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Stat 2470, Practice Exam #1, Spring 2014 Instructions: You may use your calculator for any functions the TI-84/83 model calculator is capable of using, such as probability distributions and obtaining graphs of any data. To show work on these problems, report the functions and their syntax as entered. Other things, such as integrating, must be done by hand unless specifically directed otherwise. Round means to one more place than the original data, variances and standard deviations to two more than the original data. Round probabilities to three significant figures or use exact values. In order to receive partial credit on any problem, you must show some work or I will have nothing to award partial credit on. Be sure to complete all the requested parts of each problem. 1. A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the accompanying data on time (sec) to complete the escape: 373 356 392 370 359 369 364 363 374 366 375 359 364 424 356 325 325 403 339 394 334 393 402 397 a. Construct a stem-and-leaf display of the data. How does it suggest that the sample mean and median will compare? b. Calculate the values of the sample mean and median. c. By how much could the largest time, currently 424, be increased without affecting the value of the sample median? By how much could this value be decreased without affecting the value of the sample mean? d. e. What are the values of and when the observations are re-expressed in minutes? Construct a boxplot of the data. Are there any outliers? 2. Calculate and interpret the values of the sample mean and sample standard deviation for the following observations on fracture strength. 128 131 142 168 87 93 105 114 96 98 3. At a certain gas station, 40% of the customers use regular unleaded gas ( ), 35% use extra unleaded gas ( ), and 25% use premium unleaded gas ( ). Of those customers using regular gas, only 30% fill their tanks (event B). Of those customers using extra gas, 60% fill their tanks, whereas of those using premium, 50% fill their tanks. a. What is the probability that the next customer will request extra unleaded gas and fill the tank? b. What is the probability that the next customer fills the tank? c. If the next customer fills the tank, what is the probability that regular gas is requested? Extra gas? Premium gas? 4. Two pumps connected in parallel fail independently of one another on any given day. The probability that only the older pump will fail is .15, and the probability that only the newer pump will fail is .05. What is the probability that the pumping system will fail on any given day (which happens if both pumps fail)? 5. A real estate agent is showing homes to a prospective buyer. There are ten homes in the desired price range listed in the area. The buyer has time to visit only four of them. a. In how many ways could the four homes be chosen if the order of visiting is considered? b. In how many ways could the four homes be chosen if the order is disregarded? c. If four of the homes are new and six have previously been occupied and if the four homes to visit are randomly chosen, what is the probability that all four are new? (The same answer results regardless of whether order is considered.) 6. Suppose that vehicles taking a particular freeway exit can turn right (R), turn left (L), or go straight (S). Consider observing the direction for each of three successive vehicles. a. List all outcomes in the event A that all three vehicles go in the same direction. b. List all outcomes in the event B that all three vehicles take different directions. c. List all outcomes in the event C that exactly two of the three vehicles turn right. d. List all outcomes in the even D that exactly tow vehicles go in the same direction. e. List outcomes in ,C D, and C D. 7. A geologist has collected 10 specimens of basaltic rock and 10 specimens of granite. The geologist instructs a laboratory assistant to randomly select 15 of the specimens for analysis. a. What is the pmf (pdf) of the number of granite specimens selected for analysis? b. What is the probability that all specimens of one of the two types of rock are selected for analysis? c. What is the probability that the number of granite specimens selected for analysis is within 1 standard deviation of its mean value? 8. Use the cumulative binomial probabilities table available in your text to obtain the following probabilities: a. B(4; 10, .4) b. b(4; 10, .4) c. b(6; 10, .6) d. when Bin(10,.4) e. when Bin(10,.3) f. when Bin(10,.7) g. when Bin(10,.3) 9. Automobiles arrive at a vehicle equipment inspection station according to a Poisson process with rate = 10 per hour. Suppose that with probability .5 an arriving vehicle will have no equipment violations. a. What is the probability that exactly ten arrive during the hour and all ten have no violations? b. For any fixed what is the probability that y arrive during the hour, of which ten have no violations? c. What is the probability that ten “no-violation” cars arrive during the next hour? [Hint: Sum the probabilities in part (b) from y = 10 to ] 10. Let X denote the amount of space occupied by an article placed in a The pdf of X is packing container. a. Obtain the cdf of X. b. What is c. What is d. What is the 75th percentile of the distribution? e. Compute f. What is the probability that X is within 1 standard deviation of its mean value? ? 11. In each case, determine the value of the constant c that makes the probability statement correct. a. b. c. d. e. 12. Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime X (in weeks) has a gamma distribution with mean of 40 and variance of 320. a. What is the probability that a transistor will last between 1 and 40 weeks? b. What is the probability that a transistor will last at most 40 weeks? Is the median of the lifetime distribution less than 40? Why or why not? Answers: 1. a. A. stem-and-leaf display of this data appears below: 32 33 34 35 36 37 38 39 40 41 42 55 49 stem: ones leaf: tenths 6699 34469 03345 9 2347 23 4 The display is reasonably symmetric, so the mean and median will be close. b. The sample mean is = 9638/26 = 370.7. The sample median is = (369+370)/2=369.50. c. The largest value (currently 424) could be increased by any amount. Doing so will not change the fact that the middle two observations are 369 and 170, and hence, the median will not change. However, the value x = 424 can not be changed to a number less than 370 ( a change of 424-370 = 54) since that will lower the values(s) of the two middle observations. d. Expressed in minutes, the mean is (370.7 sec)/(60 sec) = 6.18 min; the median 6.16 min. 2. The sample mean, The sample standard deviation, On average, we would expect a fracture strength of 116.2. In general, the size of a typical deviation from the sample mean (116.2) is about 25.75. Some observations may deviate from 116.2 by more than this and some by less. 3. Therefore, a. P = .21 b. P(B) = c. P( 4. Let = older pump fails, x, and x = P( ) = P( = newer pump fails, and x = P( ). The P( ) = .15 + x, P( The resulting quadratic equation, ) = .05 + has roots x = .0095 and x = .7905. Hopefully the smaller root is the actual probability of system failure. 5. a. b. (10)(9)(8)(7) = 5040 = c. P(all 4 are new) = (# of ways of visiting all new)/(# of ways of visiting) = = 6. a. Event A = {RRR, LLL, SSS} b. Event B = {RLS, RSL, LRS, LSR, SRL, SLR} c. Event C = {RRL, RRS, RLR, RSE, LRR, SRR} d. Event D = {RRL, RRS, RLR, RSE, LRR, SRR, LLR, LLS, LRL, LSL, RLL, SLL, SSR, SSL, SRS, SLS, RSS, LSS} e. Event contains outcomes where all cars go the same direction, or they all go different directions: = {RRR, LLL, SSS, RLS, RSL, LRS, LSR, SRL, SLR} Because Event D totally encloses Event C, the compound event C D = D. Therefore, C D = {RRL, RRS, RLR, RSR, LRR, SRR, LLR, LLS, LRL, LSL, RLL, SLL, SSR, SSL, SRS, SLS, RSS, LSS} Using similar reasoning, we see that the compound event C D = C. Therefore, C D = {RRL, RRS, RLR, RSR, LRR, SRR} to be more than 10 of the basaltic). 7. Following the same pattern for the other values, we arrive at the pmf, in table form below. x P(x) 5 .0163 6 .1354 7 .3483 8 .3483 9 .1354 10 .0163 b. P(all 10 of one kind or the other) = P(X = 5) + P(X = 10) = .0163 + .0163 + .0326 c. E(X) = n . ( M / N) = 15 (10 / 20) = 7.5; V(X) = (5/19) (7.5) [1 – (10/20)] = .9869; = .9934 7.5 .9934 = (6.5066, 8.4934), so we want P( X = 7) + P( X = 8) = .3483 + .3483 = .6966 8. a. B(4;10,.4) = .633 b. b(4;10,.4) = B(4;10,.4) – B(3;10,.4) = .251 c. b(6;10,.6) = B(6;10,.6) – B(5;10,.6) = .251 d. e. f. g. 9. a. = (.000977)(.125) =.000122 b. P(y arrive and exactly 10 have no violations) = P(exactly 10 have no violations / y arrive). P(y arrive) c. P(exactly 10 without a violation) = In fact, generalizing this argument shows that the number of “no-violation” arrivals within the hour has a Poisson distribution with parameter 5; the 5 results from 10. a. Therefore, b. c. d. The 75th percentile is the value of x for which F(x) = .75 e. f. =.8465 - .1602 = .6863 11. a. b. c. d. e. 12. a. b. so while the mean is 40, the median is less than 40. This is the result of the positive skew of the gamma distribution.