P. 314 #33 Small mean=1.5 stdev=.3; Large mean=2.5 stdev=.4; a) (Large – Small) has mean of 2.5 –1.5 =1.0 b) Variance of diff is sum of variances= (.3*.3)+(.4*.4)=.25; stdev is square root which is .5 c) difference is normal with mean 1.0 and stdev of .5. small has more than large when the difference is less than zero. P(difference<0) on N(1.0, .5) ? z-score is (0-1)/.5 = -2.0 look up in A-51 and see .0228 d) (Large + Small) has a mean of 2.5+1.5=4.0 and same stdev of .5 e) P(sum>4.5) has z-score of (4.5-4.0)/.5 = 1.0; A-51 shows .8413…but that is less than so we want 1-.8413=.1587 f) I think he means that we have filled the small and large bowl out of this one box, so (Box – (Large+Small)) is our concern. Mean is 16.3 – 4.0 =12.3 and variance is (.2*.2)+(.25)=.29 so stdev is square root of .29 which is .5385. #35 a) Like #33f …16.2 - 4.0 =12.2 and b) variance (.1+.1)+(.25)=.26 with square root of .51. c) P(box>13) has z-score = (13 – 12.2)/.51 = 1.57 from A-51 gives .9418…but that is less than so we want 1-.9418 = .0582.

P. 327 #15 a) binomial since 2 outcomes (in or out), independent trials, prob of success (in) is .7…all 6 in is 6C6*.7*.7*.7*.7*.7*.7=.11765 b) exactly 4 is 4C6*.7*.7*.7*.7*.3*.3= {6!/(4!2!)}*.0216= (6*5)/2*.0216=.324 c) that is 4 or 5 or 6, so .324 +( 5C6*.7*.7*.7*.7*.7*.3=

6*.05048=.3025 )+.11765=.744 d) this is 0 or 1 or 2 or 3 or 4; it is also 1 – (5 or 6) which is

1- (.3025+.11765)=.58 e) n*p=6*.7=4.2 f) square root of n*p*(1-p)=6*.7*.3=1.12 g) first success on third serve matches a geometric model so that is .3*.3*.7=.063 h) geometric mean is 1/p =1/.7 = 1.42857 i) stdev is square root of (1-p)/(p*p) = (1-.7)/(.7*.7)=.782 #17 a)n*p = 80*.7=56 b) is n*p=56>10? and n*(1-p)=80*.3=24>10? Yes c) stdev is square root of n*p*(1-p)=square root of 80*.7*.3=4.1 (round to 4.0) so we know 56+-4=(52, 60) are number of successful serves 68% of the time this player serves 80 times; 56+-(2*4)=(48, 64)…95%…;

56+-(3*4)=(44, 68) …99.7%.

P. 350 #2 a) & b) Binomial since 2 outcomes of green or not, prob green (a success) is .10, independent color on each check of an M&M, .10 does not change. Is n*p=50*.1=5>10 and n*(1-p)=50*.9=45>10. No. Not normal. C) mean for proportions is p=.1 (center) d) and stdev is square root of {p*(1-p)}/n = square root of (.1*.9)/50 = .0424 #4 see #2 and now use p=.1 and square root of {p*(1-p)}/n=.021. .1+-.021 so 68% of the time we open and count 200 M&M we will find the proportion of greens to be between .079 and .121 or 7.9% to 12.1% green ones….1+-(2*.021) so 95% of the time…..5.8% to 14.2%. c) the span, i.e. range of the percent greens we are guessing to be in a bag would narrow. #6 true proportion p is supposed to be .1 and in n=500 checkings we should get a stdev of square root of (.1*.9)/500 = .0134. The z-score for 12% or .12 is (.12-.1)/.0134 = 1.49. This is out in the more extreme area of a normal curve but not as far out as 2.0, so it is possible. #29 a) skewed right means there are a few instances where groups do big tips, so although the z-score is (20-9.60)/5.4 =1.92 and we know that the probability of a z-score greater than 2.0 is small (ie. .025) since our distribution is not normal the chance could be considerably higher than that. B) since averages for the waiter due tend to be normal (CLT) even when viewing just one tip is not we can try to do this. CLT says the average of 4 tips will still be 9.60 but the stdev is 5.4/(square root of 4) = 2.70. z-score is (15-9.60)/2.7 = 2.0. So, no c) z-score will get even larger for bigger sample size n, so No. #31 a) average of 40 tips is still 9.60 and its stdev is 5.4/(square root of 40)= .854. To make $500 he must have averaged 12.50 on each party. P(average tip>12.50)=P(z>(12.50-9.60)/.854)=3.396…way out in normal tails..virtually impossible b) what average is out in the right tail, in the upper 10%. Search table A-51 for .9000 (if left tail is this then right tail is .10), see a z-score of 1.28 (with the .8997). so (average-9.60)/.854 = 1.28 so average=9.60+(1.28*.854)=10.72. and 10.72*40=$429

P. 366 #3 a) population is all cars that could be stopped at that checkpoint and p is the proportion of these that have violations, we have n=134 that were sampled and our p-hat is 14/134=.104

b) population really is hard to determine…all those that were viewing the tv show (do the apathetic ones look like those who do bother to express their opinion?) so then p is the proportion of these folks (maybe) who favor prayer in school, the sample (and not very representative) is the 602 and p-hat is 488/602 c) the population is meant to be all parents and the p to be the proportion favoring uniforms but the sample of 345 who do return surveys only represent those who are not so busy with life that they have time to respond (how representative is that?) and their p-hat is 228/345. d) population are freshman at that school with p being the proportion who graduate in 4 years and the sample (anything unusual about picking this group to follow?) is 1632 with a p-hat of 1388/1632. #13 a) to form our CI we would do .38+-1.645*sqrt((.38*.62)/ 1012)). Then 1.645*.015 = or 2.5%

b) It is the span or range of the estimate for the percentage of people who believe in ghosts since we use .38 +-.025 which is .355 to .405 or 35.5% to 40.5% c) the more confident you want to be the wider you must make the confidence interval d) <later> e) the tighter, shorter the confidence interval (i.e. the smaller MOE) leads to less confidence.

P. 386 #11 .52+-2*sqrt((.52*.48)/881))=.52+-2*.0168=.52+-.034=(.486, .554) or 48.6% to 55.4% b) we see that in the 1960’s that 44% smoked; I assume this is a population figure, not one generated from a sample. So we wonder if H0: p = .44 or Ha:p is not .44. Based on viewing the confidence interval, the 1995 population appears to have a higher percentage of non-smokers, ie. .44 is below our interval. #20 Ho:p=.19 versus Ha:p is not .19. we will rely on a sample of 72 called for jury duty (are these off drivers’ licenses and is that representative?) and check that proportion. Thinking that there are p=.19 hispanics we know the stdev for the proportion from our sample group of 72 is sqrt of (.19*.81)/72 or .046. So in our mind’s eye we see possibilities for p-hats that center around .19 with a stdev of .046 on a normal curve. We found 9 out of 72 or .125 Hispanics. How far away is that from our .19 center. The z-score is (.125-.19)/.046=

-1.4 which is kind of far below the claimed .19. I would be inclined to question whether the jury selection process is really reflecting the ethnic composition of the county.