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Transcript
Applications of the Normal
Distribution Model
(The Confidence Interval)
©Dr. B. C. Paul 2003 revision 2009
Note – The concepts found in these slides are considered to be common
knowledge to those trained in the field of statistics. Most basic statistics texts
have coverage of these concepts. The specific table depicted in these slides is
from Fundamental Concepts in the Design of Experiments by Charles R. Hicks,
1982 published by Holt Reinhardt and Winston, although most basic statistics
texts include similar tables for a standard normal distribution.
Preparing for Rainfall

Wendy Wetone has just designed a storm sewer
system for a new housing project




Culverts and intakes will handle a 2.5 inch rainfall in
24 hours
The average big rainfall even in the area is only 1.25
inches
Wendy is ok Right?
If the roads and homes in an area are going to
wash out maybe being ready for an average rain
isn’t good enough
Reality for Major Rainfall Events

Average is 1.25 inches, but suppose there
is a 1 inch standard deviation
How would we know
Something like this?
We built a model
From weather
Records.
σ=1
μ = 1.25
Is there enough of a chance up
hear that I should be getting
heart-burn over this design?
We Know How to Solve This One

Normal Distribution is
 x   

1



fully defined by a
 2
Y  f ( x)  

e

formula
 2 
We only need to know
the average (in this
case 1.25) and the
variance (standard
deviation squared –
easy when standard
deviation is one)
2
2





What That Formula Does



Y is a probability value (chance of occurrence)
X in this case is a rainfall event
Rather obviously we are interested in rainfall
events greater than 2.5 inches


Guess that means x is 2.5
Problem – Formula gives probability for only a
discrete value – ie it will give us the probability
of a 2.5 inch rain event

We are in fact worried about any event that exceeds
our design capacity
That’s not a Problem for Us Smart
Engineers

Just Integrate the Function from 2.5
inches on up


In fact most statistical modeling is done on
cumulative probability distributions (ie
integrated areas on the probability density
function)
Just one little problem

Normal probability density function is one of
those beasts that the math teachers don’t like
to talk about – can’t get an analytical
integrated solution
That’s Only a Problem for
Mathematicians


We have numeric integration
Ok maybe that is a problem if we have to
integrate that thing

Remember – desk top computers are recent
vintage


Do you have a numeric integration package on
your computer even now?
Normal Distribution dates from 1733 so
know someone created tables of numeric
integration
Which Normal Distribution?

Probability Density function is a function of mean
and standard deviation


Tabulated Standard Normal Probability
Distribution



Infinite possibilities
Has a mean of 0
A standard Deviation of 1 (by George we got one)
Turns out there are simple formulas for
converting parameters from any normal
distribution to standard normal


Then its just a table look up
If you have the software – it may even do the look-up
Normal Distribution Table
Converting to a Value on Standard
Normal Distribution

What we want to know is what are the
chances of a rainfall event exceeding our
drainage system design


Ie what percentage of big rainstorms will
exceed 2.5 inches (on a distribution with
mean of 1.25 and standard deviation of 1)
Convert 2.5 inches to an equivalent value
on standard normal distribution

The area above that value in the curve will be
the same as our actual distribution.
Magic Conversion Formula
Z
x

2.5  1.25
1.25 
1
Now Its Look Up Time
Prob = 0.8944
Results

Table shows that from minus infinity to
1.25 there is 0.8944


Ie 0.1056 is above 1.25
English Translation


There is a 10.56% chance that a large rainfall
event will exceed the design capacity of our
drainage system
Sounds like Wendy might be doing some
design work over
Basis for Rainfall Events




10% chance called a 10 year storm (distribution
of years largest storms)
0.05% chance called a 20 year storm
0.01% chance called a 100 year storm
When say it is designed for a 100 year flood it
doesn’t mean it only happens every 100 years


It means 1% chance in any given year
Problem with other thinking is if you had a big flood 5
years ago that must mean there is no chance it will
ever happen again in your lifetime (Wrong!)
Ore Grade Control

Orville Orman is planning a truck fleet to
haul his copper ore out of his mine

Some rock will have so little copper in it that it
would cost more to process than its worth



This stuff is going to get put aside
Other pay rock will be carried to the
processing plant
Commonly have ore and waste truck fleets
but need to know how much of each type
of rock you will have to design.
Orville’s Ore




Orville knows average grade is 0.95% Cu
Standard Deviation is say 0.5% Cu
Cut-Off Grade (point at which ore costs
more to process than Cu will sell for) is
0.25%
What percentage of Orville’s ore is below
cut-off grade?
The Situation
σ = .5
How much of
My rock is
Down here?
0.25
μ = 0.95
Oh We Are Hot




Our critical x value is 0.25
We will convert this to a “Z score” from
the standard normal distribution
We will then look up in the table how
much of our distribution is from minus
infinity to our Z
We will then tell our truck planners how
much rock to prepare for
Crunch Away
0.25  0.95
 1.4 
0.5
Go to the Table
Table Says! 0.0808
About 8.1% of our Rock is Below Cut-Off
Previous Examples

Called One Tailed Tests

Our Civil Engineers were concerned about
events larger than some amount


Our Mining Engineers were concerned about
tonnage below cut-off


An upper tail test
A lower tail test
What if interest in either too much or too
little

Typical of a machine tolerance problem
Tolerance

Benjamin Bidwell would like to bid on a DOD
order for machined shafts




The spec says 1 inch +/- 0.005 inches
Benjamin knows his men and equipment can put any
chosen part size within a standard deviation of 0.0025
inches
He figures he can put in a winning bid provided no
more than 3% of the pieces he makes have to be
rejected
Can Benjamin put in a winner bid on this order?
The Situation
σ = 0.0025
How many
Products are
Out here
In the
Tails?
0.995
μ=1
1.005
We Know What to Do


Convert those limits
to Z scores
Start with the top
limit
1.005  1
2
0.0025
Table Look Up Says 0.9773 or 2.27% will be too large
Now we use our knowledge – this distribution and tolerance is
Symmetric - ie 2.27% on the bottom end
That Sucks - about 4.54% of products will be out of Spec
Summary





We represent our distribution of outcomes with a
mathematical model
Then we can easily check the limits on the model to see
if we can expect our outcomes to be acceptable
We can be doing one tailed tests to see how much will
be too much or too little
We can do two tailed tests to check tolerances
Obviously these types of models could be fit to anything



Voting polls
Drug reactions
Percentage of Criminal Rehabilitated
Your Application

You would feel deprived if you didn’t get a
chance to try this


Do Homework Unit 2 #1
The problem involves whether you need a
coal bunker to handle extra production on
those rare times when you produce more
coal than your hoist can handle.