final-practice-solutions

final-practice-solutions - 36-225 INTRODUCTION TO...

Info iconThis preview shows pages 1–11. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

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

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

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

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

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

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

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

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 36-225 INTRODUCTION TO PROBABILITY 8:: STATISTICS I '— PRACTICE FINAL You must Show your work in order to get full or be considered for partial credit. You may use two two—sided sheets of notes (8.5 by 11 inches) and a calculator. You may not share a calculator, pencil, paper or anything else during the exam. Your First Name: —_.__.._..._Your Last Name: Your Signature: Grader use: ' ‘L’hn‘ri'l' *** Do not turn this page until instructed to do so.*** Please check that you have pages 2 through 11 + three tables. 1. (4 points) If A and B are two disjoint events such that P(A) = .4 and P(B) = .3, then P(AflB) is equal to: ‘ "" LEA “2406) 5 O _ 0.12 _ 0.70 _ 0.58 .>_<0 _ None of the above 2. (4 points) If A and B are two disjoint events such that P(A) = .4 and P(B) = .3, then P(AUB) is equal to: “:52— Q HAM) 1PM) + P(B) _ 0.12 220.70 ;~A-_-_—.--_d.>58..... . ' _ H .7 .7 , _.-_ None of‘the above 3. (4 points) If A and B are two independent events such that P(A) = .4 and P(B) = .3, then P(Afl B) is equal to: fl X012 Lt) Hana :FCANWBB _ 0.70 _ 0.58 _ 0 _ None of the above 4. (4 points) lfA and B are two independent events such that P(A) = .4 and P(B) = .3. then P(AUB) is equal to: 6—3:? _0.12 L‘» HAVE) 3 WA) t W83 '?C4)P(B) __ 0.70 — " ' —— EL )5 0.58 ' '7 ' _0 _. None of the above S. (4 points) Let A and B be two independent eve If P B) = .25, what can you say about P Em _ l l l J? (5) I 1? (5 l4) _ cannot find it since P(A) is not known .._.. cannot find it since P(A n B) is not known _ cannot find it since both P(A) and P(AflB) are not known _ it is equal to .5 A it is equal to .25 6. (4 points) Let A and B be two disjoint events such that P(A) > 0 and P(B) > 0. Which of the followi ust be tru ? ’_‘ "gm 8 Pie/183w 7‘ PMDPUQ > O _ P(A|B) = P(A) ' _P(AUB)=0 :—-—> A 5 _ PUT) = P(B) __ The two events 11,13 are independent. x The two events A, B are not independent. 7. Suppose that a non—fair coin with P(Head)=.4 is tossed repeatedly. (a) (5 points) What is the probability that the first "head" occurs on the fifth toss? x: ##3ng until H want-acme tot/s“) = {NI/FY am)" Dos/m (b) (5 points) What is the probability that the third "head" occurs on the fifth toss? Y:#lmd unf‘l l“l NNB(F:%F:0'Q Moi} = (ilflwfléf :@ 8. (10 points) 0 In a certain population it is known that 10% of the individuals have a certain disease. The test for diagnosing the disease can be either positive. indicating that the disease is present. or negative, indicating that the disease is not present. This test, however is not perfect. c When the test is applied to an individual who has the disease, it is accurate (i.e.. positive) only 90% of the time 0 When the test is applied to an individual who does not have the disease, it is accurate (i.e., negative) only 80% of the time (Suggested notations: D - the person has the disease A - the test is positive). (a) The problem gives you three pieces of information (indicated by the three "bullets" (0) above. Express these three pieces of information in terms of probabilities involving event D and event A. 9 3'— 0.lO b :CL‘Ser. — j a i’lAlb) :mo A: i» .4 Myst. “El-l5) io'go (b) What is the probability that a person actually has the disease if the test indicates that he/she mm 2 womb) _ _ ?(A)b)i>(b> + HAIDBNB) l l we MW) + (hWI’fi ,O‘i ll ,M 4- ,IS \J g 9. Assume that X. the number of sales that a telemarketing salesperson makes is a random variable having a Poisson distribution with /\ = 2 per hour, and that 40% of the sales are of product A. (The remaining 60% of the sales are of other products). (a) (5 points) if the salesperson works every day from 12:00 PM to 6:00 PM (non-stop), what is the probability that in a given day the salesperson makes exactly 5 sales? Y:#Ssle5 in la baud "9535“? (352'; 5 ll) 42 g :3 Plus/3’3 : s ‘3’ Sr. (____/,_\_ (b) (5 points) What is the probability that in a given hour, the salesperson makes at least one sale of product A? Y: dis-file 6/ :4" 50 W LN" N P065640: 2’"9:'°G> W003 1 VFW—'5?) 1 (c) (5 points) Since the salesperson gets some kind of bonus for every sale of product A. a day is considered "successful" if the salesperson makes at least one sale of product A during that day. What is the probability that out of 5 randomly chosen days, exactly 2 days are "successful"? (No need to calculate a final numerical answer). Yziisf/qéf *4" I; m as NR:(>\:Z-6*.Ll :w) 10. Let the continuous random variable X have the following pdf: 3(1—fi) O<m<1 fXCE): 0 otherwise (a) (5 points) Find the expected value of X. E(X). 1 E06 : \xfsb—J’Qéx LSS(‘><"‘><I3/L)Jx D o 1 5/2 far): : -3£:<E_E§ ;3[10 .0] (b) points) Let ' Find fyfy) and identify the distribution you got. (Dhagrflh «Lath 0<X<i % 04Y<| (8] RM) 1 WW *0 =i>(f>’< 5 Y) ‘—“- @ 471(7) 1 2%; W7) —, é,- " 6176(7) azynafi 4i 7; iyiviiéyflv t : W212) 72'-l C’ 4/5“ :35 \Jj N 367%(27 2’) i’(><e 7‘3 3 @025 MW 11. The service time at a bank teller, X, has an exponential distribution with fl = 4 minutes. (a) (5 points) Suppose that a customer has been serviced for 5 minutes. What is the probability that he will be serviced for at least 7 minutes in total? . _ _3 -1 Wxsrilps) : tool) =— e V i e ’l‘ 1* mmafjlofl {Wu/l" "3;: ML eerdm'f'd Cl)? 5 l" a (b) (8 points) What is the (approximate) probability that X100, the average service time of a random sample of 100 customers. exceeds 5 minutes? Be clear about what theorem you’re using, and why you can reliably use it. Cam/Limit TWO“ ml?” Laws )4. Kmo We 13273_ ‘eé lmoo 2 3°.. [93) am =5 9 °/ \00 i 81, 3 is was”) a: We: > : we > 2.3“) :5; "5:00 a, legs a an); Win-W, ; (c) (8 points) Find E 9° x Elm "- l*"“:§-eflx " o o almosl’ 6'4( 12. (12 points) Let X and Y have the following density functions. fx(m)=3m2 O<$< 1 fy(y)=2?; 0<y<1 Assuming that X and Y are independent, find P(Y < X). XLLY i? JFK? (76,13 L 6x31 iii 0 also 13. Let X and Y be two independent random variaiales such that: 1 t2 at + 2.93“ . 3-, l 0’ '5 I . MX(t) = an? wk“ I“ 2 t 2 o .2 824—1f (a) (8 points) The distribution of X is The distribution of Y is (b) (15 points) Find E (4 6X Y2) gfleffl '— LlEkx‘ft] :{Iokp—u‘znoas qEEXJED’L] EEK] 1AA!) : 61H". : 6%. W] = WWW] -. 2+ 1;) :3 :i? EZQJYLJ T- wig/5.2:, : $6372, (c) (7 points) Find the moment generating function of X + Y. mxwrékj : mMQ/Wflfl L7 iridefafidmcg we} 4: fig— Eie Jrégc?’ " 6f 7’ e} :i e, (d) (5 points) Using the moment generating function in part (c), or otherwise, the distribution of X + Y is: 14. Let (X,Y) be jointly continuous with the following joint density: w Stay 0 < a: < 1 0 < < a: fXY($=y) = y 0 otherwise (a) (8 points) Show that the marginal distribution of X is Beta(4,1). m = - is YZO 7’er : ({4 3|-) WWI) X (MC - 8X: :2 a 0 4X4} J 0:: y¢x. - 747‘) Ll X x?" 0 else (c) (4 points) Based on your answer to part (b), are X and Y independent? YES (d) (7 points) Find P(Y > % IX = V?— Hz. J “£7”le :87 “"4” Win-5*} —: 3ny7 and 0 Us" Y”: 1’3 71V; (e) (8 points) Find I x I l‘ x 7' 8 3 1:. S 8:36;;(79‘74; : 83 87 (1)5 : ‘B—JXG‘X “so 720 )(I—D 150 x;o i —,. ZXY ) __ ’2: ’3— 3 10 15. Assume that: (3 )1 - EM "'9 W: a? I 3 (a) (6 points) Find E(Y). I l L a ,— X _ _, :. Eff] : EEI‘I’M] 1 i:[3::i L 1; EB] — z. 4‘ 2.. (b) (8 points) Find V(Y). (c) (12 points) It can be shown that in this case E(XY) = 19.5. Use this information and the results you obtained in the previous two parts to find: i. 00V(X,Y)- 2'. E LX «.23 —— E [x] EU] 2H5“— 6‘3 1 "f ...
View Full Document

This note was uploaded on 05/16/2011 for the course STATISTICS 225 taught by Professor Finegold during the Spring '11 term at Carnegie Mellon.

Page1 / 11

final-practice-solutions - 36-225 INTRODUCTION TO...

This preview shows document pages 1 - 11. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online