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# e71_hw2__scan - Stats for Strategy Whitten HW 2 Solution A...

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Unformatted text preview: Stats for Strategy Whitten HW 2 Solution A. Exercises 4.86 a: = one student’s ACT score 11 = 18.6 0 = 5.9 (31) Find P(:L‘ 2 21) . a: — ,1 _ 21 — 18.6 2: =0“ or 5.9 P(Z 2 0.41) = 1 — Table(0.41) -Z 0 OJ“ = 1 — .6591 = .3409 ui=M=- (the mean of the bell curve for 5?: is still [1, also the mean for an individual score as) 0' 5.9 a— = — = —— = 0.8344 (c) Find 13(5 2 21) . 5 — ,1 21 — 18.6 2 = = —— = 2. 055 0.8344 88 E P(Z 2 2.88) = 1 — Table(2.88) O 8.28 = 1 — .9980 = .0020 (d) The mean score for 50 students will be 21 or higher in about 2 out of every 1000 50—student exams. 4.88 x = one year’s return on common stock, in percent p=13% a=17% 71:45 (3.) Find 13(2 > 15) . 02 = «H = 73 = 2.5342 _ (E — ,u _ 15 — 13 _ Z _ of _ 2.5342 _ 0'79 Z P(Z > 0.79) = 1 — Table(0.79) 0 0.19 = 1 — .7852 = .2148 (b) Find P(5 < 7) . zi—M_ 7—13 05 "2.5342 E P(Z < —2.37) = Table(—2.37) = = —2.37 4.9.51 o 4.95 :1: = amount ﬁlled into a single bottle, in m1 )1. = 298 a = 3 (a) Find P(a: < 295) . w—p_295—298_ a _ 3 __ 3Z- P(Z < —1.0) = Tab1e(—1.00) = Z = —1.0 'l.0 o (b) Find P(5 < 295) . a 3 5,- — 0 295 — 298 = = —— = ~2.4 i2 Z 03-: 1.2247 5 -9.~Is o P(Z < —2.45) = Table(——2.45) = (c) About 1587 out of every 10,000 bottles of coke contain less than 295 ml. (d) About 71 out of every 10,000 slx—packs of coke average less than 295 ml. 4.96 .7: = a single measurement of Shelia’s glucose level, in mg/dl p = 125 a = 10 (3.) Find P(a: > 140) . 9: — )1 _ 140 — 125 z = a 10 = 1.50 :2 P(Z > 1.50) = 1 — Tab1e(l.50) 0 "5° = 1 — .9332 = .0668 (b) Find P(5 > 140) . 0'— — .3.— — ﬂ — 5 I W x/Z Z = m _ ,u = M = 3'0 0'5- 5 Z P(Z > 3.00) = 1 — Table(3.00) 0 3.00 = 1 — .9987 = (c) In Shelia’s case, it’s deﬁnitely better to base doctors’ diagnosis on four measurements instead of one measurement since the chances of avoiding expensive (and unnecessary) medical treatment increase from 93.32% to 99.87%, an improvement of 6.55%. 4.98 a: = # of accidents in a single week 11 = 2.2 cr = 1.4 n = 52 weeks (a) Since n = 52 Z 30, :i' has a bell curve with mean ui=u=- 1.4 standard deviation ai = — = — = 0.1941 n «5—2 (2. (b) Find P(§: < 2) . Zza—M 2—2.2 ;_L P(Z < —1.03) = Table(-1.03) = = —1.03 -‘ LOS 0 (C) P(tota.l # accidents < 100) = P(:1‘c < 100/52) = 13(5: < 1.9231) Z = \$0511 2 19391194122 : _1.43 Z P(Z < —1.43) = Table(—1.43) = - l.‘-|3 O 4.99 w = one week’s postage expense, in \$ 11 = 312 a = 58 n = 52 weeks (a) Find P(2 > 400) . 05 = 5—; = 3—583 = 8.0432 _ :E-p __ 400—312 _ Z ’ 0i ‘ 8.0432 ‘ 10'94 P(Z > 10.94) m E -3 O 3 1°31"! (b) Since 71 = 1 < 30 for one week, we cannot use the formula z=\$_” a to ﬁnd the answer unless we know that :1; is normally distributed. 4.110 .21: # of people 1n a car which enters the interchange ,u— — 1. 5 o— —- 0. 75 (a) The variable 3: cannot be exactly normal since a: is a. discrete variable while the bell curve is used for continuous variables. (b) We have n = 700 cars. Since 71 = 700 2 30, :2 is approximately normally distributed with mean #5 = u = a 0.75 standard de iation — = — = — = 0.0283 v Um ﬁ ,—700 -— (C) P(t0tal # people > 1075) = P(§: > 1075/700) = P(f: > 1.5357) :7: — ,u 1.5357 — 1.5 = : — = .26 Z 05 0.0283 1 2 P(Z > 1.26) = 1 — Table(1.26) 0 1:34: 21—.89622- (d) More than 1075 people in cars will pass through the interchange in about 104 out of every 1000 rush hours. 5.41 p = probability or proportion of correct answers = 0.75 (.75)(.25) _ =p(1‘/ T _ .0433 p—p_ .70— .75 = _ —— = —1.1 Z- 0,, .0433 5 P(Z g —1.15) = Table(-1.15) = .1251 (b) n = 250 Find Pg? 3 .70) . =p_(__/ 1— (.75)(.25) _ 2 = p_p:M=_1_82 a; 0274 P(Z g —1.82) = Table(——1.82) = .0344 (a) n = 100 Find P033 .70) . "lols O ““1082- O (c) A larger sample size (n = 250 vs. n = 100) shrinks the standard deviation 0,; and so tends to move if closer to the mean value p = 0.75. Since 5 _<_ .70 represents area below (away from) the mean p = 0.70, it’s probability is reduced. 2 /\ "-1 .75 F 5.42 ' O (b) P(70 or more home runs) = P(§Z 70/509) _ —(P (p_ > 0.1375) =p(/ (.116)(.884) — = . 142 509 0 p—p_ .1375 — .116 = —— =1.51 z 05 0.142 O "5' P(Z 2 1.51) = 1 — 155141.51) = 1 —- .9345 = .0655 5.43 (C) P(70 or fewer respond) = P(1’)‘S 70/150) =P(p<0.%m) =1/p( =1/(51)5(05) =.0408 — p__ .4667 — .5 = _ ———-——— = —0. 82 Z 105,, .0408 ~o.82 O P(Z g —0.82) = Table(—0.82) = .2061 5.44 p = proportion of a_ll orders which ship on time = 0.90 (a) P(86 or fewer orders on time) P(p__ < 0.86) =p___/ (1— (.90)(.10) _ 100 “ 03 = p—p :M = -133 2 0'5 .03 - 1553 0 P(Z g —1.33) = Table(—1.33) = .0918 b The roportion {5 of orders shi ed on time in a. sample isn’t a perfect reﬂection of the P PP proportion p of a_ll orders shipped on time. In fact, if p = 0.90, there is still about a. 9% chance that 86 or fewer out of 100 sampled orders ship on time. 5.45 P(sample contains 170 or fewer blacks) 2: P015 170/1500) _ -—P( (p_ < 0.1133) /p(1—10/_8___0.8)12=0084 — p__ 1133 — 1.2 = _ —— = —0 80 2 pop .0084 P(Z g —0.80) = Tab1e(——0.80) = .2119 (c) If blacks are fairly represented in the selection process, then about 21 out of every 100 samples of 1500 persons will contain 170 or fewer blacks. —O§%0 O B. Calculating with MINITAB 1. Exercise 4.98 (b) .1514 (c) .0768 2. Exercise 4.1 10 .1036 3. Exercise 5.41 (a) .1241 (b) .0340 ...
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