Discrete Random Variable Practice

Transcription

1 IB Math High Level Year Discrete Probability Distributions - MarkScheme Discrete Random Variable Practice. A biased die with four faces is used in a game. A player pays 0 counters to roll the die. The table below shows the possible scores on the die, the probability of each score and the number of counters the player receives in return for each score. Score Probability Number of counters player receives 5 5 n Find the value of n in order for the player to get an expected return of 9 counters per roll. Answer:.. (Total marks). The quality control department of a company making computer chips knows that % of the chips are defective. Use the normal approximation to the binomial probability distribution, with a continuity correction, to find the probability that, in a batch containing 000 chips, between 0 and 0 chips (inclusive) are defective. (Total 7 marks). A supplier of copper wire looks for flaws before dispatching it to customers. It is known that the number of flaws follow a Poisson probability distribution with a mean of. flaws per metre. (a) Determine the probability that there are exactly flaws in metre of the wire. () (b) Determine the probability that there is at least one flaw in metres of the wire. () (Total 6 marks). In a game a player rolls a biased tetrahedral (four-faced) die. The probability of each possible score is shown below. Score Probability x Find the probability of a total score of six after two rolls. Answer:... (Total marks) C:\Users\Bob\Documents\Dropbox\Desert\HL\6StatProb\Practice\HL.DiscreteRandomVarPractice.docx on 0/6/06 at : PM Page of 5

2 IB Math High Level Year Discrete Probability Distributions - MarkScheme 5. The probability distribution of a discrete random variable X is given by x P(X = x) = k, for x = 0,,,... Find the value of k. Answer:... (Total marks) 6. A satellite relies on solar cells for its power and will operate provided that at least one of the cells is working. Cells fail independently of each other, and the probability that an individual cell fails within one year is 0.8. (a) For a satellite with ten solar cells, find the probability that all ten cells fail within one year. (b) For a satellite with ten solar cells, find the probability that the satellite is still operating at the end of one year. (c) For a satellite with n solar cells, write down the probability that the satellite is still operating at the end of one year. Hence, find the smallest number of solar cells required so that the probability of the satellite still operating at the end of one year is at least (5) (Total 9 marks) 7. In a school, of the students travel to school by bus. Five students are chosen at random. Find the probability that exactly of them travel to school by bus. Answer:.. () () (Total marks) 8. X is a binomial random variable, where the number of trials is 5 and the probability of success of each trial is p. Find the values of p if P(X = ) = 0.. Answer:... (Total marks) C:\Users\Bob\Documents\Dropbox\Desert\HL\6StatProb\Practice\HL.DiscreteRandomVarPractice.docx on 0/6/06 at : PM Page of 5

3 IB Math High Level Year Discrete Probability Distributions - MarkScheme 9. (a) Patients arrive at random at an emergency room in a hospital at the rate of 5 per hour throughout the day. Find the probability that 6 patients will arrive at the emergency room between 08:00 and 08:5. (b) The emergency room switchboard has two operators. One operator answers calls for doctors and the other deals with enquiries about patients. The first operator fails to answer % of her calls and the second operator fails to answer % of his calls. On a typical day, the first and second telephone operators receive 0 and 0 calls respectively during an afternoon session. Using the Poisson distribution find the probability that, between them, the two operators fail to answer two or more calls during an afternoon session. (5) (Total 8 marks) 0. A coin is biased so that when it is tossed the probability of obtaining heads is. The coin is tossed 800 times. Let X be the number of heads obtained. Find (a) the mean of X; (b) the standard deviation of X. () Answers: (a)... (b)... (Total marks). When John throws a stone at a target, the probability that he hits the target is 0.. He throws a stone 6 times. (a) Find the probability that he hits the target exactly times. (b) Find the probability that he hits the target for the first time on his third throw. Answers: (a)... (b)... (Total 6 marks) C:\Users\Bob\Documents\Dropbox\Desert\HL\6StatProb\Practice\HL.DiscreteRandomVarPractice.docx on 0/6/06 at : PM Page of 5

4 IB Math High Level Year Discrete Probability Distributions - MarkScheme. Two children, Alan and Belle, each throw two fair cubical dice simultaneously. The score for each child is the sum of the two numbers shown on their respective dice. (a) (i) Calculate the probability that Alan obtains a score of 9. (ii) Calculate the probability that Alan and Belle both obtain a score of 9. (b) (i) Calculate the probability that Alan and Belle obtain the same score, (ii) Deduce the probability that Alan s score exceeds Belle s score. (c) Let X denote the largest number shown on the four dice. (i) x Show that for P(X x) =, for x =,, (ii) Copy and complete the following probability distribution table. x 5 6 P(X = x) (iii) Calculate E(X). (7) (Total marks). The random variable X is Poisson distributed with mean µ and satisfies P(X = ) = P(X = 0) + P(X = ). (a) Find the value of µ, correct to four decimal places. () (b) For this value of µ evaluate P( X ). () (Total 6 marks). Give your answers to four significant figures. A machine produces cloth with some minor faults. The number of faults per metre is a random variable following a Poisson distribution with a mean. Calculate the probability that a metre of the cloth contains five or more faults. (Total marks) () () 5. When a boy plays a game at a fair, the probability that he wins a prize is 0.5. He plays the game 0 times. Let X denote the total number of prizes that he wins. Assuming that the games are independent, find (a) E(X) (b) P(X ). Answers: (a)... (b)... (Total 6 marks) C:\Users\Bob\Documents\Dropbox\Desert\HL\6StatProb\Practice\HL.DiscreteRandomVarPractice.docx on 0/6/06 at : PM Page of 5

5 IB Math High Level Year Discrete Probability Distributions - MarkScheme 6. Give all numerical answers to this question correct to three significant figures. Two typists were given a series of tests to complete. On average, Mr Brown made.7 mistakes per test while Mr Smith made.5 mistakes per test. Assume that the number of mistakes made by any typist follows a Poisson distribution. (a) Calculate the probability that, in a particular test, (i) Mr Brown made two mistakes; (ii) Mr Smith made three mistakes; (iii) Mr Brown made two mistakes and Mr Smith made three mistakes. (b) In another test, Mr Brown and Mr Smith made a combined total of five mistakes. Calculate the probability that Mr Brown made fewer mistakes than Mr Smith. (5) (Total marks) 7. On a television channel the news is shown at the same time each day. The probability that Alice watches the news on a given day is 0.. Calculate the probability that on five consecutive days, she watches the news on at most three days. (6) Answer:... (Total 6 marks) 8. The random variable X has a Poisson distribution with mean λ. (a) Given that P(X = ) = P(X = ) + P(X = ), find the value of λ. () (b) Given that λ =., find the value of (i) P(X ); (ii) P(X X ). (5) (Total 8 marks) 9. The random variable X has a Poisson distribution with mean λ. Let p be the probability that X takes the value or. (a) Write down an expression for p. () (b) Sketch the graph of p for 0 λ. () (c) Find the exact value of λ for which p is a maximum. 0. Let X be a random variable with a Poisson distribution such that Var(X) = (E(X)) 6. (a) Show that the mean of the distribution is. (b) Find P(X ). Let Y be another random variable, independent of X, with a Poisson distribution such that E(Y) =. (c) Find P(X + Y < ). (d) Let U = X + Y. (i) Find the mean and variance of U. (ii) State with a reason whether or not U has a Poisson distribution. (5) (Total 7 marks) () () () () (Total 0 marks) C:\Users\Bob\Documents\Dropbox\Desert\HL\6StatProb\Practice\HL.DiscreteRandomVarPractice.docx on 0/6/06 at : PM Page 5 of 5

6 IB Math High Level Year Discrete Probability Distributions - MarkScheme Discrete Random Variable Practice - Markscheme. Let X be the number of counters the player receives in return. E(X) = p(x) x = n = (A) n = 0 n = 0 (A) (C). Let n = number of chips = 000, p = the probability that a randomly chosen chip is defective = 0.0. Hence, the mean np = (000) (0.0) = 0 and the variance = np( p) = (000) (0.0) (0.98) = 9.6. Suppose X is the normal random variable that approximates the binomial distribution. The X N (0, 9.6) Thus p(9.5 X 0.5) = p Z (A)(A) = p( 0. Z.7) = 0.59 (A) Note: Line before last should be p( 0. Z.7) = Accept 0.55 or If student s work is not shown but there is evidence that he/she used the calculator to find the answer, accept the answer.. Note: Throughout the whole question, students may be using their graphic display calculators and should not be penalized if they do not show as much work as the marking scheme. (a) Let X denote the number of flaws in one metre of the wire. Then E(X) =. flaws and P(X = ) = e. (.)! = (A) Note: Award (C) for a correct answer from a graphic display calculator. [] [7]. (b) Let Y denote the number of flaws in two metres of wire. Then Y has a Poisson distribution with mean E(Y) =. =.6 flaws for metres. Hence, P(Y ) = P(Y = 0) = e.6 = ( sf) (A) Note: Accept e.6. P X( = x) = a ll x Therefore, x = Therefore, x = (A) 0 P(scoring six after two rolls) = [6] C:\Users\Bob\Documents\Dropbox\Desert\HL\6StatProb\Practice\HL.DiscreteRandomVarPractice.docx on 0/6/06 at : PM Page of 7

7 IB Math High Level Year Discrete Probability Distributions - MarkScheme = (A) (C) 5. Since X is a random variable, P X( = x) = Therefore, k + k + k + k +... = k = k = (A) (C) 6. (a) P(all ten cells fail) = = (A) (b) P(satellite is still operating at the end of one year) = P (all ten cells fail within one year) = 0.07 = (A) (c) P(satellite is still operating at the end of one year) = 0.8 n. (C) We require the smallest n for which 0.8 n Thus, 0.8 n 0.05 n 5 0 log0 n =. log.5 (A) Therefore, solar cells are needed. (C) 5 7. Let p be the probability of choosing a student who travels to school by bus. Let X be the random variable the number of students who travel to school by bus. Then X ~ B(n, p) with n = 5 and p = 5 Therefore P(X = ) = (using formulae and statistical tables) (A) 0 = or 0.65 (A) a ll x [] [] [9] [] 8. If X ~ Bin (5, p) and P(X = ) = 0. then 5 p ( p) = 0. 5p 5 5p + 0. = 0 (A) p = 0.59 ( s.f ) or 0.97 ( s.f) (G) (C) [] C:\Users\Bob\Documents\Dropbox\Desert\HL\6StatProb\Practice\HL.DiscreteRandomVarPractice.docx on 0/6/06 at : PM Page of 7

8 IB Math High Level Year Discrete Probability Distributions - MarkScheme 9. (a) Let X be the number of patients arriving at the emergency room in 5 a 5 minute period. Rate of arrival in a 5 minute period = = (.75) P(X = 6) = e.75 6! = (A) P(6 patients) = (G) (b) Let F, F be random variables which represent the number of failures to answer telephone calls by the first and the second operator, respectively. F ~ P 0 (0.0 0) = P 0 (0.). (A) F ~ P 0 (0.0 0) = P 0 (.). (A) Since F and F are independent F + F ~ P 0 (0. +.) = P 0 (.) P(F + F ) = P(F + F = 0) P(F + F = ) = e. (.)e. = 0.08 (A) P(F + F ) = 0.08 (M0)(G) 5 0. n = 800, p = (a) E(X) = np = 00 (A) (C) (b) SD(X) = n ( p p) = 0= 0 (A)(C) [8] []. (a) 6 Probability = (0.) (0.6) (A) = 0.8 a c c e o 0 p r. t 5 (A)(C) (b) Probability = (0.6) 8 0. = 0. or 5 (A) (C). (a) (i) P(Alan scores 9) = (= 0.) 9 (A) (ii) P(Alan scores 9 and Belle scores 9) = = 9 8 (= 0.0) (A) [6] 6 (b) (i) P(Same score) = C:\Users\Bob\Documents\Dropbox\Desert\HL\6StatProb\Practice\HL.DiscreteRandomVarPractice.docx on 0/6/06 at : PM Page of 7

9 IB Math High Level Year Discrete Probability Distributions - MarkScheme 7 = (= 0.) (A) 68 (ii) 7 P(A>B) = = 96 (= 0.) (A) (c) (i) x P(One number x) = (with some explanation) 6 (R) x P(X x) = P(All four numbers x) = 6 (AG) (ii) x P(X = x) = P(X x) P(X x l) = 6 x 6 x 5 6 P(X = x) (A)(A)(A) Note: Award (A) if table is not completed but calculation of E(X) in part (iii) is correct (iii) E(X) = = (= 5.) (A) (a) The given condition implies that µ µ µ µ e = e + µ e 6 µ 6µ 6 = 0 (A) µ.87 (G) (b) P( X ) = P(X = ) + P(X = ) + P(X = ) µ e µ e P(X = ) = = 0.5, P(X = ) = = 0., 6 µ e P(X = ) = = 0.59 (G) Hence P( X ) = 0.67 (A) P( X ) = P(X ) P(X ) = (G) = 0.67 (G). Note: Accept answers to an accuracy of at least significant figures do not apply AP. P(X = 0) = , P(X = ) = 0.96, P(X = ) = 0.0, P(X = ) = 0.0, P(X = ) = [] [6] C:\Users\Bob\Documents\Dropbox\Desert\HL\6StatProb\Practice\HL.DiscreteRandomVarPractice.docx on 0/6/06 at : PM Page of 7

10 IB Math High Level Year Discrete Probability Distributions - MarkScheme Sum = P(X ) = 0.85 (accept 0.85) Hence, P(X 5) = l 0.85 = 0.87 (accept 0.88) P(X 5) = P(X ) = 0.87 (A) (G) (A) 5. (a) X is B(0, 0.5) (seen or implied) (R) so E(X) = =.5 (A)(C) (b) P(X ) = (0.75) (0.75) 9 0 (0.5) + (0.75) 8 (0.5) (A) = 0.56 (A) (C) 6. B ~ P(.7), S ~ P(.5) ( ) (G) (a) (i).7 e.7 P(B = ) = = 0.5 (A) (ii).5 e (.5) P(S = ) = 6 = 0. (A) (iii) The two events are independent. P((B = ) (S = )) = P(B = ) P(S = ) = = 0.05 (A) 6 (b) 5. 5 e (5.) P(B + S = 5) = (A) e.5 (.7) e.5 P((B = ) (S = )) = = e (.7) e (.5) P((B = ) (S = )) = = 0.0 (A) e 0.5 (.7) 5 e.5 P((B = 0) (S = 5)) = = (A) P(B < S B + S =5) = = = 0.6 (or 0.6) (A) 5 7. METHOD X is Binomial n = 5 p = 0. (A)(A) P(X ) = P(X = ) P(X = 5) = (A)(A) = (0.9 to sf) (A) (C6) [] [6] [] METHOD P(X ) = P(X = 0) + P(X = ) + P(X = ) + P(X = ) = (A) = (0.9 to sf) (A) (C6) [6] C:\Users\Bob\Documents\Dropbox\Desert\HL\6StatProb\Practice\HL.DiscreteRandomVarPractice.docx on 0/6/06 at : PM Page 5 of 7

11 IB Math High Level Year Discrete Probability Distributions - MarkScheme λ λ λ 8. (a) e λ e λ e λ = +!!! λ λ = 0 λ = 6 (A)(A) (b) (i) P(X ) = e. e.. = 0.89 (A) P(X ) = 0.89 (G) (ii) P ( X ) P(X X ) = P( X ).. e. e. + = 6 (A)..e = 0.50 (A) P(X X ) = 0.50 (G) 5 9. (a) p = λe λ λ + e λ (A) [8] (b) φ 0 λ (A) Note: Award (A) for a maximum point in [0, ]; sketch need not be accurate. (c) = e λ + λ( e λ ) + e λ dp λ λ + ( e ) dλ (A) = e λ λ (A) dp λ p max when = 0 = 0 dλ λ = λ = (do not accept.) (A) 5 Note: If no working shown, award (A) for an answer of. obviously obtained from a GDC. [7] C:\Users\Bob\Documents\Dropbox\Desert\HL\6StatProb\Practice\HL.DiscreteRandomVarPractice.docx on 0/6/06 at : PM Page 6 of 7

12 IB Math High Level Year Discrete Probability Distributions - MarkScheme 0. Note: In this question do not penalize answers given to more than three significant figures. (a) Let λ = E(X) = Var(X). Then λ = λ 6 ± 5 Therefore λ = and since λ must be positive λ =. (A) (b) Then P(X ) = 0.67 (A) (N) (c) E(X + Y) = + = 5. Since X and Y are independent X + Y has a Poisson distribution with mean = 5. Hence P(X + Y < ) = (A) (N) Note: Award (N0) if P (X + Y ) is given instead. (d) (i) E(U) = E(X) + E(Y) = 7 (A) Var(U) = Var(X) + Var(Y) = (A) (ii) U does not have a Poisson distribution (A) because Var(U) E(U). (R) [0] C:\Users\Bob\Documents\Dropbox\Desert\HL\6StatProb\Practice\HL.DiscreteRandomVarPractice.docx on 0/6/06 at : PM Page 7 of 7