MATH 251, Probability and Statistics I, Fall 2007, Wed. Oct. 24 Day 26.after class.

Sec. 5.1  X Binomial   B(n, p) , continued 
  Mean and s.d. : Binomial X = S₁+S₂+...+S_n where each of the S_i's is a Bernoulli trial, 1 if Success, 0 if Failure
   s     0     1
   prob  1-p   p    Mean = p (p. 340)   Variance = p(1-p) (Homework)
Then  µ_x = np,   sigma_x² = np(1-p) by algebra of means and variances.
Die:  "Success" is getting a 3.  p = 1/6.  Roll 36 dice.
   Mean number of 3's = 36 ·1/6=6.  Variance = 36 ·1/6·5/6=5  S.d. = sqrt(5) = 2.24
(do this Friday:) Notice standard deviation grows like square root of n, not as fast as the n possibilities. Applet (normal approx to binom)--try with n= 10, n = 100

Formula:  P(X = k)  p. 349
   For any branch with k successes, there are exactly (n-k) failures
    So the probability of that single branch is p^k(1-p)^(n-k)   .
  But (except for k = 0 and k = n) there are several branches with exactly k successes.  How many?   Each is a list of  k S's and (n-k) F's.  How many different lists are there?  Choose which k places in the list of n places get the S's; the others get the F's.
Call the number of different lists "n choose k", written nCk or, the "binomial coefficient", the "number of combinations of n things taken k at a time".
We see from the tree that 3C0 = 3C3 = 1, 3C1 =3C2 = 3.  The case of flipping a coin 4 times (fig. p.280) shows that 4C0 = 4C4 = 1,  4C1 = 4C3 = 4, 4C2 = 6.
Each of the branches with k successes has the same probability, so we can multiply that probability by how many there are.
So P(X = k) = p^k(1-p)^(n-k)
How to calculate nCk without actually writing down all the possibilities? 
Since n! = n(n-1)(n-2).....3·2·1, nCk "simplifies" (when you write it down and do a lot of canceling) to n(n-1)(n-2)...(n-k+1)/k! (k terms on top and k on the bottom).  (The study of "counting" or combinatorics, of which these are a part, is very rich.  More in Math 300)
   Note: 0! = 1, which makes nC0 and nCn work out to = 1, as they should.

.Start here Friday. Look back at mean and s.d. of binomial.
Often we're interested in the sample proportion -- p-hat =X/n  , where X is B(n, p)  p. 341-3
  Mean of p-hat: = mean of (X/n) = (np)/n = p
     Sample proportion is an "unbiased estimator" of p
   S.d. =  (Homework)
  Applies to results of SRS from population, approximately, if (rule of thumb) Pop. size N  > 20·n. (in fact, if sample "uses up" a good chunk of the population, the s.d. only gets smaller than this number, so some authors use 10·n.) Mean still works, no matter how big the sample.)  The exact s.d. is  the above, times sqrt [(N-n)/(N-1)], the "Finite population correction."  You can see that this is always <1, and if N is very large compared to n, it's very close to 1.

Normal approximation:  Both X and p-hat are approximately Normal, for large n, using their means and s.d.
  Important for theory.  And for quick and dirty calculations.
  Rule of thumb:  Normal approx. is barely OK if both np and n(1-p) > 10; better the farther from 10 they are..

e.g. Flip a fair coin  100 times  B(100, .5).  X =  # of heads, p-hat = X/n = proportion of heads.  np=n(1-p) = 50 >>10.
   µ_x = 50, sigma_x = sqrt(100·.5·.5) =  sqrt(25) = 5   X is approximately N(50, 5)
 Probability that # of heads X is between 40 and 60 =  P(40 < X < 60)
            = P[(40-50)/5 < Z < (60-50)/5]  = P(-2 < Z < 2) = 95%, approximately.

  µ_p-hat = .5, sigma_p-hat = sqrt(.5·.5/100) =  sqrt(.0025) = .05   p-hat is approximately N(.5,.05) 
Probability that the Proportion of heads p-hat is between 40% and  60% =  P(.4 < p-hat < .6)
            = P[(.4-.5)/.05 < Z < (.6-.5)/.05]  = P(-2 < Z < 2) = 95%, approximately.

Approximating a discrete distribution by a continuous one--the ">" vs. ">" issue gets lost.  ("Continuity correction," pp. 347-8 helps with this; but we don't usually bother to do it with p-hats, so we'll skip it.  If you must be more accurate, use SPSS or Excel or some other computational aid.)

Next: 5.2, Sampling dist. of the mean of an SRS