Math 151 , Fall 2008  Wed. Day 26, Oct. 29Hit reload...After class.

HW: Review Normal distribution. (Re)read rest of Ch. 10 (Note Normal distribution is back) . Check 10.21, 28 Read ahead, Ch. 11.
Hand in  Monday ..
= = = = = Probability , Ch. 10, discrete,.
p. 261, 10.16 Grades RV
p. 252, 10.6 and 10.7 D&D, 4-sided dice (These are "just like" the 6-sided dice in examples 10.4-5, with a little added twist.)

On a separate sheet, simulations using http://www.whfreeman.com/bps4e  "Probability " applet:  If you do it jointly, one sheet for both people (I'll aggregate the results)
p. 270, 10.55 runs of free throws
p. 270, 10.56 a.  For b, do 20 people 10 times, but do 320 people only twice.  Record not only the proportion (.63 or whatever) but the fraction (like 201/320.  208/320 = .65 exactly)

Postpone the rest: Continuous sample spaces:
***For A and B, Use  Densities Handout, from Day 5. Answers to old questions 
Change language from "description of a population of data" to "pick an individual from the population, call the value X"
A. ("Uniform")   X = number the spinner points to.
a) (example)  The probability that the spinner points to a number less than .6 = P( X < .6) = .6 .
b) P (.2 < X < .6) = ?   Say it in words: ?
c) For what x is there probability .4 of being greater than x ?      (In notation: P(X > x) = .4.  Find x)
B.  Y = (number you get from) the sum of two spinners. ("Triangular")
  This is the same random variable as Y in 10.14, p. 259!
a) The probability that the sum is a number less than .6  =  P(       ?        ) =.18     P(Y > .6) =  ?  
b) P(Y < 1.6)  =          P (Y < 1) =   ?             P( 1 < Y < 1.6) =  ?
c)  P(Y > x) = .92  Find x:  ?   (Hint:  P(Y<x) = .08)
***
p. 259, 10.13 uniform, 0-1 (Note, this is distribution A on the handout)
p. 259, 10.14  sum of two uniform (Note, this is distribution B, "Triangular", on the handout)
pp. 236-7 10.48 and 10.50 uniform on 0-2 (This corresponds to a single spinner with its edge labeled with values going from 0 to 2, rather than the 0 to 1 we used in our homework up to now. Use the area-of-rectangles formula to find the probabilities.)

Ahead: Normal distribution:  Restate each problem from "The probability that X is..." to "The proportion of the population of x's that ...." and use your old techniques.
p. 259, 10.15  Iowa Test Scores
p. 269, 10.49 Did you vote?
p. 269,  10.51 NAEP scores
p. 269,  10.53 Friends
Read, to discuss


Optional 



Exam this Friday (next class) Oct 31 (Day 27). Sign up today, as usual.   Bring one sheet of notes.  Chapters 8 and 9 all, Some  of Ch 10 --through HW assigned today. All but 7d are "on" the exam.
   Sample exam available today in class, or in white folder outside my door.  Solutions Here.
Note that there is no problem on the sample involving two (or more)" factors" or expressly listing "factors" and "treatments", but such questions could be on the exam. 
I have been known to ask questions on the exam specifically on the "outside" reading, such as the Placebo Effect articles; Bradley effect.

p. 249, 10.3
  Please add your proportion of "heads" = "0's" from 200 repetitions ("tosses") to the list and  dotplot circulating!  "Expect" 1/10 should be "heads". I did it 4 times (four sets of 200 tosses) , got .06, .09, .07, .14The list/dotplot


Chapter 10, Probability (intro)  continued  Details Day 24

Bare Bones recap:  Chance  behavior (a random phenomenon): Unpredictable in the short run,  predictable regular pattern in the long run.   "Probability" of particular something happening: proportion of times it would happen in a very long series of (independent) repetitions of the phenomenon.   Applet:  Probability

Probability Models : (p. 250-256)
   Random phenomenon,
described by
    Sample space S: 
set of all possible outcomes (no overlap of descriptions)

    Event:  any outcome or set of outcomes
    Probability model: S, and a way of assigning a probability to each event.
Sample space depends on what you want to know:
Phenomenon: Flip coin twice.
    S1 = {HH, HT, TH, TT}     S2 = {0, 1, 2} number of heads   S3 = {Y, N} both are heads?

We looked at the probabilities for these, implicitly using the "common sense" rules for proportions just below.

Probability rules:  pp. 253, in words, then in notation.
A an event in sample space S, P(A) is "the probability that  A occurs"
    These rules are all true for proportions in long run (Probabilities), proportion of counts, proportions of areas.
    1.  0 < P(A) < 1   (any probability is a number between 0 and 1. )
    2. P(S) = 1         (all the outcomes together have total probability 1)
    3.  A and B are  disjoint if they have no outcomes in common (can't happen simultaneously.)
        If A and B are disjoint, their probabilities add:  P(A or B) = P(A) + P(B)
   
4. For any event A, P(A does not occur) = 1 - P(A)

Discrete models: Assign a probability to each outcome (>0) so they add to 1.  Prob. of an event is sum of prob's of its outcomes.

Homework questions? Day 25
Questions on exam material?

New material (not on exam:)
Random Variable: Day 24 (X, Y, Z...) Variable whose value is a numerical outcome of a random phenomenon.
 Probability distribution of X tells us what values X can take and how to assign probabilities to them.
    If X has a finite number of possible values (Discrete distributions), nothing new except notation.
     P(X < 2) is "Prob. that X is less than 2."
Probabilities follow the "common sense" rule for proportions of a whole.  Same rules for proportions of areas, proportions of counts, proportions in histograms, proportions of times in the long run something would happen. 
Two principles for assigning probabilities:
--Sometimes, a properly chosen sample space will have equally likely outcomes. You can use this to find other probabilities.
--If you pick an individual at random from a population, the probability that one individual will be XYZ is the same as the proportion of XYZ's in the population.

Example of an equally likely sample space, which can be used to find probabilities for another: p.251, Examples 10.4, 5, Two dice.

Another probability example: Recall exercise 8.10, sample of managers.
How many East Asians did you get in a sample of size 6? Day 25

Start here Monday.
This was a Teaser for Ch. 11: 
We know that a sample from a population will not exactly represent the population.  If we take a random sample, the behavior of samples will not be individually predictable, but there will be predictable pattern in many random samples from the same population.  Knowing the pattern (here the probability distribution) will be  as good as we can do.

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So far, everything was for Discrete sample spaces (you can list the outcomes)  Now:

Looking ahead (back)
Random variables with intervals of outcomes ("continuous") Ch.10 (p. 256 on) 
If the sample space is an interval of values (or the whole line), the way we assign probabilities to events is with a density curve (Ch. 3, cf. Day 5 on) (remember density curves were idealizations of histograms--of repeating the "experiment" many many times)
  P(a <  X < b) = the probability that X is between a and is the area under the density curve, between a and b.
We declare P (X = a) = 0 , so P(a <  X < b) = P(a < X < b)
Notation: Use capital letter for the random variable, the "label" of the phenomenon.  Use small letters for particular values it can have.  But this rule is often broken--Moore uses x-bar where many (I) would use X-bar.
  HW: For A and B, Use Densities Handout, from Day 5,7. Answers to old questions 
Change language from "description of a population of data" to "pick an individual from the population, call the value X"
A. ("Uniform")   X = number the spinner points to.
a) (example)  The probability that the spinner points to a number less than .6 = P( X < .6) = .6 .
b) P (.2 < X < .6) = ?   Say it in words: ?
c) For what x is there probability .4 of being greater than x ?      (In notation: P(X > x) = .4.  Find x)
B.  Y = (number you get from) the sum of two spinners. ("Triangular")
  This is the same random variable as Y in 10.14, p. 259!
a) The probability that the sum is a number less than .6  =  P(       ?        ) =.18     P(Y > .6) =  ?  
b) P(Y < 1.6)  =          P (Y < 1) =   ?             P( 1 < Y < 1.6) =  ?
c)  P(Y > x) = .92  Find x:  ?   (Hint:  P(Y<x) = .08)
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Our most important probability model: NORMAL DISTRIBUTION family.  Same techniques as before, only we ask "probability that one chosen at random..." instead of "proportion of all..."  Review Normal techniques: Day 7Day 8 cover it all.
 Take a random sample of size 1 from a population which is N(110, 25). 
(Give an individual, chosen at random, the "Classic IQ test", which has a normal distribution, mean 110, s.d. 25.   X is the score on the test.)
Find P(100 < X < 145), prob. that individual gets between 100 and 145. (in class?)   Work is on Day 8, what proportion.


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