regardless of which station they work at. There is too great of a risk that you’d be
paying their salary for their idle time, or time that they aren’t even working.
Case 6.3 discusses American Oil, a company that uses advanced equipment to
detect oil under the earth’s surface. Chad Williams is a field geologist who has
recently learned that a design-engineering group in the company has come up with
an enhancement to their current equipment that will greatly improve their ability to
detect and find oil. However, this enhancement requires the use of 800 electrical
capacitors. These capacitors must operate within 0.5 microns from the standard of
12 microns. This causes a problem for Chad, because he must order these
capacitors, and the supplier can only provide capacitors that operate at a normal
distribution, with a mean of 12 microns, and a standard deviation of one micron.
Since these capacitors are very expensive, American Oil wants to take a 98% chance
that they will receive 800 capacitors that fit their needs for the enhancement, and
Chad must figure out how many of them to order.
To do this, you can use z score to find out the area of the normally distributed
bell curve that will fit American Oil’s needs. The top of the bell curve is centered on
the mean, which is 12 microns. Knowing this, you will use the standardized normal z
value formula. This is your value you are finding minus the mean and divided by the
standard deviation. You will do this for both plus and minus 0.5 microns. Doing this
will give you a z-value of 0.5, and -0.5. You will then take these z-values and look
them up in the normal distribution table. You will find out that a z score of 0.5 will
produce .1915. Since both sides of the bell curve have the same z score, then they
both have the area of .1915. Adding these two together gives you .383, or the total
area of capacitors that the producer can give that will be useful to American Oil.
Since American Oil needs their capacitors to be within 0.5 microns from the
standard mean of 12 microns and the producer can only provide capacitors with a
standard deviation of one micron from the standard mean, American Oil must order
more than they need. They want to choose a value high enough so that 38.3 percent
of what is ordered is equal to 800. To find out how many capacitors are needed, you
will take the amount needed (800) and divide it by 38.3%. Doing this will yield
2088.77. Since it is impossible to have 0.77 of a capacitor, you will round up to 2089
capacitors. This amount guarantees that American Oil will receive 800 capacitors
that work to their specifications. However, since the capacitors are very expensive,
American Oil is willing to take a 98% chance that they will receive their 800
capacitors. Figuring this number out is as easy as multiplying the total needed
(2089) by 98%. This will give you an answer of 2047.22. Again, since you cannot
have 0.22 of a capacitor, you would round up to 2048.
American Oil is a company that needs capacitors that operate within 0.5
microns from the standard mean of 12 microns. The producers can only provide
American Oil with capacitors that operate with a mean of 12 microns and a standard
deviation of one. American Oil needs 800 capacitors that fit their needs. They are
willing to take a 98% chance that they will receive enough capacitors to fulfill their
needs. This is because of the high price for the capacitors. Using the basic statistical
technique of finding the z-score, we found that American Oil needs to order at least
2048 capacitors to fit all of the criteria.
In case 7.2, Truck Safety Inspection, the Idaho Department of Law Enforcement
began a truck inspection program to reduce the number of trucks with safety defects
operating in Idaho. Jane Lund the head of the programs would like to know what
statistically sampling plan would be beneficial when estimating the number of currently
defective trucks using eight weigh stations and a limited number of time, money and
resources.
In order to find a reasonable estimate of the defective trucks, random sampling
would be the best statistical plan. We as a group decided to use 30 trucks as the sample
size representing n because according to the business statistics text book and the central
limit theorem, the larger the sample size, the better approximation to the normal
distribution. We would then use sample proportion from each of the eight weigh stations
considering that the x value is the defective number of trucks. After obtaining the
proportions of all eight weigh stations, you can then calculate the mean of the sample
proportions to find an accurate representation of the population proportion represented by
pi. Once you have the pi value you can then find the standard error where n is 240 and
pi is the mean. n would be 240 because it is the total number of trucks (30) at all eight
weigh stations. Now that you have the pi value and the standard deviation you can
display the data into a normal distribution. This distribution should be viewed as the
distribution of the population of tucks in Idaho. In conclusion, after the program is
implemented, Jane Lund can repeat the entire process keeping the sample sizes, sampling
method and weight stations used constant. She wants to repeat this process to see if there
is a change in the number of defective trucks operating and to see if the plan is effective.
According to Joan Joseph Castillo from Explorable.com, random sampling is one
the most popular types of random or probability sampling. The advantage of random
sampling is that it is a good and fair way of selecting a sample from a given population
because all members of the population are given equal opportunities of being selected. As
a group we decided random sampling was the best choice for finding a plan that would be
practical for estimating defective trucks. With this random sampling we were able to find
the standard of error. According to investopedia.com, standard error is a statistical term
that measures accuracy where a sample represents a population. It is also the standard
deviation of the sampling distribution. We used a large sample because according to
investopedia.com, the larger the sample size the smaller the standard error because the
statistic will approach the actual value. Both these methods allowed us as a group to
accurately find a practical sampling plan to reach Jane Lund’s objective.
Works Cited (By Cases)
Case 5.3:
Brooks, Bruce E. "Statistics | The Poisson Distribution." Statistics | The Poisson
Distribution. UMass, 24 Aug. 2007. Web. 4 Nov. 2012.
<http://www.umass.edu/wsp/statistics/lessons/poisson/index.html>.
DiRaimondo, Tommy, Rob Carr, Marc Palmer, and Matt Pickvet. "Discrete
Distributions: Hypergeometric, Binomial, and Poisson." - ControlsWiki. UMich,
14 Nov. 2007. Web. 4 Nov. 2012.
<https://controls.engin.umich.edu/wiki/index.php/Discrete_Distributions:_hyperg
eometric,_binomial,_and_poisson>.
Groebner, David F. Business Statistics: A Decision-making Approach. 8th ed. Upper
Saddle River, NJ: Prentice Hall/Pearson, 2011. Pages 192, 213-17.
Case 6.3:
Groebner, David F. Business Statistics: A Decision-making Approach. 8th ed. Upper
Saddle River, NJ: Prentice Hall/Pearson, 2011. Pages 234-242.
"The Z-score Statistics." The Z-score Statistics. N.p., n.d.
Web. <http://www.pindling.org/Math/Statistics/Textbook/Chapter6_Normal_Dist/z_scor
e.htm>.
"Standard Normal Distribution." Standard Normal Distribution. N.p., n.d. Web.
<http://www.oswego.edu/~srp/stats/z.htm>.
Case 7.2:
Castillo, Joan Joseph. "Random Sampling." - Probability Sampling. Explorable,
2009. Web. 20 Oct. 2012. <http://explorable.com/simple-randomsampling.html>.
Groebner, David F. Business Statistics: A Decision-making Approach. 8th ed. Upper
Saddle River, NJ: Prentice Hall/Pearson, 2011. Pages 290-293.
"Standard Error." Investopedia. N.p., 2012. Web. 20 Oct. 2012.
<http://www.investopedia.com/terms/s/standard-error.asp#axzz2Da7Osnen>