Download Consider the attributes that you might use to describe a rock such as

tom.h.wilson
[email protected]
Department of Geology and Geography
West Virginia University
Morgantown, WV
Rock property assessment
Consider the attributes that you
might use to describe a rock such as
grain size, porosity, composition
(percent quartz, orthoclase, …), dip,
etc.
How are these different
attributes obtained?
How reliable are the values that
are reported?
Concepts and terminology -
Specimen - a part of a whole or one
individual of a group.
Sample - several specimens
Population - all members of the
group, all possible specimens from
the group
Concepts and terminology The attributes derived form the
sample are referred to as
statistics.
The attributes derived from the
entire population of specimens are
referred to as parameters.
Consider your grade in a class Let’s say that your semester grade
is based on the following 4 test
scores.
85, 80, 70 and 95.
What is your grade for the semester?
You grade is the average of
these 4 test scores or 82.5
The average is often used to represent
the most likely value to be encountered
in a sample or population.
Is the average grade of 82.5
a statistic or a parameter?
Since the entire population of grades
for the student consists of just
those 4 test scores, their average
score is a parameter.
Pebble Masses
At left is a table of
masses (in grams) of
100 pebbles taken
from a beach.
The average
mass of these
pebbles is
350.18 grams
This average is a
... statistic
374
389
358
395
371
334
224
335
256
340
374
423
338
373
342
242
318
454
346
408
403
384
397
307
409
294
256
359
352
330
269
355
283
301
346
393
386
338
380
357
326
403
317
301
394
407
350
375
303
384
284
403
341
435
307
420
342
331
331
331
290
383
370
302
394
329
324
283
355
311
265
364
322
283
367
287
340
401
422
369
379
432
368
338
327
433
370
343
450
318
384
355
366
324
353
277
359
400
314
389
Computation of the mean or average
1N 
m    mi 
N  i 1 
In this equation
mi is the mass of pebble i
N is the total number of specimens
i ranges from 1 to N
m = the average mass of all the
pebbles in the sample.
If we draw smaller samples at random from our
original sample of 100 specimens and then
compute their averages, we begin to appreciate
that the statistical average is only an estimate
of the population average.Recall that the mean
estimated from 100 samples was 350.18.
Specimen Sample 1 Sample 2 Sample 3 Sample 4 Sample 5
1
340
359
383
394
401
2
374
352
370
407
422
3
423
330
302
350
369
4
338
269
394
375
379
5
373
355
329
303
432
6
342
283
324
384
368
7
242
301
283
284
338
8
318
346
355
403
327
9
454
393
311
341
433
10
346
386
265
435
370
Average
355
337.4
331.6
367.6
383.9
<average>
= 355.1g
Other measures of the most common
value in a population include the
median and the mode.
If we sort measured values (for example
pebble mass) in increasing order, from the
lightest pebble to the heaviest and look at
the mass of the center pebble, that value
is the median mass.
In the sample (1,2,3,4,5) 3 is the center or median
value of the sample.
In the sample 1,2,3,4 we have
an even number of observations
and in this case the median is
taken as the average of the
middle two values or 2.5.
At right, the pebble
mass has been sorted
in ascending order
from the lightest to
heaviest pebbles in
the sample.
224
242
256
256
265
269
277
283
283
283
284
287
290
294
301
301
302
303
307
307
311
314
317
318
318
322
324
324
326
327
329
330
331
331
331
334
335
338
338
338
340
340
341
342
342
343
346
346
350
352
There are an even
number of specimens
in this sample, so the
median must be
determined from the
average of specimens
50 and 51
Median =352.5 g
353
355
355
355
357
358
359
359
364
366
367
368
369
370
370
371
373
374
374
375
379
380
383
384
384
384
386
389
389
393
394
394
395
397
400
401
403
403
403
407
408
409
420
422
423
432
433
435
450
454
The mode is the value that occurs
most frequently. For example, in the
following sample, (1,2,2,3,3,3,4,5), 3
occurs most frequently and would be
the mode of this sample.
In the sample of
rock masses,
283, 331, 338,
355 and 403 all
occur 3 times.
We cannot define
a single mode.
224
322
353
384
242
324
355
386
256
324
355
389
256
326
355
389
265
327
357
393
269
329
358
394
277
330
359
394
283
331
359
395
283
331
364
397
283
331
366
400
284
334
367
401
287
335
368
403
290
338
369
403
294
338
370
403
301
338
370
407
301
340
371
408
302
340
373
409
303
341
374
420
307
342
374
422
307
342
375
423
311
343
379
432
314
346
380
433
317
346
383
435
318
350
384
450
318
352
384
454
- a graphical display of the
distribution of values.
Histogram of pebble mass
Number of occurences
10
8
6
4
2
0
200
250
300
350
Mass
400
450
500
You might group your
samples into 25 gram
ranges extending
from 226 through
250, 251 through
275 and so on.
Another histogram
30
25
Number of occurrences
The appearance of
the histogram will
vary depending on the
specified range you
use to subdivide the
sample values.
20
15
10
5
0
200
250
300
350
400
Mass (grams)
450
500
Or 50 gram intervals The median is 352.5 grams
and the mean 350.18 grams
Location of the mean
and median values
Histogram of Pebble Masses
40
Number of specimens
35
30
25
20
15
10
5
0
150
200
250
300
350
Mass
400
450
500
550
Histogram of pebble mass (Beach B)
10
10
8
8
Number of occurrences
Number of occurences
Histogram of pebble mass
6
4
2
0
200
6
4
2
250
300
350
Mass
400
450
500
0
200
250
300
350
400
450
Mass
The distribution of masses from beach B
is similar in shape to that from our first
beach, but its range is much smaller.
Both samples have the same mean.
500
Histogram of pebble mass (Beach B)
Histogram of pebble mass
10
A
8
8
Number of occurrences
Number of occurences
10
6
4
2
0
200
6
4
2
250
300
350
Mass
400
450
500
0
200
250
300
350
400
450
500
Mass
The pebbles on beach B are much better
sorted than those on beach A. There is not
as much variation in pebble mass on beach B.
Sample B is better sorted than our first sample. The
mass distribution below has the same range as
distribution B and nearly the same mean (348 grams),
but its shape is very different. This distribution is
much more irregularly distributed across the range i.e. there’s not a preferred value.
Another distribution
Number of occurrences
12
10
8
6
4
2
0
280
300
320
340
360
Mass
380
400
420
A parameter that describes the spread
or dispersion in the values of a population
is its variance.
  (mass  average mass)
2
2
Note the brackets indicate
that we are taking the mean
or average of this quantity.
2 is used to represent
the population variance
The statistic used to quantify the spread
or dispersion in the values of a sample is
the sample variance.
The sample variance is computed
in the following way 1N
2
s    (mi  mi ) 
N  i 1

2
s2 represents the sample variance
The standard deviation of the
sample values is a statistic that
is also often used to describe
the degree of variation in values
of a sample.
s
1N
2
  (mi  mi ) 
N  i 1

The standard deviation is just
the square root of the variance.
Histogram of pebble mass (Beach B)
Another distribution
10
10
Number of occurrences
Number of occurrences
12
8
6
4
6
4
2
2
0
280
8
300
320
340
360
380
400
420
Mass
Standard deviation =32.37
0
280
300
320
340
360
380
400
420
Mass
Standard deviation = 23.83
While these two distributions have similar
means, the one on the right is better
sorted and, the standard deviation - not
the range - reveals this difference.
The standard deviation describes
geological differences in the
sample that are not apparent in
their means.
One sample is better sorted - has
smaller standard deviation than
the other, which is less sorted and
has higher standard deviation.
The sample variance is considered
to be an underestimate of the
population variance or actual
variance of the parent population.
To compensate for that, and to obtain
an estimate of the population variance
which is considered more accurate,
the sample variance is corrected to
form an “unbiased ” estimate of the
population variance.
The following equation is
used to compute the
unbiased estimate of the
population variance -
 N  2
sˆ  
s

 N  1
2
Probability
Probability can be thought of as
describing that fraction of all possible
values that a specific value or range of
values will be observed out of the total
number of observations or specimens in
a sample.
Just as with the average and the
standard deviation, there is a
distinction between the probabilities
associated with a sample and those of
the parent population….
The probabilities observed in the
sample give one an estimate of the
probability or likelihood of occurrence
in the parent population.
Probabilities are used to
make predictions as well
as for characterization.
Note that the probability of
individual occurrences may
vary. The probability that
you will get a pebble with a
mass of 242 grams is one in
a hundred.
The probability of picking up
a pebble that has a 283
gram mass is 3 in 100
(0.03), etc.
224
242
256
256
265
269
277
283
283
283
284
287
290
294
301
301
302
303
307
307
311
314
317
318
318
322
324
324
326
327
329
330
331
331
331
334
335
338
338
338
340
340
341
342
342
343
346
346
350
352
353
355
355
355
357
358
359
359
364
366
367
368
369
370
370
371
373
374
374
375
379
380
383
384
384
384
386
389
389
393
394
394
395
397
400
401
403
403
403
407
408
409
420
422
423
432
433
435
450
454
The probability
that a pebble will
have a mass
somewhere in the
range 300 to 350
will be 35 out of
100 or 0.35.
224
242
256
256
265
269
277
283
283
283
284
287
290
294
301
301
302
303
307
307
311
314
317
318
318
322
324
324
326
327
329
330
331
331
331
334
335
338
338
338
340
340
341
342
342
343
346
346
350
352
353
355
355
355
357
358
359
359
364
366
367
368
369
370
370
371
373
374
374
375
379
380
383
384
384
384
386
389
389
393
394
394
395
397
400
401
403
403
403
407
408
409
420
422
423
432
433
435
450
454
Probability Distribution
0.40
0.35
PROB
0.30
0.25
0.20
0.15
0.10
0.05
0.00
150
200
250
300
350
400
450
500
550
BinM
These probabilities are the
probabilities that individual values in a
sample will fall in a 50 gram range,
and thus represent the integral of
individual probability over the range.
Probability Distribution
0.40
0.35
PROB
0.30
0.25
0.20
0.15
0.10
0.05
0.00
150
200
250
300
350
400
450
500
550
BinM
Probability distributions with a shape
similar to the above example are quite
common. They are nearly symmetrical and
values near the average are more probable
than those further from the average.
Distributions of this type are often
referred to as Gaussian or normal
distributions.
p( x) 
1
2 2
[  ( x  x ) 2 / 2 2 ]
e
And we can estimate the probability
distribution of the parent population
from statistical estimates of the
mean and variance.
p ( x) 
1
2sˆ 2
[  ( x  x ) 2 / 2 sˆ 2 ]
e
This expression is often simplified
by substituting Z for (x-x)/. Z is
referred to as the standard normal
deviate, and the Gaussian
distribution is rewritten as p ( x) 
1
2 2
[  z 2 / 2]
e
Note that (x-x)/ represents the
number of standard deviations the
value x is from the mean value.
Thus a value x corresponding to a z of
2 would be located two standard
deviations from the mean in the
positive direction.
Using the pebble mass statistics,
<x>=350.18 and s=48. Thus a specimen
with z of 2 implies that its mass x =
446 grams.
Probability Distribution of Pebble M asses
0.01
Probability
0.008
0.006
0.004
0.002
0
0
100
200
300
400
500
600
700
Pebble Mass (grams)
The Gaussian (normal) distribution of pebble
masses looks a bit different from the
probability distribution we derived directly from
the sample.
1
[  ( m  m ) 2 / 2 sˆ 2 ]
p ( m) 
e
2sˆ 2
Pebble masses collected from beach A
0.40
0.35
Probability
0.30
0.25
0.20
0.15
0.10
0.05
0.00
150
200
250
300
350
400
450
500
550
Mass (grams)
The probability of occurrence of
specific values in a sample often takes
on a bell-shaped appearance as in the
case of our pebble mass distribution.
Probability/area tables … an example
Number of
standard
deviations
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
P(a) 
1
2
Number of
standard
deviations
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
Area
0.000
0.080
0.159
0.236
0.311
0.383
0.451
0.516
0.576
0.632
0.683
a
[  z 2 / 2]
e


a
2
dz
Area
0.729
0.770
0.806
0.838
0.866
0.890
0.911
0.928
0.943
0.954
Number of
standard
deviations
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
Area
.964
.972
.979
.984
.988
.991
.993
.995
.996
.997
Number of
standard
deviations
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Area
0.000
0.080
0.159
0.236
0.311
0.383
0.451
0.516
0.576
0.632
0.683
Number of
standard
deviations
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
Area
0.729
0.770
0.806
0.838
0.866
0.890
0.911
0.928
0.943
0.954
Number of
standard
deviations
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
Area
.964
.972
.979
.984
.988
.991
.993
.995
.996
.997
Number of
standard
deviations
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Area
0.000
0.080
0.159
0.236
0.311
0.383
0.451
0.516
0.576
0.632
0.683
Number of
standard
deviations
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
Area
0.729
0.770
0.806
0.838
0.866
0.890
0.911
0.928
0.943
0.954
Number of
standard
deviations
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
Area
.964
.972
.979
.984
.988
.991
.993
.995
.996
.997
What is the equivalent probability that a
pebble having a mass somewhere between
Pebble masses collected from beach A
401 and 450 grams
will be drawn from a
normal distribution
having the same mean
and standard deviation
as the sample.
0.40
0.35
Probability
0.30
0.25
0.20
0.15
0.10
0.05
0.00
150
200
250
300
350
400
Mass (grams)
450
500
550
Note that 401 grams lies (401-350)/48
or +1.06 standard deviations from the
mean.
450 grams lies (450-350)/48 or
+2.08 standard deviations from the
mean value.
1.06 and 2.08 are z-values or the
standard normal representation of
the mass data.
How can we estimate the area
between p (z = 1.06) and
p(z=2.08) from the area table?
Note that area shown above corresponds to the
probability that a sample drawn at random from
this population will have a value somewhere
between 401 and 450 grams.
This
Area
Note that we can express the area (or
probability) as one half the difference
of areas cited in the table.
That yields -
Let’s take a simpler example and
determine the probability that a sample
will have a value that falls between 1 and
2 standard deviations from the mean.
Number of
standard
deviations
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Area
0.000
0.080
0.159
0.236
0.311
0.383
0.451
0.516
0.576
0.632
0.683
Number of
standard
deviations
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
Area
0.729
0.770
0.806
0.838
0.866
0.890
0.911
0.928
0.943
0.954
Number of
standard
deviations
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
Area
.964
.972
.979
.984
.988
.991
.993
.995
.996
.997
First read the areas beneath the
normal distribution for ± 1 and ±2
standard deviations from the mean.
0.954-0.683 = 0.271
The difference equals the sum of two areas, one
between -1 and -2 standard deviations from the
mean and the other between 1 and 2 standard
deviations between the mean.
We’re only after the area on the positive
side of the bell between 1 and 2
standard deviations - so take 1/2 the
difference.
The format of Davis’s
table makes it a little
easier; but, similar
principles apply.
The probability that a
pebble has a mass lying
between 1 and 2 standard
deviations from the mean
is 0.9722 – 0.8413 or
0.136.
Use linear interpolation to estimate the
area under the curve between + and - 1.06
standard deviations. Confirm that the area
under the curve is _____0.71
Confirm for yourself that the area out to +
0.962
and - 2.08 is _____
0.252
The difference is _____
0.126
Now take one-half of that to get _____
Let’s use Davis’s table A.1
and answer this question.
What is the area under the
curve from 1.06 to 2.08
standard deviations from
the mean – i.e. what is the
probability that a pebble
will have a mass that falls
in this range?
Confirm for yourself that this is ~0.1259
0.126 is the normal probability of
obtaining a pebble with mass between 401
and 450 grams from the beach under
investigation.
Pebble masses collected from beach A
0.40
0.30
Probability
Note that the value
derived from the normal
distribution compares
nicely with that
observed in the sample
(0.126 vs. 0.14).
0.35
0.25
0.20
0.15
0.10
0.05
0.00
150
200
250
300
350
400
Mass (grams)
450
500
550
Analyze the eruption
recurrence times for Mt. Aso
in Japan.
Bring to class next Tuesday
for discussion