20.1	_# tosses_  _E(#H)_  _SE(#H)_  _E(%H)_  _SE(%H)_
	     100	50	  5	   50	     5
	    2500      1250	 25        50        1
           10000      5000       50        50       .5  
         1000000    500000      500        50       .05

20.3	a) 50,000
	b) 0 or 1
	c) False, SD = .4  (using the quick method)
	d) True
	e) SE of # of incomes greater than $50K is 
			(900)^.5 * .4  =  12.   (or 1.3%)
	   So, plus or minus 1% is (1/1.3)=.75 standard units.
		The chance is around 55%.

20.5	The appropriate box model has a mean of 150 lbs and a
	SD of 35 lbs.  The expected value for the sum of 50
	draws is 7500 pounds, and the SE is 
	   (50)^.5 * 35 = 250lbs, so 8000 lbs is +2 standard
	units.  There is a 2% chance of failure.
	  (By the way, elevators are not designed like this!)

20.6	(ii) is best.  Because California is more populous, its 
	sample will be larger, and more accurate for estimating
	a percentage value (see section 4).

20.7	a) True.  Expected values have no associated error.
	b) False. see a)
	c) True.
	d) False.
	e) True.  (You can count the tickets!)
	f) False.

20.9	200, with SE of about (200)^.5 * (1/3*2/3)^.5 = 12.

20.11	a) observed = 357, expected = 340
	b) observed = 71.4%, expected = 68%


21.1	a) The same is like 500 draws from a box with 25,000 tickets;
	   each ticket is marked 1 (has computer) or 0 (does not have
	   a computer).  The number of sample households with
	   computers is like the sum of the draws.

	   The fraction of 1's in the box is unknown, but can be 
	   estimated by the fraction in the sample, as 79/500=0.158.
	   On the basis, the SD of the box is estimated as
	   sqrt(0.158*0.842) ~= 0.36.  The SE for the number of sample
	   households with computers is estimated as sqrt(500)*0.36 ~= 8, 
	   and 8 out of 500 is 1.6%.  The percentage of households in 
	   the town with computers is estimated as 15.8%, and the estimate 
	   is likely to be off by 1.6% or so.

	b) The 95% confidence interval is 15.8% +- 3.2%.

	(NB: You need to estimate the percentage for the town from
		data for the town:  the national figures may not
		apply.  In this case, the town seems pretty close to
		the national average.)

21.2	a) 99.6%, 0.3 of 1%
	b) Can't be done:  the box is so lopsided that the normal 
	   approximation won't work.  See exercises 5-6 on p 324;
	   exercises 3-4 on p. 383.

21.4	a) The box has millions of tickets, one for each 17-year-old
	   in school that year.  Tickets are marked 1 for those who 
	   knew that Chaucer wrote The Canterbury Tales, and 0 for
	   the others.  The data are like 6000 draws from the box, and
	   the number of students in the sample who know the answer is
	   like the sum of the draws.  The fraction of 1's in the box
	   can be estimated from the sample as 0.361.  On this basis,
	   the SD of the box is estimate at sqrt(0.361*0.639) ~= 0.48.
	   The SE for the number of students in the sample who know 
	   the answer is estimated as sqrt(6000) * 0.48 ~= 37.  The SE
	   for the percentage is 37/6000, which is about 0.6 of 1%.  The
	   percentage of students in the population who know the answer
	   is estimated at 36.1%, give or take 0.6 of 1% or so.  The 95%
	   confidence interval is 36.1% +- 1.2%.

	b) 95.2% +- 0.6 of 1%

21.8	This is not the right SE.  The bank has mixed up 73 cents with
	73%.  (The right way to figure the SE for an average is
	explained in Chapter 23).

21.10	710/100=7.1, so th two options describe the same event in
	different words.  They are the same.

21.11	Option (ii) is it.  For example, about 95% of the estimates 
	will be right to within 2 SEs, about 99.7% of them will be 
	right within 3 SEs, and so forth.

21.12	(i) is irrelevant, (ii) is a histogram for the numbers drawn,
	and (iii) is a probability histogram for the sum.  Reason:
	(iii) looks like the normal; (ii) looks like a histogram for
	the contents of the box; (i) would allow 3 and 4 among the 
	draws.

21.13	In all three cases, the expected value is 500 and the SE is
	about 16.  You know what is in the box, the expected value and
	SE can be computed exactly and do not depend on the data.  The three
	chance errors are 29, -16, and 14.