MATH 251, Probability and Statistics I, Fall 2001, Fri. Oct. 26, Day 24 Final version

Continuing the algebra of means and variances.
 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)
IF you were absent Friday, read carefully example 4.23, pp. 334-5, and examples 4.25 and 26, pp. 338-9.  We spent the class working problems of this sort.
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Will begin Monday:
Sec. 5.1  Counts and Proportions ("Binomial Distribution")

Read Sec. 5.1.  First I will cover pp. 372-379, then mean and s.d. pp. 380, then  the binomial formula pp. 387-9, Memorize the binomial formula and the mean and s.d. for a binomial distribution. Then pp. 381-385 (proportions and normal approx.)  Skip "continuity correction" pp 386-87.
QUIZ WEDNESDAY?:  Computing mean and s.d., using algebra of means and variances (pp.334, 337)

Hand in:  Sec.5.1:  A and B use just the algebra rules; you can do them even if you don't "understand" the binomial, as long as you accept X=S1+S2+...+Sn., p.380  A.)  Let S be a random variable with this distribution: s     0     1  prob  p    1-p Calculate the variance of S, using the definition of variance (p. 336) .  (They calculate the mean of S on p. 380). This is the fill-in for the "hole" in the  text p. 380 --- "Similarly [?] the definition of the variance shows that sigma2S = p(1-p) " Read and understand the rest of the derivation of the mean and s.d. of the binomial, X, p. 380. B) Write up the derivations of the mean and s.d. of the sample proportion p-hat (p.382), from the mean and s.d. of the binomial X given on p. 380, by using the fact that p-hat = X/n.  This fills in the "we can obtain" , top of p.382.  The following will be assigned Monday 5.1, 2, 3(briefly) for understanding the binomial setting.  For the following, define carefully what a "success" is, and its probability, and n, the number of observations or trials.  Then the rest is routine.  5.11 married? Use formula and table C for a; table C for b&c.  5.12 players graduate? Use formula and table C for a; table C for b. To use the table, redefine "success" to mean "don't get degree" and restate the probabilities.  5.15 ESP For parts a, b, c, assume the experimenter has 40 cards, ten of each kind.  She shuffles (thoroughly) all 40, and chooses one.  Then puts it back and shuffles again, repeating 20 times.  In part d, the deck is shuffled only in the beginning, then cards are taken off the top one at a time and looked at.  5.17 lie detector 5.23 binomial coefficient facts

Read, discuss

Optional

Restating probabilities, exchanging role of success and failure (e.g. 5.12) I use a system like this:  For n = 20, X = # who get degree.  Let Y = # who don't get degree.  X+Y = n.  "p" = .8 for X.  "p" = .2 for Y.  Make a table:
        Possible values
X:  0  1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20
Y: 20 19 18 17 16 15 14 13 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 <  1 = {0, 1}in the x line, = {20,19}in the y line, = Y > 19.)
Algebra should  work:  P(X <  1) = P(20 - Y <  1) since X = 20 -Y;  = P( 20 <  1 + Y) =  P( 19 <  Y)   But watch the direction of the inequalities.