Cheat sheet

pdf

School

Chandler-Gilbert Community College *

*We aren’t endorsed by this school

Course

MAT122

Subject

Electrical Engineering

Date

Feb 20, 2024

Type

pdf

Pages

Uploaded by AdmiralGrasshopper2240

DisCvete (on+nUuowy ;'vx PmF pty) P(x>3) beomes - ple=3) PLOD - 1-P(xed) TIPS: - For a discrete pmf, the = matters. ) P (x> 2) Shay Plx=) ' abwag 2670 P(X>3) becomes 1-P(X<=2) But for a continuous pdf, the = does not matter. P(X<3) stays P(X<3) - In discrete: P(X>=3) becomes 1-P(X<3) which becomes 1-P(X<=2). But for continuous, P(X>=3) stays. - A pmf (discrete) is referred to as p(x), and pdf (continuous) is referred to as f(x). Both are the probabilities for each value of x in a range. It is denoted as P(X=x). However, for a continuous distribution, P(X=x) will always be 0. - A cdf (both discrete & continuous) is the cumulative distributive function. It is all the probabilities added up (meaning, f(x1) + f(x2) + f(x3)...+ f(xn)). It is denoted as P(X<=x). For discrete, it will only be a cdf if it is P(X<=x). P(X<x) is not a cdf for discrete. But for continuous, P(X<=x) is same as P(X<x). It is denotes F(x) BAYES' THEOREM P certain region of the northwest, it is sunny: 66%, rainy: 17%, and foggy: 17%. In condmons, the proboblllty of an acc:dent with accident with death: 1/12.1n Given that an accident without death occurred, what is the conditional probability that it was foggy? Ans: We are given the probability fhat it is foggy [P(F)] we are given that the en what is the probability that death occurred We are asked to find the probclblllty that it was foggy given that an accident without death has . occurs given it was foggy [P(D|F)]. B [P(D)]. occurred [P(F|D P\D"|F)P(F P(F|D) =—( ' ),)( ) P(D') = P(D’IF) P(F) + P(D’|R) PR) + P(D'|S) P 5)(0.17) + (1-1/12)(0.17) + (1- 1/24)(0 66) = |"(D'IF) = (1-1/5) P(F) = 0.17 P(D'|F)P(F) P(Dr) T _ (1-1/5)(0.17) P(FID) = - 0.924 death: 1/24. In rainy conditions, the probablllty of an foggy conditions, the probability of an accident with death: 1/5. = Spi - Vavante - lven thar event A has CONDITIONAL PROBABILITY ppere, vt v ~ |coating | | _ happens? Here, information on both e Surface High Low are given. High 17 i Formula for conditional probak Low 99 26 P(B|A)= P(ANB) A) If the coating is high, what’s the probability that surface is high? P(A) _pPlens) . G P(SIC) = P(C) ~ (17+99)/156 =0.147 S: event that surface B) If the surface is high, what’s the probability that coating is high? s high C: event that coating p(cns) _ (156) i PSS} = Ps) - (reiayise . O0o48 is high C) If the surface is low, what'’s the probability that coating is low? P(Srnc - (156) BEIS) S (5" (99+26)/156 =0:208 PROBABILITY MASS FUNCTION (DISCRETE) p (‘F’D‘) Q: Verify the function is a probability mass function and find the probabiliti p(x) = (512/73)(0)%, x = {1, 2, 3} wa c\f‘(& va\dwlll we are not Total probability rule always comes in the denominator 0.924 ~, The first part of the total proboblllty / always be on the numerato = 0.147 at death Is it a pmf2 Idea: all the p(x)’s need to add up to 1. ploiy) PP p(1) + p(2) + p(3) = 1. Yes, it is a pmf vy, A) P(X<=1) = p(1) = 0.8767 Lo’ B) P(X>1) = 1 = P(X<=1) = 1-p(1) = 0.1233 C) P(X>=3) = 1-P(X<3) = 1-P(X<=2) = 1-p(2) cule il D) P(2<X<6) = P(3<=X<=5) since the range is 3, we would interpret this c P(X=3) = p(3) = 0.0137

PMF CONTINUED Q: An assembly consists of 2 components. The probability that the 1" component meets specifications is 0.90, and 2" is 0.92. Assume that the components are independent. Determine the pmf and the mean and variance for X. X is the number of components out of 2 that meets the specifications P(X=0) = (1-0.90)(1-0.92) = 0.008 mean - P(X=1) = (1-0.90)(0.92) + (0.90)(1-0.92) = 0.164 vavion(e - P(X=2) = (0.90)(0.92) = 0.828 Mean: i = x1p(x1) + x2p(x2) + x3p(x3) plw) - 090 = 0(0.008) + 1((0.164) + 2(0.828) Ple) =091 Variance: 0% = xlzp(xl) + x22 p(x2) + x32p(x3) - = 0% (0.008) + 1% (0.164) + 2% (0.828) - =0 C(DF CONTINUED X is the number of samples out of 16 that are mutated Q:F(x) =0 x<I > F(1) 0.8 1<=x<4 ->F(2) 0.9 4<=x<7 ->F(3) 1 x>=7 > F(4) P(X=0) = B) What is the probability that at most one sample is mutated? A) P(X<=4) = F(3) = X Fb‘\) | \MlP(’") " yL1 (L) (x/'):\f-o.qo P(x') ,r\~0'0\')_ A) What is the probability that no samples are mutated? binompdf(16,0.03,0) = 0.61 CDF (DISCRETE) lvent ( Sk plx) = (27/13) D%, x = {1, 2, 3} . Pl A)F(1) = p(1) = 9/13 e 2 B) F(2) = p(1) + p(2) = 12/13 C) F(3) = p(1) + p(2) + p(3) = 1 would evaluate F(1) E) P(X<=3) =F3) =1 F) P(X>2) = 1-F(2)=1/13 0-02D X o= (or) P(X<2.5) - P(X<=1) = binomcdf(16,0.03,1) = 0.92 D) P(X<=1.5) = P(1<=X<=1.5) = F(1) added up until 1.5 and since the range B) P(1<X<2.5) = flz's Q: Evaluate the CDF, F, for the random - CDF (DIS(RETE) Q: Evaluate the CDF, F, for the random variable, X, at specified values. p(x) = (27/13) D%, x = {1, 2, 3} A)F(1)=p(1)=9/13 B) F(2) = p(1) + p(2) =12/13 Q) F3)=p(1) +p(2) +p(3) =1 D) P(X<=1.5) = P(1<=X<=1.5)=F(1)=9/13 s asking for all the probabllmes added up un'rll l 5 and since the range starts at ] wn'rh the next value bemg 2 and it is discrete, E) P(X<=3) = F(3) = 1 F) P(X>2) = 1-F(2) = 1/13 G)P(1<X<=2) = P(2<=X<=2) = P(X=2) = p(2) = 3/13 (OR) P(X<=2) - P(X<=1) = F(2) - F(1) = 3/13 G)P(1<X<=2) = P(2<=X<=2) = P(X=2) = p(2) = 3/13 (OR) P(X<=2) — P(X<=1) = F(2) = F(1) = 3/13 BERNOULLI TRIALS AND BINOMIAL PMF CONT. Q: Samples of rejuvenated mitochondria are mutated in 3% of cases. Suppose n = 16 and independent. 1) Suppose that f(x) = e ™™ for x>0 A) P(X>1) = 1-P(X<=1) = 1-f e *dx = 0.3679 e X*dx = 0.2858 P(X<1) C) What is th bability that half rated? C) P(X=3) = O because it is a continuous distribution B) P(X>7) = 1-P(X<=7) = 1-F(7) = 1-1 = 0 ) at is the probability that half are mutated? P Half = 16/2 = 8 D) P(X<4) = [ e *dx = 0.9817 C DF C) P(X<=5) = \ eqaHon P(X>8) = 1-P(X<=7) = 1-binomcdf(8,0.03,7) =1-1 =0 E) P(X<=4) = fo e *dx =0.9817 D) P(X>4) = 1-P(X<=4) = 1-F(3) = 1-0.9 = 0.1 E) P(X<=2) = F(2) = CONTINUQUS DISTRIBUTION CDF Suppose that X has CDF: Q: F(x)=0 x<-0.8 -> F(1) 0.25x+0.2 -0.8<=x<3.2 -> F(2) 1 x>=3.2 -> F(3) A) P(X<1.8) = F(2) = 0.25(1.8)+0.2 = 0.65 B) P(X>-1.5) = 1-P(X<-1.5) = 1-F(1) = 1-0 = 1 C) P(X<-2) = D) P(-1<X<1) = P(X<1)-P(X<-1) = (0.25(1)+0.2) = 0 = 0.45 1. You are not to post any question(s) from this exam to any online site, such as Chegg, Course Hero, Quizlet, Study Soup, or any other site. You are not to post any question(s) from this exam on a discord, Ed Discussion, Snapchat, WhatsApp, text message, or any other similar medium. A. I understand that if I do this, my exam score will be reduced to 0 and I will be reported to the Dean s Office for an Academic Integrity Violation. (There is one answer choice because there 1 wer Bubble in A on the Scantron sheet.) 2. If you want to graph the pdf of X, what will the axes labels be using Dr. C.’s notation? A and +o) A. x and p(x) yN one of these other answers is correct Copyright 2021 Arizona State University Exam R & i 3. Determine the value of (nr_ll) N \ —-’"fifi . ) (Y R A.0 n,\y\"‘ $ B.1 { : “n D.n+1 E.n-1 Copyright 2021 Arigona State University Exam 4. A standard normal random variable has a standard deviation equal to . Select the correct answer. A.0 £lzy-0 V()= B.1/4 c.11/2 \y = N E. Cannot be determined from the information given. 5. If random variable, X, consists of M possibl es (M is an integer > 1) that are equally likely, what is the name of the probability distribution of thls random variable? A. Bernoulli B. Binomial C. Poisson S Discrete Uniform E. Exponential Copyright 2021 Arizona State University Exam F) P(X>=3) = 1-P(X<3) = 1- f03 e *dx = 0.0498 EXPONENTIAL DISTRIBUTION Q: The time to failure is modeled by exponential distribution with lambda = 0.0004 A) What is the probability that the laser will last at least 20056 hours? P(X>=20056) — e—lambda(x) — e—0.00004(20056) = 0.448 (heotlet B) What is the probability that the laser will last at most 30073 hours? P(X<=30073) =1- e—lambda(x) =1- e—0.0004(30073) = 0.7 C) What is the probability that the laser will last between 20056 hours and 30073 hours? P(20056 < X < 3007 3) = P(X<30073)-P(X<20056) = 0.148 upper _ (or) [, ey lambda x e~amPA2X) gy > pdf 30073 —0.0004(x) X ' = sanee U0004 % e dx = 0.148 —— @ | RePost: How to know which distribution to use? a# 4 Linda Chattin A common difficulty with probability is determining what pmf or pdf to use in a2 word problem. Here are some Llips: Discrele Bernoulli Trial «- when the wording indicates a situation of two oulcomes. one with probability "p® and the other "1 - Binomial «- when the wording indicates a count of "successes” in n Bernoulli Lrials Discrete Uniform «- when the wording indicates a finile sel ol values, each equally likely Poisson «- you have Lo be lold in the problem that the Poisson pml is applicable Conlinuous Conlinuous U llll()l'lll <= when the wording of the problem says the probability is conslant on an interval [a. b] or (a, h) Normal and Standard Normal «- vou have to be told in the problem that Normal or Standard Normal pdf is applicable I \pom ntial «- you have o be told in the problem that the E xponential pdfl is applicable cement is closed for comments

6. To get the pdf from a CDF for a continuous random variable, we need to do 9. The probability that the volume of fluid produced by this chemical process is more than 4 mlis ____. Select the A. an integration ,(:(" N - _i ¥ (,1) best answer and/or the answer closest to yours. B. an additon d* C. a multiplication A.05 72 differentiation B. 0.4375 E. an infinite sum M) 5625 Copyright 2021 Arizona State University ixam E None of these other answers is correct Questions 7 and 8 are based on the following: A commercially produced table is inspected to see if it has flaws or iYL ans ik URiiad. e t. Th bability th: h. ne o e L ‘“Eﬂsse e ais i 10. (3 pts) X is the number of defects on a small mirror that is produced for automobiles. Historical data show that F n X has the pmf below. What is the numerical value of k that makes X a valid probability mass function 7. If you wanted to determine the probability that fewer than seven of the tables have flaws, what would the best - k =0 definition of the random variable X be in words? PO = *- () l"\ A = 2k x=1 A. X 1s the probability that fewer than seven of the tables out of nine have flaws - ok 2 k 4 al L4 Y‘( =1 B. X is a Bernoulli trial with a probability equal to 12.03% - 0.125 X is the number of tables out of nine that have flaws n= a {O.ZSO g'fi = ’ D. X is a Poisson random variable with A = 0.1203 L0 : - . C. 0.750 < E. X has the normal pdf with a mean equal to 0.1203. P - O \ B D.1 K: )3 E.8 Copyright 2021 Arizona State University Exam Copyright 2021 Arizona State University Exam 8. The standard deviation of the number of tables with flaws is . Select the answer closest to yours and/or the best answer. 11. According to the text, wh1ch probabl.l.\ty distribution is often used as a model of the number of events in an r interval and has a countabl = < - \V A. 0.9525 tables? ny (-p) L A. Bernoulli B. 0.9759 flaws? B. Binomial C. 0.9525 flaws m.9759 tables . ' }Ddla)lscrete Uniform E. Cannot be det d from t f ion given E. Exponential Copyright 2021 Arizona State University Exam Copyright 2021 Arizona State University Exam Questions 9 and 10 are based on the following: The volume of fluid produced in a chemical process, X, has the 12. X is the number of cars that pass through an mtersecuon and Xis known to have the Poisson distribution. The following CDF: mean number of cars that pass through tt period is equal to 37.80 cars. What is E(X) = for x < 3 ml for a 2.5-minute period? _ x%-9 Fz) = 16 m”ﬂ/-i—x <5ml /(64.5 cars = i oy 0 =1 for x > 5 ml. B. 15.12 minutes o= 33 C. 44.3 intersections - D. 15.12 cars £ ., =7 iy E. None of these other answers is correct Q‘S % 9-80%2°3 5 A 3 >H | Loy A th Copyright 2021 Ari State University Exam l‘, JT\L to complete, start to finish. X is known to have an gxponential pdf. E(X) = 3 minutes. VS =7 — 13. The probability that it takes this reaction ew\nes to compTeLe is . Nt vas Student 12:59PM Thu Feb 8 TCBAO5% % = Zwen 4 Pl m*“* . Formulae X |IEE380+Exam+1+R... +380+Practi. I Cheat sheet B. 0.2636 ) T =uUmrn >\ = '/‘gmm C.0.4724 N=0 D. 0.8428 w‘ E. 1 Copyright 2021 Arizona State University Exam 14. The range of X is __. Select the best answer. A. (-0, +0) Questions 13, 14, and 15 are based on the following: Random variable X is the time it takes for a chemical reaction h ) B. (-0, 0) o complete, start to finish. X is known to have an exponential pdf. I:(X) 3 minutes. e L0, +0) 13. The probability that it takes this reaction e):fi]y_ﬂ#utes to comp ete is 2 Nt D.0,1,2,3, ... % = Bmin [ E0123 .un T S et B. 0.2636 / T =Umr )\ = '/ 3”‘0 Copyright 2021 Arizona State University Exam C.0.4724 o »=9° . B . _ ' . D. 0.8428 15. What is the probability that this cl 1 I han 2 minutes to complete? Select the answer closest to yours. A\ Copyright 2021 Arizona State University Exam P(x> a)= 1-plx£2) (- e 14. The range of Xis ______ . Select the best answer. B. 0.4866 I~ F(2'> _3fy V7 C.05134 - (1-€ ) A. (-0, +a0) D. 0.7769 B. (-0, 0) _e0, +w) Copyright 2021 Arizona State University Exam D.0,1;2,3; .:: 16. The probability that a student earns a C or better in Chetrustry is equa] to O 70. The probability that this student earns a C or better in Biology is cqual to 0 75: C or better in both Copyright 2021 Arizona State University Exam Chenust:y and Bnology is 0. 82 et A be h C or better in Chemistry. Let B be the C or better in Biology. Select the TRUE statement. 15. What is the probability that this chemical reaction takes more than 2 minutes to complete? Select the answer ’N’ closest to yours. 2 \ ) A. A and B are independent ?((‘)) =070 ¢ p(x>a)= 1-plx£2) it B.P(A UB) = 0.82 -~ <25 . |- F‘J-) \ - L " A and B are not independent L( N -0 * 0 B. 0.4866 _33 D.PAUB) =145 ? — C.0.5134 |- l |- ) E. None of these other choices is TRUE D. 0.7769 ye 79 F FIAn® Copyright 2021 Arizona State University Exam 16. The probability that a student earns a C or better in Chemistry is equal to 0.70. The probability that this student earns 2 C or better in Blolog)v is equal to O 75. The ptobablhty that this student earns a C or better in both Chen-ustry and Blolog) is 0.82. Let A be C or better in Chemistry. Let B be the Questions C or better in Biology. Select the TRUE statement. 17,18, and 19 are based on the following: Customers are used to evaluate preliminary product designs. In the past, 94% of highly successful products received good reviews, 60% of moderately successful products received y {am ood reviews, and 11% of poor products received good reviews. In addition, 40% of products have been highl A. A and B are independ =070 - ¢ & B. P(;nu B)are idependeat 9(()) - < successful, 35% have been moderately successful and 25% have been poor products. 8o 52 Aand B depend - 0% ° . ng(i{‘ ;re r1lot B 4 L5 e 17. What is the probability that a product is highly successful? Select the best answer. E. None of these other choices is TRUE yo) o F FIAn® A.0.06 E. None of these other choices is correct Copyright 2021 Arizona State University Exam Copyright 2021 Ari State University Exam 18. What is the probability that a randomly selected product gets a good review? 76 ] Ao4127 P(q) = Plajn)ptn) + P(GIm)Pi) + 7o) PO .6135 C. 0.5557 D. 1.6500 E. 0.1650 Copyright 2021 Arizona State University Exam 19. If we randomly select a product with a good review, what is the probability that the product is moderately e plmje) = PlalmPm) . By C.0.3423 D. 0.6723 E. 0.8650 Copyright 2021 Arizona State University Exam 20. The weight of a package of frozen peas has the normal distribution with a mean equal to 16.0 oz. and a variance equal to 2 oz”. If we randomly select a package of these peas, the probability that it weighs 15.2 oz or less is Select the answer closest to yours. 0 wean = 16-02 ?U _us-oL) = N, B.0.8924 A C. 03446 NovianC 0.2858 E. 0.0236

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version