- random error theory
- probability, experiments, plots of combining observations
- normal distribution curve,
- properties of the normal distribution function
- properties of the standard error
- linear error probable
- linear interpolation
2. Probability
• The ratio of the number of times that an
event should occur to the total number of
possibilities
– If event A can occur m ways and fails to occur
n ways, then the probability of event A
occurring is m/(m + n)
• Probability of throwing a 1 on a fair die
– Should occur one time
– Will fail to occur 5 times
– Probability is 1/(1+5) = 1/6
• Probability of an event failing to occur is
1 − m/(m + n)
3. Probability
• Probability is always between 0 and 1
– 0 means there is no chance of occurrence
– 1 means that the event will absolutely occur
– The probability of all event occurring is 1
4. Compound Events
• Compound event: The simultaneous
occurrence of two or more events.
• The probability of a compound event is the
product of each probabilities of each
individual event occurring.
P = P1 ∙ P2 ∙ … ∙ Pn
5. Experiment
• Assume that we have 1 red ball in box 1 with 4
balls total and 2 red balls in box 2 with 5 balls
total. What is probability that 2 red balls will be
drawn when one ball is randomly selected from
each box?
7. Experiment
Drawing Box 1 Box 2
1 White Ball 1 Red Ball 1
2 White Ball 1 Red Ball 2
3 White Ball 1 White Ball 3
4 White Ball 1 White Ball 4
5 White Ball 1 White Ball 5
6 Red Ball 2 Red Ball 1
7 Red Ball 2 Red Ball 2
8 Red Ball 2 White Ball 3
9 Red Ball 2 White Ball 4
10, and so on to 20 Red Ball 2 White Ball 5
8. Experiment
• There are 20 total events with only two producing
two red balls, so the probability is 2/20 = 0.1
• By formula
– Probability of drawing a red ball from Box 1 is 1/4
– Probability of drawing a red ball from Box 2 is 2/5
– Probability of drawing two red balls is
1 2 2
0.1
4 5 20
P
9. Application
• Least squares adjustments are the
probability of the unknowns (usually
coordinates) being a certain value based
on the simultaneous occurrence of all of
the observations.
– Distances, angles, azimuths, elevation
differences, etc.
10. Another Example
• Assume that a tape exist that can only
make a +1 ft or −1 ft when taping a
distance.
– Let t be the number of ways that each error
can occur
– And T be the total number of possibilities
– Then the possible occurrence of these
random errors is
11. Example
• For a distance of 1 tape length
– Only one +1 error or one −1 can occur
• For a distance of 2 tape lengths
– +1 and +1 occurs once with a resulting combined
error (ce) of +2
– +1 and −1 can occur twice (ce = 0)
– −1 and −1 occurs once (ce = −2)
• For a distance of 3 tape lengths
– +1, +1, +1 error occurs once (ce = +3)
– −1, +1, +1; +1, −1, +1; +1, +1, −1 (ce = +1) occurs 3 times
– −1, −1, +1; −1, +1, −1; +1, −1, −1, (ce = −1) occurs 3 times
– −1, −1, −1; (ce = −3) occurs 1
• And so on…
12. Number of
combining obs
Value of
resulting error
Frequency, t Total
possibilities
Probability
1 +1
−1
1
1
2 1/2
1/2
2 +2
0
−2
1
2
1
4 1/4
1/2
1/4
3 +3
+1
−1
−3
1
3
3
1
8 1/3
3/8
3/8
1/3
4 +4
+2
0
−2
−4
1
4
6
4
1
16 1/16
1/4
6/16
1/4
1/16
5 +5
+3
+1
And so on
1
5
10
32 1/32
5/32
10/32
13. Plots of Combining Observations
• Note the progression and fundamental shape of
the histograms
14. Shoot the Star-Out Game
• If you aimed at the center of the star, the
following 2D pattern of shots may occur.
15. Shoot the Star-Out Game
• Now
– Draw a single line across the target tabulating
the number of BBs that pierced the target
along the line
– Create a frequency plot of the hits versus
distance from the center of the target
• It may look as follows
16. Normal Distribution Curve
• Note that the previous
histograms were approaching
this shape.
• This is the normal distribution
curve
– So named b/c it occurs
frequently in nature
– AKA probability density curve
of a normal random variable
• The area under the curve
always equals 1.
17. Normal Distribution Curve
• Gauss derived the function that defines this curve
(see Appendix D) as
2
2
2
1
2
x
f x e
18. Properties of the Normal
Distribution Curve
• The area under the normal distribution curve
provides the probability of the simultaneous
occurrence of errors.
– Mathematically speaking
2
2
2
area under curve
1
2
x
t
P
e dx
19. Probability of an Event
• If we wish to find the probability of a single
discrete event occurring at x
– Define a infinitesimally small region under the event
and compute the area of the region
2
2
2
1
( )
2
x
P x y dx e dx
20. Properties of the Normal
Distribution Function
• From calculus
So the first and second derivative are
2
2
where
2
u
u
de du x
e u
dx dx
and
2
2
2
2 2
2
2 2 2
x
dy x x
dx
d y x dy y
dx dx
e y
21. Properties of the Normal
Distribution Function
• Substituting the first derivative equation into the
second derivative equation yields
2 2
2 4 2
2
2 2
1
1
d y x
y y
dx
y x
22. Properties
• From calculus, we know that
– First derivative = slope of the line at
any point
– So dy/dx = 0 when either
• x = 0 or at y = 0
• So curve
– Is asymptotic to x axis where y = 0
– Peaks at x = 0
2
dy x
y
dx
23. Properties
• Second derivative provides
– The rate of change of the slope at a point
– Points of inflection can be found by setting the second
derivative equal to 0.
– Occurs when y = 0 (never happens) or
2 2
2 2 2
1 0
d y y x
dx
2
2
2
2 2
2
1 0 or
1 or
x
x
x x
68% of area
Inflection point Inflection point
σ +σ
24. Properties
• By observation, the maximum value for y occurs
when x = 0.
– Setting x = 0 in original function yields
– Note: As σ goes to 0, y becomes larger (i.e. higher
probability). So smaller σ means more precise data
and higher probability that sample mean is a good
estimate for the population mean!
68% of area
Inflection point Inflection point
σ +σ
2
2
2
0
1
2
1
2
1
2
x
y e
e
25. Standard Normal Distribution Table
• Note:
– Impossible to create a table for every value of μ and σ
– Thus a table is only developed for a population having
a mean of zero (μ = 0) and variance of 1 (σ2 = 1)
• Known as the standard normal variable
– Integration of normal distribution cannot be carried out in
closed form and thus is developed numerically.
– Table of values known as Standard Normal Distribution
Table (Table D.1) since μ = 0 and σ2 = 1. (pages 562–563)
26. Using Table D.1
• Table D.1 gives the value of t along the x axis
where the area under the curve is tabulated
going from −∞ to t.
– t is known as the critical value
– Thus area from −∞ to 1.68 is 0.95352
27. But What About My Values?
• Any normal random variable y with
variance σ2 can become a standard
normal random variable, z, with variance
σ2 = 1 by
– Computing z = (y – μ)/σ
– This will be used in SUR 441 to
• Analyze the statistical significance of variables
• Check data for blunders/outliers/mistakes
28. Probability Computations
• For any value z, the probability of
occurrence is
P(z < t) = Nz(t)
– To determine if z is between, say, a and b, we
write
P(a < z < b) = Nz(b) − Nz(a)
– Where Nz(t) is from Table D.1
29. Probability of the Standard Error
• AKA – Probable Error
• Since the variance = 1, σ = ±1
• So we need to know the probability that
• Table D.1
P(−1 < z < 1) = Nz(1) − Nz(−1)
=0.84134 − 0.15866
= 0.68268
• Note that about 68.3% of data is between ±σ
30. Properties of Distribution
• P(−t < z < t) = P(|z| < t)
= Nz(t) − Nz(−t)
– From the symmetry of the normal distribution, P(z > t)
= P(z < −t)
– Thus the 1 − Nz(t) = Nz(−t)
– So, P(|z| < t) = Nz(t) − [1 − Nz(t)]
= 2Nz(t) − 1
-z +z
-t +t
31. Linear Error Probable
• 50% Probable Error
P(|z| < t) = 0.50 = 2Nz(t) − 1
• So 1.5 = 2Nz(t)
• Thus 0.75 = Nz(t)
• We need to interpolate 0.75 from Table
D.1
32. Linear Error Probable (50%)
• From Table D.1
– Nz(0.67) = 0.7486 and Nz(0.68) = 0.7517
– So t = 0.67 + 0.0045 = 0.6745
0.75 0.7486
0.68 0.67 0.7517 0.7486
0.75 0.7486
0.01
0.7517 0.7486
0.004516
t
t
33. Linear Interpolation
• Method of determining a value between
two tabulated values.
• Given:
• Find n as
t1 n1
tn n
t2 n2
1
1 2 1
2 1
( )
n
n n
t t t t
n n
34. Example
• Recall the example from Chapter 2 with mean = 23.5 ±1.37
20.1 20.5 21.2 21.7 21.8
21.9 22.0 22.2 22.3 22.3
22.5 22.6 22.6 22.7 22.8
22.8 22.9 22.9 23.0 23.1
23.1 23.2 23.2 23.3 23.4
23.5 23.6 23.7 23.8 23.8
23.8 23.9 24.0 24.1 24.1
24.2 24.3 24.4 24.6 24.7
24.8 25.0 25.2 25.3 25.3
25.4 25.5 25.9 25.9 26.1
• 50% of observations should be between
23.5 − 0.6745(1.37) = 22.58 < z < 24.42 = 23.5 + 0.6745(1.37)
• By inspection, 27 of 50 are between these values or 54%. So close
for a sample set. Why is it not exactly 50%?
36. 95% Probable Error
• No surveying business is willing to reject 50% of
their observations as blunders.
• 95% is typically used in all surveying standards
such as ALTA-ACSM
• Determine t for 95% probable error
0.95 = P(|z| < t) = 2Nz(t) − 1
So 1.95 = 2Nz(t)
Nz(t) = 0.975
37. 95% Probable Error
• From Table D.1
• Nz(1.96) = 0.9750
– no interpolation required!
38. Example
• Again using example from Chapter 2 with mean = 23.5 ±1.37
20.1 20.5 21.2 21.7 21.8
21.9 22.0 22.2 22.3 22.3
22.5 22.6 22.6 22.7 22.8
22.8 22.9 22.9 23.0 23.1
23.1 23.2 23.2 23.3 23.4
23.5 23.6 23.7 23.8 23.8
23.8 23.9 24.0 24.1 24.1
24.2 24.3 24.4 24.6 24.7
24.8 25.0 25.2 25.3 25.3
25.4 25.5 25.9 25.9 26.1
• 95% of observations should be between
23.5 − 1.96(1.37) = 20.81 < z < 26.18 = 23.5 + 1.96(1.37)
• By inspection, 48 of 50 are between these values or 96%. So close
for a sample set. Is there reason to be concerned about data set?
