Geology 351 - Geomath Chapter 7 - Statistics tom.h.wilson wilson@geo.wvu.edu Department of Geology and Geography West Virginia University Morgantown, WV

Probability Probability can be thought of as describing the fraction of the sample that contains a specific value or range of observed values. Just as with the average and the standard deviation, there is a distinction between the probabilities observed in a sample and those of the parent population.

But the idea is that 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.

Spreadsheet interactions ... The probability that a specimen will have a certain value or range of values is expressed as the fraction of the total observations (specimens) that have that value or range of values. Spreadsheet interactions ... Compute the frequency of occurrence for each range of masses used to construct the histogram.

Note that the probability of individual occurrences may vary 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.

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.

Spreadsheet interactions ... Convert number of occurrences to probabilities and construct the probability distribution histogram.

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 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. Population And we can estimate the probability distribution of the parent population from statistical estimates of the mean and variance. Statistic

This expression is often simplified by substituting Z for (x-x)/ This expression is often simplified by substituting Z for (x-x)/. Z is referred to as the standardized variable or the standard normal deviate, and the Gaussian distribution is rewritten as - 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, the z of 2 implies x (for z of 2) = 446 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.

The Gaussian or normal distribution p(x) is a mathematical representation of this bell-shaped characteristic. This mathematical representation yields a bell shaped curve of probabilities whose form and extent is uniquely defined by the mean and variance derived from a sample.

The Gaussian (normal) distribution of pebble masses looks a bit different from the probability distribution we derived directly from the sample.

35 values fall in the >300 to 350 range

In the probability histogram, each bar represents a discrete sum of masses over a 50 gram range divided by the total number of the pebbles.

What do the areas in the above table represent?

Consider question 7.4 (page 124) of Waltham - In this question Waltham evaluates the equivalent probability that a pebble having a mass somewhere between 401 (i.e. >400) and 450 grams would be drawn from a normal distribution having the same mean and standard deviation as the sample. Not from the sample distribution – we already know what that probability is.

Note that 401grams lies (401-350)/48 (i. e. +1 Note that 401grams lies (401-350)/48 (i.e. +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 standard normal representations of the pebble masses associated with this sample. Note that Waltham rounds off the mean and standard deviation to 350 and 48, respectively.

How can we estimate the area between p (z = 1.06) and p (z=2.08)? Note that this area corresponds to the probability that a pebble drawn at random from this ideal parent population will have a value somewhere between 401 and 450 grams.

The areas we get from tables The area we want to find This Area The areas we get from tables Note that we can express that area as one half the difference of the two areas shown above

... one half the combined areas. yields - X

As an example, from our table, we have the areas between  1 and 2 standard deviations from the mean.

Areas taken from the table

0.954-0.683 = 0.271 This difference equals the sum of two remaining 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. ...Of course, it will work for either side.

Confirm for yourself that the area out to + and - 2.08 is 0.962. Question 7.4 is a little more complicated. We no longer have numbers listed in the table. Waltham goes through a weighted average determination of the area under the curve between + and - 1.06 standard deviations. He obtains the area 0.71. Confirm for yourself that the area out to + and - 2.08 is 0.962. The difference is 0.252. Now we take one-half of that to get 0.126

This method of linear interpolation assumes linearity in the curve between 1 and 1.1 P(1.06) is six-tenths of the way from P(1) and P(1.1) or 0.683 plus 0.6 times the difference (0.046) =0.683 + 0.0276 = 0.71

P(2)=0.954 P(2.1)=0.964 P(2.08) is eight-tenths of the way from P(2) and P(2.1) or 0.954 plus 0.8 times the difference (0.01) =0.954 + 0.008 = 0.962

Be sure you understand the difference 0.126 is the normal probability of obtaining a pebble with mass between 401 and 450 grams from the beach under investigation. Note that the value derived from the normal distribution compares nicely with that observed in the sample (0.126 vs. 0.14). Be sure you understand the difference

Be sure to finish reading Chapter 7 Be sure to finish reading Chapter 7. On Tuesday we will consider the idea of statistical difference. Look at problem 7.13. We will discuss this in class on Tuesday as well. Hand in the PsiPlot histogram and probability distribution. Your effort will be graded so be sure to put you name on the plot

We may begin discussion of derivatives next Tuesday Start Reading Chapter 8 We may begin discussion of derivatives next Tuesday