ࡱ>  @ bjbj5*5* W@W@<64  | V :Y:Y:Y:Y\YV v[([[[`]_6__ $R5| 9dbo_"_efx\][mmmf8\]m9dmmR؅]j[ 8iuJ:Yh̆4 ͔0j< t J $_V=am1bbD___V V M:YmV V :YCOB 191 Statistical Methods Sample Exam 2 (S. Stevens) DO NOT TURN TO THE NEXT PAGE UNTIL YOU ARE INSTRUCTED TO DO SO! The following exam consists of 25 questions, each worth 4 points. You will have 50 minutes to complete the test. This means that you have, on average, about 2 minutes per question. Record your answer to each question on the scantron sheet provided. You are welcome to write on this exam, but your scantron will record your graded answer. Read carefully, and check your answers. Dont let yourself write nonsense. Keep your eyes on your own paper. If you believe that someone sitting near you is cheating, raise your hand and quietly inform me of this. I'll keep an eye peeled, and your anonymity will be respected. If any question seems unclear or ambiguous to you, raise your hand, and I will attempt to clarify it. Be sure your correctly record your student number on your scantron, and blacken in the corresponding digits. Failure to do so will cost you 10 points on this exam! Pledge: On my honor as a JMU student, I pledge that I have neither given nor received unauthorized assistance on this examination. Signature ______________________________________ Excel Reminders: = BINOMDIST ( successes, trials, prob of success, cumulative) = POISSON ( x value, mean, cumulative) = NORMSDIST( z value) = NORMSINV ( probability) = NORMDIST ( x value, mu, sigma, TRUE) = NORMINV (probability, mu, sigma) Put your cursor over the yellow blocks to see answers/comments! Questions 1-17 deal with the scenario below. Beautiful U, Inc., offers a one month weight loss program. According to their flyers, a weight student (a person enrolled in the Beautiful U program) is almost certain to lose weight over the month. The small print on their brochures clarifies this claim: A student enrolled for the full month in the Beautiful U program is at least 80% likely to lose weight. You work for a government agency responsible for identifying possible advertising fraud, and you have decided to look into the Beautiful U claims. Youve pulled the records of 40 randomly selected Beautiful U graduates (people who have completed the one month program), and compared their weights at the time that they entered the program to their weights one month later. The data on these 40 graduates is presented on the last page of this test. For each graduate, you are shown the initial weight, final weight (after 1 month in the program), and amount of weight lost during the month. All of these figures are in pounds, measured to the nearest 0.5 pounds. Note that some students actually gained weight over the month; these students have a negative weight loss. For your convenience, there is also a column recording whether a student lost weight (0 = no weight lost, 1 = some weight lost). The pulled records reveal that only 24 on the 40 graduates achieved any weight loss during the month. That means that only 60% of the records pulled demonstrate any weight loss. The question you must now consider is: is this evidence strong enough to justify an accusation of false advertising by Beautiful U? Your approach will be as follows. Imagine a hypothetical weight loss program for which the Beautiful U claim is true, but just barely. This is, in the hypothetical program, each student has an 80% chance, exactly, of losing weight. Your question then is: THE QUESTION: If every student in a weight loss program has an 80% chance of losing weight, how likely is it that 60% or less of a 40 student sample will succeed in losing weight? Questions 1-8 all deal with THE QUESTION described in the approach above. They also assume that a student is a success if and only if he or she loses weight during their month as a Beautiful U student. THE QUESTION involves the sample proportion, ps. In this problem, ps represents the fraction of the Beautiful U graduates that were surveyed. the fraction of the surveyed graduates who lost weight. the fraction of all of Beautiful U s graduates who lost weight. the fraction of unsurveyed graduates who lost weight. the number of graduates in the sample who lost weight. The population in THE QUESTION can be summarized by the values N = 40,  = 0.8 N = 40,  = 0.6 N = 24,  = 0.8 N = 24,  = 0.6 N unknown,  = 0.8 We will not find the exact answer to THE QUESTION. Instead, we will find an approximate answer by recognizing that the sampling distribution for this problem is approximately normal. Questions 3-9 all assume that we are making this normal approximation. We may approximate the sampling distribution of the proportion with a normal distribution only if certain conditions are met. In THE QUESTION, they are. Which of the following three statements indicate necessary checks that we must make in this problem? With a success rate of 80%, wed expect 32 people to lose weight; thats at least 5. With a success rate of 80%, wed expect 8 people to not lose weight; thats at least 5. With a sample of size 40, we need that the population itself is symmetrically distributed. It is. a) I only b) II only c) III only d) I and II only e) I, II and III In answering THE QUESTION, we must decide whether to use the finite population modifier (FPM). Which of the following statements best summarizes that decision process? We will not use the FPM, because we dont know the size of the population. We will not use the FPM, because it would increase the size of the standard error. We will use the FPM because the FPM must be used on all proportion problems. We will use the FPM because only a finite number of people have graduated from the weight loss program. We will use the FPM because the sample size is 30 or more. Consider again THE QUESTION. In symbols, its answer would be represented as a) P(( < 24) b) P(p < 0.6) c) P( = 0.8) d) P(0.6 <  < 0.8) e) P( < 0.6) For this question only, pretend that p = 0.8. Compute (p, the standard error for this problem, also called the deviation of the sampling distribution of the proportion. DO NOT INCLUDE THE FINITE POPULATION MODIFIER IN YOUR CALCULATIONS. The value of (p is about a) 0.004 b) 0.0632 c) 0.16 d) 0.2 e) 0.2821 (You may find it useful to draw a picture for this one.) In answering THE QUESTION, we find that a value of 0.6 in the sampling distribution of the sample proportion has a z-score of 3.16. It follows that, if z is standard normal variable (as usual) then P(p < 0.6) = P(z < -3.16) P(p < 0.6) = P(z = -3.16) P(p < z) = -3.16 P(p < -3.16) = z P(z < -3.16) = p Considering the size of the z score of about -3 found in the previous problem, which statement is the best conclusion of this analysis? If each graduate of a weight loss program has an 80% chance of losing weight, then the chance of having less than 25 successful students in a group of 40 randomly selected students is zero; it simply is not possible. less than 1%. about 10%. about 20%. about 30%. Questions 3-8 are based on the normal approximation to the binomial distribution. To find the exact answer to this question, we would have to use the binomial distribution itself. The calculation =BINOMDIST(24, 40, 0.8, TRUE) results in the value 0.002936. The calculation of =BINOMDIST(24,40,0.6,TRUE) result in the value of 0.55978. Which of the following statements follows from this information? The chance of Beautiful Us claim being true is 0.002936. The chance of Beautiful Us claim being true is 0.55978. If each student has an 80% chance of losing weight, then to see less than 25 successes in 40 randomly selected students is very unlikely (less than 3 times in 1000). If each student has a 60% chance of losing weight, then in a group of 40 randomly selected students, seeing exactly 24 successes is very common. It happens about 56% of the time. Both answer a and answer d are correct. Up to this point, we have been addressing THE QUESTION, above. For questions 10-17, we investigate THE SECOND QUESTION: THE SECOND QUESTION: Suppose that the average weight lost by a student in the program is 5 pounds, with a standard deviation of 20 pounds. How likely is it that the average weight loss in a random sample of 40 students will be 8.45 pounds or more? Our task to find the probability that something is greater than or equal to 8.45. What is that something? a) z b) ( c) s d) ( e)  EMBED Equation.3 (x-bar) Our sample size is 40, so well be able to apply our 191 techniques to answer THE SECOND QUESTION. This is because the histogram of a 40 student sample suggests that the population is a) perfectly normal b) roughly normal c) roughly symmetric d) roughly binomial e) bimodal  Question #12 says that well be able to apply our 191 techniques to answer THE SECOND QUESTION. This is because the observation that we made in Question #12 allows us to conclude that the population is approximately normal ( is approximately normal the sampling distribution of the mean is approximately normal s is a good approximation for ( ( is a good approximation for EMBED Equation.3 . To answer THE SECOND QUESTION, we need to find the standard deviation of the sampling distribution of the mean. It is equal to 20 20/40 = 0.5 (8.45 - 5)/20 = 0.1725 SQRT(8.45 ( (20 8.45)/40) = 1.562 20/SQRT(40) = 3.162 We also need to know the mean of the sampling distribution of the mean. In this problem, this is =NORMSINV(0.1725) = -0.9443 =NORMSINV(1-0.1725) = 0.9443 5 =SQRT(40) = 6.325 8.45 Which calculation would give the probability that one student chosen at random from the Beautiful U graduates would have lost 8.45 pounds or more? (Assume, for this problem only, that weight loss is normally distributed.) = NORMDIST( 5, 8.45, 20, TRUE) = NORMDIST( 5, 8.45, 3.162, TRUE) = NORMDIST( 8.45, 5, 20, TRUE) =1 - NORMDIST( 8.45, 5, 20, TRUE) =1 - NORMDIST( 8.45, 5, 3.162, TRUE) Suppose that the answer to THE SECOND QUESTION is 0.1376. What does this mean? About 14% of all Beautiful U students lose 5 pounds or more. About 14% of all Beautiful U students lose 8.45 pounds or more. About 14% of the students in a sample of 40 students would be expected to lose 5 pounds or more. About 14% of the students in a sample of 40 students would be expected to lose 8.45 pounds or more. None of these interpretations is correct. Suppose that the answer to THE SECOND QUESTION is 0.1376. Use this information to answer this question: How likely is it that the average weight loss in a random sample of 40 Beautiful U students is between 1.55 pounds and 5 pounds? (Hint: Note that 1.55 + 3.45 = 5, and that 5 + 3.45 = 8.45. Draw a picture.) a) About 14%. b) About 21% c) About 29% d) About 36% e) About 86%  Use the excerpt from Table E.2b appearing on the last page of this test to compute P(-0.23 < z < 0.58). Its value is a) 0.2888 b) 0.310 0 c) 0.690 0 d) 0.7112 e) 1.128 A fair coin is flipped 6 times. What is the probability that it comes up heads in exactly 4 out of the 6 times? =BINOMDIST(4, 6, 0.5, TRUE) =BINOMDIST(4, 6, 0.5, FALSE) =BINOMDIST(6, 4, 2/3, TRUE) =BINOMDIST(4, 6, 2/3, TRUE) =BINOMDIST(6, 4, 2/3, FALSE) Generally, increasing the sample size will decrease the standard deviation of the population. decrease the standard error of the mean. decrease the standard error of the proportion. All of the above (a-c) are true. b and c are true, but a is not. Consider these six values: -3, -, 0, , 1, 3. Let B be a binomially distributed random variable. It is impossible for B to take on some of these values. How many of the six values are impossible values for B? a) 1 b) 2 c) 3 d) 4 e) 5 Which of the following random variables is most likely to be reasonably approximated by a Poisson distribution? HELP!! Whether a person rolls doubles on a pair of dice (0 = no, 1 = yes) The number of times a person rolls doubles in 10 throws of a pair of dice. The number of times a person must roll a pair of dice before throwing doubles. The time between when one customer enters a casino and when the next customer enters the casino. The number of customers who enter the casino between 4 and 5 PM. Consider these Excel expressions: =NORMSINV(NORMSDIST(2)) =NORMSDIST(NORMSINV(2)). Which of the following statements is true about the values that Excel assigns to these expressions? (You may wish to draw a picture.) both expressions equal 2. I equals 2, II is undefined. I is undefined, II equals 2. both expressions are undefined. the expressions are equal, but the value is probably not 2. Consider these Excel expressions: =NORMSINV(0.3) =NORMSINV(0.7). Which of the following statements is true about the values that Excel assigns to these expressions? (Hint: draw a picture.) both expressions give the same value (i.e., I = II) the value assigned to the two expressions add to one (i.e., I + II = 1) the value assigned to the second expression is the negative of the value assigned to the first (i.e., II = - I). The difference of the two values would equal =NORMSINV(0.4) (i.e., II I = NORMSINV(0.4)) The difference of the two values would equal =NORMSINV(0.5) (i.e., II I = NORMSINV(0.5)) Which of the following Excel expressions would give exactly the same value as =NORMSDIST(0.6)? =NORMDIST(0.6, 1, 0, TRUE) =NORMDIST(0.6, 0, 1, TRUE) =NORMDIST(0.6, 1, 1, TRUE) =NORMDIST(0.6, 1, 0, FALSE) =NORMSINV(0.4) weight before programweight after programlost weight?weight loss189.5198.00-8.5186208.00-22.0175.5176.00-0.5210174.0136.0182156.5125.5178172.515.5214188.5125.5196158.0138.0160155.514.5184191.50-7.5184.5148.5136.0171.5123.0148.5176.5151.5125.0160.5169.00-8.5161.5138.0123.5188177.0111.0180.5153.0127.5198192.016.0175180.50-5.5175.5179.00-3.5171168.512.5170181.00-11.0165.5139.5126.0192189.512.5157165.50-8.5185.5175.0110.5183192.00-9.0213.5187.5126.0162189.50-27.5207209.50-2.5184162.5121.5221.5239.00-17.5163.5157.516.0158176.50-18.5158.5179.00-20.5230.5219.5111.0165137.5127.5177178.50-1.5171.5138.5133.0174.5143.0131.5average loss= 8.45 pounds# losing wt = 24 students z0.000.010.020.030.040.050.060.070.080.09-0.50.30850.30500.30150.29810.29460.29120.28770.28430.28100.2776-0.40.34460.34090.33720.33360.33000.32640.32280.31920.31560.3121-0.30.38210.37830.37450.37070.36690.36320.35940.35570.35200.3483-0.20.42070.41680.41290.40900.40520.40130.39740.39360.38970.3859-0.10.46020.45620.45220.44830.44430.44040.43640.43250.42860.424700.50000.50400.50800.51200.51600.51990.52390.52790.53190.53590.10.53980.54380.54780.55170.55570.55960.56360.56750.57140.57530.20.57930.58320.58710.59100.59480.59870.60260.60640.61030.61410.30.61790.62170.62550.62930.63310.63680.64060.64430.64800.65170.40.65540.65910.66280.66640.67000.67360.67720.68080.68440.68790.50.69150.69500.69850.70190.70540.70880.71230.71570.71900.7224Entry represents area under the cumulative standardized normal distribution from -infinity to Z PAGE \# "'Page: '#' '" Ill give you this same information on your test. PAGE \# "'Page: '#' '" B. Answer C is the value of p. E would be useful if we were doing the problem as a binomial, where it would be the number of successes. PAGE \# "'Page: '#' '" E. The population is all Beautiful U graduates, and we dont know how many there are. n = 40 and p = 0.8 would be correct, but n = 40 is information about the sample, not the population. We don t know what the size of the population, N, may be. PAGE \# "'Page: '#' '" D. The approximation of a proportion by the normal distribution requires that n and n(1-) each be at least 5. Answer III is wrong on almost all counts. Proportion problems dont care about the sample size per se, and the population (consisting of only two values, success and failure) is a LONG way from normal! PAGE \# "'Page: '#' '"  Ive put this question in pink because I wont ask you to use the FPM on your test. Knowing about it is another matter, however. The FPM reduces the size of the standard error for mean or proportion problems, but the reduction is significant only if the sample is at least 10% of the population. In this problem, we dont know the population size, so we cant use the FPM anyway. The answer is thus A. PAGE \# "'Page: '#' '" B. Answer A could be credible of the mu were replaced by x-bar, but even then it wouldnt be consistent with our decision to approach this as a proportion problem. The last three answers are all wrong, since they are hypotheses about the population proportion, . We are given in this problem that  = 0.8. In later chapters we ll build confidence intervals for p, but for the current material, population parameters are always given. PAGE \# "'Page: '#' '" Note, by the way, that this isn t much of an assumption, since  is 0.8 in this entire scenario! PAGE \# "'Page: '#' '" SQRT((1-)/n) = SQRT(0.8(0.2)/40) = 0.4/SQRT(40) = 0.0632. The answer is B. PAGE \# "'Page: '#' '" A. Saying that s has a z score of  3.16 is the same as saying this: the location of s on the normal curve representing the sampling distribution of the proportion corresponds to the same location as  3.16 on the standard normal curve. Saying it another way, 0.6 is 3.16 standard deviations below the mean of its normal curve. PAGE \# "'Page: '#' '" B. It is certainly possible, by random chance, to get any number of successes in the sample, even 0. The z-score of 3.16, though, means that getting 60% successes or less is as likely as getting a z score of 3.16 or less on a standard normal curve. We know that 99.9% of all observations lie within 3 standard deviations of the mean, so a z score of less than 3 will occur about 0.05% of the time, or about one time in 2000. PAGE \# "'Page: '#' '" Only the first BINOMDIST is needed, since we are talking about a population in which each element has an 80% chance of being a success. We know, then, that the chance of 24 successes or less in a set of 40 trials is 0.002936. The answer is thus C. Note that answers A and B are out of the running, since the calculation was done by assuming that the 80% figure was correct. D is wrong because it says exactly 24 successes. For this interpretation, wed need =BINOMDIST(24,40,0.6,FALSE) to be about 56%. PAGE \# "'Page: '#' '" E. Again, recall that mu, the population mean, is given as being 5. PAGE \# "'Page: '#' '" C. With a sample size of 40, we can be confident that the sampling distribution of the mean is essentially normal, provided that the population itself is roughly symmetric. The histogram on the data page suggests that this is the case. Certainly, that histogram doesnt suggest a perfectly normal or even roughly normal population. Answer D is irrelevant, and the curve doesnt look binomial anyway (binomials are unimodal). E is irrelevant. PAGE \# "'Page: '#' '" C. Dont get suckered into answer A! The central limit theorem assures us that the sampling distribution of the mean is essentially normal for a large enough sample size. It doesnt care about the population. Answer B is meaningless (mu is a number!). D and E are true statements for large enough samples, but these facts are not the reasons we can proceed with THE SECOND QUESTION. Make sure you answer the question posed! PAGE \# "'Page: '#' '" Sigma divided by the square root of the sample size = 20/SQRT(40), or answer E. Note that this looks different from the formula for standard error of the proportion, SQRT(p(1-p)/n). PAGE \# "'Page: '#' '" The mean of a sampling distribution for the mean or for a proportion is always equal to the mean of the population, so the answer is 5answer C. PAGE \# "'Page: '#' '" D. Lets see why. First, since were taking only one student, the distribution of relevance is that of the population, not the sample of size 40. (This is the reason that we had to assume a normal population for this problem.) Since it is a or more problem, were going to need to approach it by finding the probability of the corresponding or less, then subtracting the result from 1. The probability of a single student losing 8.45 points or less is =NORMDIST(8.45, 5, 20, TRUE). As an aside, E is the answer to THE SECOND QUESTION, as you should be able to verify. PAGE \# "'Page: '#' '" Answer E! A nd B are out, since our conclusion cant be about individuals in the population. (THE SECOND QUESTION deals with a sample of 40 students. The calculations we did, using NORMDIST, dont even make sense if they arent applied to a normal distribution, and the population might not be normal.) C and D try to use P(x-bar > 8.45) as the proportion of students in the sample whose loss exceeds some threshold. Our work here is about means, not proportions. The correct interpretation of the value: A randomly selected sample of 40 graduates has about a 14% chance of having its average weight loss be 8.45 pounds or more. (This assumes, of course, that the values of mu and sigma given are correct.) PAGE \# "'Page: '#' '" It is, by the way. PAGE \# "'Page: '#' '" Sketch a normal curve with a mean of 5, and mark 8.45 on the axis. The part of the curve to the right of this 8.45 cutoff is about 14%. That means that the part of the curve below 1.55 is also about 14%, since the distance of 1.55 from 5 is the same as the distance of 8.45 from 5, and the normal curve is symmetric around its mean. Now 50% of the normal curve lies to the left of 5 (again, since 5 is the mean), so the area we want (between 1.55 and 5) is 50% - 14%, or about 36%. The answer is D. Answer E is crazy, since it says that more than half of the time, the mean of the sample will lie between 1.55 and 5. The mean of the sample will lie between infinity and 5 only one-half of the time! PAGE \# "'Page: '#' '" This question is in pink because I wont make you use z-tables. The equivalent kind of problem for you would provide you with =NORMSDIST values. Your answer would then be =NORMSDIST(0.58) NORMSDIST(-0.23) PAGE \# "'Page: '#' '" The TRUE answers are all going to be wrong, since they deal with the likelihood of k successes or less. Answer B is right; 4 successes in 6 trials with a 50% chance of success each time. Answer E is impossible, with 6 successes in 4 trials. The probability of success is also wrong for a fair coin. Note that its a bad idea to try to find the best answer to one of my questions by finding the most common elements of the five answers. If you tried that here, youd get D, which is dead wrong. PAGE \# "'Page: '#' '" E. The populations standard deviation indeed, any characteristic of the populationhas nothing to do with a sample. B and C are both true, since the standard errors are sigma/sqrt(n) and sqrt(p(1-p)/n), respectively. As n increases, these decrease. PAGE \# "'Page: '#' '" Binomials count number of successes, so only give integer values between 0 and the number of trials. Thus 0 and 3 are possible (assuming we have at least 3 trials), but the rest are not. The answer is D. PAGE \# "'Page: '#' '" Poisson distributions arise this way. We have an interval of interestusually a time interval, but it can be an interval of space as well. We break this interval down into tiny parts, all the same size. In any of these parts, we could have a success, although the chance of success in one of these tiny parts is small. Its small enough, in fact, that the chance of more than one success in a single tiny part is essentially 0. Were going to count the total number of successes in the whole interval. We expect this count to be Poisson distributed provided that two additional conditions are met. First, the chance of a success has to be the same for each of the tiny parts. Second, a success in any one part has to be independent from successes in any other part. For example, imagine popcorn in a popper, and lets see if its reasonable to expect the number of pops in a minute to be Poisson distributed. Well break the minute into tiny parts, say 0.01 second each. This is small enough that the chance of a success (a pop) in the 0.01 seconds is small, and the chance of two pops in the same 0.01 second is essentially zero. So far so good. Now lets check the other assumptions. Can we assume that each 0.01 second has the same chance of a pop as any other? Well, that depends. If we choose our minute to be toward the beginning of the popping, probably not. A later second will have more chance of a pop than an early second. So lets restrict our attention to a minute in the middle of the popping frenzy. The assumption is probably pretty well met in this case. How about the other assumption, of independence? This means that the chance of a pop in any 0.01 seconds is not affected by whether or not a pop occurred in any other 0.01 second. With thousands of kernels popping in the popper, this assumption is very close to true. So, during the middle of the popping, we expect the number of kernels popping per minute to be very nearly Poisson distributed .Note that a Poisson distribution can in theory take on any of the values 0,1,2,3, and so on forever. The fact that there is no theoretical upper limit is one thing that can help you distinguish it from a binomial. PAGE \# "'Page: '#' '"  A is a Bernoulli trial, with only two possible outcomes. B is a binomial. Binomials and Poissons can look a lot alike, but only when the number of trials is huge and the probability of success on any given trial is quite small (less than 5%, say). This fact is not one that we discussed in class, and youre not responsible for it. It does mean though, that we cant weasel B into a right answer. C is a distribution we never discussed, the geometric. It isnt Poisson because a success (a roll) during each little part (opportunity to roll) depends on the successes in other parts. Specifically, if I know that a person doesnt take a fourth roll, then they got doubles in one of their first three rolls, so I also know that he or she doesnt take a fifth roll. Failure in the fourth part guarantees failure in the fifth part. 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The correct answer is E. Itd be wise to see what assumptions underlie this choice! PAGE \# "'Page: '#' '" NORMSDIST and NORMSINV are inverses. That is, p = NORMSDIST(z) if and only if z = NORMSINV(p). (You should be clear on what these statements say in terms of a sketch of a standard normal curve.) So if p = NORMSDIST(2), then 2 = NORMSINV(p). Putting these together, NORMSINV(NORMSDIST(2)) = NORMSINV(p) = 2. So expression I is just 2. Expression II has a problem, however, since NORMSINV(2) doesnt mean anything. It would be the z value, the area to the left of which is 2. Since the entire normal curve only has an area of 1, this makes no sense. It says, in different words, 200% of the standard normal curve is to the left of what value?. Answer B is thus correct. Note that to get a question like this correct, you need to understand what these functions are telling you! PAGE \# "'Page: '#' '" The first value gives the z value which is the cutoff for the leftmost 30% of the standard normal curve. The second value gives the z values which is the cutoff for the leftmost 70% of the standard normal curve, or, equivalently, the rightmost 30% of the standard normal curve. Since these cutoffs isolate the lower 30% of the normal and the upper 30% of the normal, respectively, they must be the same distance from the mean of the standard normal distribution, which is 0. They are therefore negatives of one another. The answer is C. (This is really pretty easy to see with a sketch.) PAGE \# "'Page: '#' '" The Excel given represents the fraction of the standard normal curve that lies to the left of z = 0.6. The standard normal has mean 0 and standard deviation 1, so answer B represents exactly this quantity. Answers A, D, and E are silly: standard deviations of 0 mean no normal curve, and NORMSINV gives a (negative) z value, not a probability.  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