MATH 251, Probability and Statistics I, Fall 2005, Fri. Oct. 21, Day 24

Read Sec. 5.1. (Note, p. 334)  First I will cover pp. 334-41, then mean and s.d. pp. 340-1, then  the binomial formula pp. 348-50, Memorize the binomial formula and the mean and s.d. for a binomial distribution. Then pp. 341-347 (proportions and normal approx.)  Skip "continuity correction" pp 347-48.
QUIZ WEDNESDAY or FRIDAY:  Computing mean and s.d., using algebra of means and variances (pp.298, 300)
Hand in: Sec. 4.4, p. 306 ff   (From Wed.)
Handout on Algebra of means and variances:  B, C, D. 
4.64 typing errors again.  Use your results from 4.60 
4.69, 4.70 object length, distribution
B) X and Y are independent, µx = 1  µY = 4, sigmax = 2,  sigmaY =4
 W = 3X - 2Y + 1.  Find mean and s.d.
4.77 special case of corr. = 1.  (Lecture did an example of this. Now use algebra on the general rule 3)
4.73, 75  position + attach 
4.76  average two measurements (Pay attention to this; we'll use this pattern again) 

also 4.83, 84 portfolio analysis.
   Also,C) Why are people who can't tolerate much loss (e.g. close to retirement) counseled to use a mix of stocks and bonds?   Look at the correlation between stocks and bonds and fill in the ( ) correctly:  When stocks go (low/high), bonds go (low/high). 

Sec. 5.1 p. 351ff.  Use table C to find probabilities.
5.1, 5.2 Is it Binomial?
5.9a  Random digits as binomial  (Note, P("at least one") = 1 - P("none")--check it out on a list of possibilities for X)
5.3 proofreading again.
5.7  "P(X > m) no larger than .05" means "P(X > m) < .05"
A)  You simulated flipping a coin with P(Heads) = .2, 20 times.  This was B(20, .2).  Using Table C, sketch the probability histogram for B(20, .2). 
You also simulated flipping a coin with P(Heads) = .6, 20 times.  This was B(20, .6).  Table C doesn't have .6.  Make a table column for B(20, .6) by exchanging the role of success and failure, and using Table C. Sketch the probability histogram for B(20, .6). 

 Read, discuss 
4.81, 4.82 insurance		 Optional 
4.85 more portfolio.
 
Homework questions?
Sec. 4.4  Algebra of Mean and variance of a probability distribution of R.V. X, continued: Day 23
  X + Y: the means always add .  Variances add only when X and Y independent.
We can combine the rules:  W = a + bX + cY:  mean, variance?  ("affine" transformation--linear + shift)
W = a + bX + cY + dV + eU etc.

Sec. 5.1 Counts and proportions (Binomial distribution)
Roll a single die: Does it come up "3"? Y/N
Drawing a person at random from a (big) population.  Is it a green-eyed person? Y/N
   This is a basic building block callled a "Bernoulli trial"--outcomes are "Success" or "Failure". (2 outcomes--bi-nomial)
Repeat n independent times, count the number of successes, in the n "trials".
X = # of successes in n independent "trials"; probability of success in each single trial is p:
X has the Binomial distribution B(n, p)

Drawing without replacement from a (big) population= almost independent.  (Population > 20x sample).  Then trial = "observation" and sampling distribution of a count X of  "successes" in a SRS of size n  is  (almost) Binomial.

What do they look like?  Mound shape; if p close to 0 or 1, somewhat skewed.  Applet "Normal approximation to Binomial"

Finding probabilities:  Table C:  P(X = x).   (Many books + SPSS:  P(X < x):  Transform>Compute: CDF.BINOM(x,n,p))
(Where is Table C?  Back of book, Blue-edged pages, T-6; just before solutions.)
Table C: only p's < .5.  But:  Your Success is my Failure: restate problem in terms of the "other" outcome.
Restating probabilities, exchanging role of success and failure (e.g. ex.  5.6).   I use a system like this:  For n = 12, X = # of free throws missed.  Let Y = # of free throws made.  X+Y = n.    "p" = .75 for Y.  "p" = .25  for X.  Make a table:
        Possible values
X:  0  1  2  3  4  5  6  7  8  9 10 11 12
Y: 12 11 10  9  8  7  6  5  4  3  2  1  0
Then circle the values in the x-line corresponding to your probability statement; the values in the y-line below these are the ones for the Y variable probability statement with the same probability.
(e.g. X >  5 = {5, 6, ...12}in the x line, = {7, 6, 5, ..0}in the y line, = Y < 7.)
Algebra should  work:  P(X >  5) = P(12 - Y >  5) since X = 12 -Y;  = P( 12 >  5 + Y) =  P( 7 >  Y)   But watch the direction of the inequalities.

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