ME 322: Instrumentation
Lecture 2
January 23, 2015
Professor Miles Greiner
Quad measurement calculations and results for
Lab 2, Probability Distribution Functions,
Examples (symmetric, one-sided)
Announcements
• Lab 1 work sheet due now
• HW 1 due Monday
– Use ME 322 ID number (from WebCampus), not name
• If you would like to move to a different section
– Find someone who is willing to trade with you
– Both of you must send an email to Marissa Tsugawa
• tsugawam@nevada.unr.edu
– If she approves then you may switch
Results of the Quad Measurements
• Data is on Lab 2 website
• How to process that data
– You will repeat and present this in Lab 2, Analysis of Quad
Measurement
• Spreadsheet has Measured Data and room for Calculations
–
–
–
–
–
H, DC , NCi, Calculate: NCA, F = DC/NCA
NSi, Calculate: NSA, S=F*NSA
NLi, Calculate: NLA, L=F*NLA
Calculate: A = LS
Calculate: C = A*($3.49/200 ft2)
• How to Plot
– F versus H; L versus S (scatter plots)
– Cost Estimate Histogram: To install analysis ToolPack: File, options,
add-in, Manage: Excel Add-ins, Go, Analysis ToolPack)
Questions
• Is stride length F “highly correlated” with height, H?
• Does the distribution of the cost estimates look the way
you expect?
– Do any of the cost estimates look “out of place?”
• How can you interpret the sample mean and standard
deviation of the cost estimates in terms of the
distribution?
• Are the measured values of L and S “highly” correlated?
– Should they be? What does this mean?
• If you budget the amount of your cost estimate, then
– You are only 50% sure to have enough to cover quad (be
above the average value, which we assume is the most
accurate estimate)
• How much money should you budget to be 90% sure to
have enough
Randomly Varying Processes
• The output of a measurement instrument is affected by the
measurand (the quantity being measured) and many
uncontrolled, undesired and randomly varying factors
• Consider a process (such as a measurement) that has a very
large number of factors that can, independently, increase or
decrease the value of the outcome
• Its not likely that all or a large majority of the factors will push
the outcome in the same direction.
• Its more likely that “most” of the factors will cancel each
other, and push the outcome only “slightly” in one direction
or the other.
• This describes how uncontrolled factors affect the output or
reading of measurement systems
• Each time you make a measurement, the reading can be
different
Gaussian (Normal)
Probability Distribution Function
𝑓 𝑥; 𝜎, 𝜇 =
1
𝜎 2
𝑒
1𝑥−𝜇
(−2 𝜎 )2
• Has a “bell” shape
• Describes Randomly Varying Processes
• Looks like the pattern observed from the cost (area)
estimate histogram in Lab 2
– We were able to estimate m ~ 𝐶 and s ~ 𝑠𝐶 for that data
How can we use this?
• If a sample is very large, and
• If the process variations are “normally” distributed,
– Then expect sample histogram to form a bell shape,
– And, if we know s and m, then the probability that the next
measurement 𝑥 will be in the range 𝑥1 < 𝑥 < 𝑥2 is
𝑃 𝑥1 < 𝑥 < 𝑥2 =
𝑓 𝑥; 𝜎, 𝜇 =
𝑥2
𝑓(𝑥 ; 𝜎, 𝜇 ) 𝑑𝑥
𝑥1
1
𝜎 2
𝑒
(−
≤1
1𝑥−𝜇 2
)
2 𝜎
Number of standard deviations
𝑥 is above the mean
• Note, for any s and m :
𝑓 𝑥
𝑥1
∞
𝑓(𝑥; 𝜎, 𝜇) 𝑑𝑥 = ?
−∞
𝑥2
Non-Dimensionalization
• Define 𝑧 =
𝑥−𝜇
𝜎
– Number of standard deviations 𝑥 is above the mean
• We can show that the probability that the next
measurement is between z1 and z2 is:
– Where
Graphical Representation of I(z)
Area from center (z = 0) to z
For z > 0 :
Note:
This integral is tabulated on page 146, for z > 0
I(z)
For negative values of z
If z1 < 0 :
𝐼 𝑧1 = −𝐼(−𝑧1 )
This will be a
Positive number
𝑃 𝑧1 < 𝑧 < 𝑧2 = 𝐼 𝑧2 − 𝐼(𝑧1 )
= 𝐼 𝑧2 − [−𝐼 −𝑧1 ]
= 𝐼 𝑧2 + 𝐼 −𝑧1
Symmetric Example
Find the Probability the next measurement will be
within one standard deviation (s) of the mean (m).
𝑥1 = 𝜇 − 𝜎
𝑥2 = 𝜇 + 𝜎
𝑧1 =
𝑧2 =
𝑥1 −𝜇
(𝜇−𝜎)−𝜇
=
=
𝜎
𝜎
𝑥2 −𝜇
(𝜇+𝜎)−𝜇
=
=
𝜎
𝜎
-1
𝑃 −1 < 𝑧 < 1 = 𝐼 1 − 𝐼 −1
= 𝐼 1 − −𝐼 1
1
= 2𝐼(1)
Page 146
𝑃 −1 < 𝑧 < 1 = 2𝐼 1 = 2 ∗ 0.3413 = 0.6826
= 68.26%
Next measurement is within 2s or 3s
of the mean
𝑃 −2 < 𝑧 < 2 = 2𝐼 2 = 2 .4772 = 95.44%
𝑃 −3 < 𝑧 < 3 = 2𝐼 3 = 2 .4987 = 99.74%
One-sided example
• From Lab 2, what seed cost will cover (be greater than) 90%
of all future estimates?
•
•
•
•
•
•
•
•
One-sided example
P = 0.9 = I(z2) - I(z1)
z2 = ?
But z1-∞, so I(-∞)= -I(∞)=-0.5
So P = 0.9 = I(z2) – [-0.5]
0.4 = I(z2)
𝑥−𝜇
𝑧=
Interpolate between z2 = 1.28 and 1.29
𝜎
Get z2 = 1.2817
𝐶 = 𝐶 + 𝑧1 𝑠𝐶 = $980 + 1.2817 ∗ $330 = $1403
Lab 2
• If you make a measurement and cost
estimate, there is a 50% likelihood it is below
the mean (best) value.
• How much should you add to your best cost
estimate to be 90% you are above the mean?
• Answer:
– 1.282 standard deviations above your best cost
estimate
Extra Slides
Area of UNR Quad
• Find Short Side (S)
– NSi
– NSA
• Find Long Side (L)
– NLi
– NLA