{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern