INFORMATION ABOUT EXAM A n - New York Universitypeople.stern.nyu.edu/.../Exams/E1infoSpring2012.pdf · 2012. 2. 9. · INFORMATION ABOUT EXAM A OF 2012.FEB23 ... The horizontal axis

INFORMATION ABOUT EXAM A OF 2012.FEB23 n n n n n n n n n n n n n n n n n n n n n n n n n n n n

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The exam of 2012.FE B.23 will be of the open-book variety. You may also be asked to examine output from the program Minitab. You may find it helpful to have a functioning calculator. Here is a set of sample problems, including problems from old exams. These should be similar in spirit to problems that will actually appear on the exam of February 23. Solutions follow, beginning on page 13. This set of sample problems was pieced together from previous exams. Missing sequence numbers (S15, S16, and many more) correspond to problems that do not match the material for the February 24 exam. Problems marked with ♦ are very good exercises, but they are not likely candidates for the exam. The ♦ problems were either too complicated, too confusing, or too computationally intensive. S1. For this set of values, give the median, lower quartile, and upper quartile: 37.2 29.2 30.6 34.4 28.7 31.1 35.4 S2. Guess the standard deviation of the ages of the students in NYU’s undergraduate population. S3. The following histogram was based on the weekly receipts at a hardware store for 100 business weeks. The horizontal axis is thousands of dollars; for instance, 20 refers to $20,000.

(a) For how many weeks (out of the 100) were receipts in the range $10,000

to $12,000? (b) For how many weeks (out of the 100) were receipts in the range $16,000

to $20,000? (c) For how many weeks (out of the 100) were receipts in excess of $20,000?

302010

20

10

0

C1

Freq

uenc

y


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S4. For each of the following situations, indicate whether list P or list Q has the smaller sample standard deviation. You need not justify your solution, nor do you need to actually compute any standard deviations.

(a) List P has 100 values. The values are the integers 1, 2, 3, ..., 99, 100. List Q has 100 values. The values are 2, 4, 6, 8, ..., 198, 200.

(b) List P has 50 values.

The number 180 appears 10 times. The number 200 appears 30 times. The number 220 appears 10 times.

List Q has 50 values. The number 180 appears 5 times. The number 200 appears 40 times. The number 220 appears 5 times.

(c) List P contains 500 numbers.

The value 18 appears 100 times. The value 19 appears 100 times. The value 20 appears 100 times. The value 21 appears 100 times. The value 22 appears 100 times.

List Q contains 500 numbers. The value 18 appears 150 times. The value 20 appears 200 times. The value 22 appears 150 times.

S5. You have just calculated the average and standard deviation of a list of 101 numbers, finding an average of 240.0 and a standard deviation of 25.88. Unfortunately, a check of the list of numbers uncovers two errors.

A number originally listed as 230 should be 200. A number originally listed as 250 should be 280.

(a) After you make the corrections, what will be the average of the list? (b) What will be the standard deviation, after you make the correction?


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S6. A financial magazine has undertaken a survey of selling prices of three-bedroom homes in the suburbs of two market areas, Indianapolis and Denver. These are summarized in the box plots below: INDIANAPOLIS : ┌────────┬───────────┐ ─────────┤ │ ├─────────────────────── └────────┴───────────┘ DENVER : ┌────────┬─────────────┐ ─────────────┤ │ ├──────────────────────────── └────────┴─────────────┘ --------+---------+---------+---------+---------+---------+---------+-- 80 100 120 140 160 180 200

Cost in thousands of dollars For each of the following questions, answer either T (true) or F (false) or N (no way to tell).

(a) The average home cost is higher in Denver. (b) The median home cost is larger in Denver. (c) The IQR is larger in Denver. (d) If you combined the data from the two cities, the median would exceed

$110,000. (e) The midrange is higher in Indianapolis. (f) The lowest recorded home cost in this survey comes from Denver. (g) The largest home cost in this data base exceeds $200,000. (h) More homes were surveyed in Denver than in Indianapolis. (i) In Indianapolis about half the home costs were above the local median,

while in Denver about half the home costs were below the local median. (j) The generally lower home costs in Indianapolis were a consequence of

lower construction costs. S7. List A has 20 entries, with an average value of 510, and with a standard deviation

of 76. List B has 15 entries, with an average value of 580, and with a standard deviation

of 88. (a) If the lists are combined to produce a single list of 35 entries, what would be the

average? (b)♦ For the combined list of 35 entries, what will be the standard deviation?


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S8. The following stem-and-leaf display shows the dollar amounts of the adjustments on a set of invoices: -1⏐10 -0⏐9976665 NOTE: 1⏐8 means $180. -0⏐321 0⏐12344 Value of leaf position is $10. 0⏐578 1⏐ 1⏐8 2⏐2

(a) How many invoices were used in the construction of this stem-and-leaf display?

(b) What is the smallest dollar amount among the invoices? (c) What is the median amount? (d) What is the lower quartile? (e) Why do the rows with negative stem values list the leaves in apparent

descending order?

S10. Supply a numeric answer for each of the situations given below. In each case, the calculation is simple. You may use “cannot be determined from given information” as an answer. (a) The average inventory of flour, determined at 25 inventory audits, at Beck’s

Pretzels was 4,650 pounds. If all 25 values were converted to kilograms, what would be the average in kilograms? (Note: 2.2 pounds = 1 kilogram.)

(b) The standard deviation of amounts for the 25 values in the previous statement was 660 pounds. What would be the standard deviation after the conversion to kilograms?

(c) A random sample of 11 supermarkets in Blair County found that the median size of the on-shelf stock of Bananabits breakfast cereal was 26 boxes. A similar sample of 17 supermarkets in adjoining Cosco County had a median of 31 boxes. If the data are combined to get a sample of 11+17=28 supermarkets, what would be the median?

S9. (a) A list of 20 values x1, x2, …, x20 has an average x = 135 and a sample standard

deviation s = 14. It is discovered that there is a 21st data value; it was missing when we formed this list. This value, x21 , happens to be numerically equal to 135. What is the average of the whole list of 21 values?

(b)♦ For the situation described in part (a), what is the standard deviation of the whole list of 21 values?


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(d) For each of a sample 15 New York law firms, the annual salary for newly-appointed partners was determined. The average of these 15 values was $140,000. A similar survey was done for 10 law firms in Boston, obtaining $100,000. What is the average for all 25 firms?

S11. (a) Exhibit a set of three numbers in which the mean exceeds the median. (b) Exhibit a set of three numbers in which the mean equals the median. (c) Exhibit a set of three numbers in which the mean is smaller than the median. S12. A set of 140 data values has minimum 115 pounds, maximum 220 pounds, median 150 pounds, and quartiles at 125 pounds and 170 pounds. Give a box-and-whisker plot for these data, using the scale below. Be sure to provide labels for this scale! ----+----+----+----+----+----+----+----+----+----+----+----+----+--- S13. Each of the following questions can be answered as true or false. Indicate your solutions on the blank spaces below. Your answers do not have to be explained. _____(a) The median of a set of numbers is always larger than the mean. _____(b) The upper quartile of a set of numbers is greater than or equal to the

median of the set of numbers. _____(c) The interquartile range of a set of numbers must be at least as large as the

median of the set of numbers. _____(d) Negative values of the standard deviation indicate that the set of values is

even less dispersed than would be expected by chance alone. _____(e) The standard deviation is commonly used because of its ease of

calculation using only paper and pencil. _____(f) If the median salary of all male employees in a firm exceeds $26,000, and

if the median salary of all female employees in the firm exceeds $25,000, then it is certain that the median salary of all employees exceeds $25,000.

_____(g) It happens that the average pay of a floor supervisor exceeds the average pay of a line worker in factory Q, and the same statement also holds in factory R, If the data for the two factories are combined, then the average pay of all floor supervisors is guaranteed to exceed the average pay of all line workers.

_____(h) The mean is used because it is a commonly-understood simple statistical summary of a set of numbers.


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S17. Six people are to be seated randomly at a round dinner table. Find the probability that Anna will be sitting next to Joe. S18. There are 620 students who live on campus at Bracken College. Some of these students also keep automobiles on campus, according to this breakdown:

Grade Level Car on campus? NO

Car on campus? YES Total

Freshman 145 30 175 Upperclass 265 180 445

Total 410 210 620

(a) If a student is to be selected randomly for an article on student life in the campus

newspaper, find the probability that this student will be a freshman. (b) If a student is to be selected randomly, find the probability that this student will be

a freshman, given that this student keeps a car on campus. (c) If a student is to be selected randomly, find the probability that this student keeps

a car on campus, given that he or she is a freshman. (d) If two students are to be selected randomly, find the probability that both students

will be freshmen. S19. The probability of winning a certain casino game is 0.28. If you play this game ten times, find the probability of winning at least once. You need not evaluate the numeric

expression, but you should show it in final form; that is, an answer in forms like 56

18FHIK or

1 0 10− .42 , would be acceptable. S20♦. Five cards are dealt from a standard deck. Find the probability that these five cards include exactly two hearts, given that they include exactly two spades.

S14. Prepare a stem-and-leaf display from the following list of values:

21.4 30.6 29.7 28.8 35.9 19.4 31.1 28.0 28.4 22.7 18.9 24.5 27.2 20.7 24.6 25.5 23.4 27.1 28.8 20.7


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S22. If a fair coin is to be flipped five times, find the probability that there will be at least three heads. S26. In each of the situations below, a set of data is described. Circle the value which, you believe, could be the standard deviation.

(b) An auditor examined the numbers of cars handled by individual toll booth

attendants at the Triborough Bridge during rush hours. The standard deviation of the cars-per-hour for these attendants could be

3.5 35 350 3,500 (c) The standard deviation of the heights in inches of the female members of

the Stern class of 2013 could be 0.3 inch 3 inches 9 inches 62 inches

(d) An accounting firm took a survey of its entry-level clerk-typists who have

rental apartments in Manhattan. The standard deviation of the monthly rents could be

$20 $200 $2,000 $8,000 (e) For the purchase amounts by customers at the Campus Eatery, the

standard deviation could be 12 cents 25 cents $1.20 $12.00

S21. Consider a two-stage game. At the first stage, we flip a coin. If the coin comes up heads, we select one ball from Urn 1, and if the coin comes up tails, we select one ball from Urn 2. The Urns have the following composition:

Urn 1 has 5 green balls

10 yellow ballsRST Urn 2 has

20 green balls10 yellow ballsRST

You arrived late for this game, and you missed the coin flip. However, you did get to witness the selection of a yellow ball. What is the probability that this came from Urn 1?

(a) Ms. Rivera’s third grade class, containing 29 students, was asked about daily milk consumption. The standard deviation of this set of 29 values could be (approximate metric equivalents are shown)

16 oz 32 oz 64 oz 128 oz 480 cc 960 cc 1.9 l 3.8 l


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S27. The Pareto chart below tabulates major movie releases for 1995-1996 according to the studios that release them.

How many studios do you need to include in order to capture 50% of the total number of releases? S28. The following list of 62 numbers has ix∑ = 869.04 and 2

ix∑ = 12,929. (This uses x1, x2, …, x62 for the data.)

17.14 16.28 16.30 7.70 15.35 14.02 14.98 14.51 17.66 9.74 10.00 16.22 14.10 9.37 13.59 13.91 13.61 12.33 11.32 14.80 17.98 13.48 19.67 14.22 13.96 5.66 11.54 8.41 17.99 9.28 11.55 19.01 19.27 15.37 16.72 12.98 15.95 16.24 16.38 16.74 9.69 8.99 15.82 4.63 14.42 10.93 14.79 11.11 10.94 15.03 18.75 17.64 19.25 20.77 15.05 13.34 18.61 13.58 9.98 14.50 13.71 12.18

Based on this information, find the mean and the standard deviation.

Norstar

FineLine

New_Line

Goldwyn

SonyClassic s

Oc toberMGM

OrionSavoy

Gramercy

Miramax

NewLine

MGM_UA

Univ ersal

ParamountFox

WarnerBros

Sony

BuenaV ista

1 1 1 3 3 3 3 51011212323303030384748

0 0 0 1 1 1 1 2 3 3 6 7 7 9 9 9111415

100100 99 99 98 97 96 95 94 91 88 81 74 67 58 49 40 29 15

300

200

100

0

100

80

60

40

20

0

Defect

CountPercentCum %

Perc

ent

Cou

nt

Major movie releases 1995-1996, by studio


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S29. The display below is a stem-and-leaf display. Based on this information, find the lower quartile, the median, and the upper quartile.

Stem-and-leaf of C1 N = 58 Leaf Unit = 0.10 2 5 67 3 6 5 8 7 23557 17 8 012345667 24 9 0036789 28 10 1124 (8) 11 03556779 22 12 01456779 14 13 1149 10 14 0445 6 15 4479 2 16 0 1 17 1 18 8

S30. In the game of roulette, the probability is 1838

= 919

≈ 0.4737 that a black number

will come up. Suppose that Carla plays this game exactly four times, and that she bets on black for each of the four turns. What is the probability that a black number will come up at least once in the four turns? S31. Al, Betty, Cindy, and Don have lunch at a diner that specializes in individual pizzas. Al and Betty order garlic pizzas, Cindy orders pepperoni pizza, and Don orders vegetable pizza. The waiter returns with the pizzas, but he has forgotten which pizza was ordered by which person. He randomly sets the four pizzas before the four diners. Find

(a) the probability that all four people get correct orders; (b) the probability that exactly three people get correct orders; (c) the probability that exactly two people get correct orders; (d) the probability that exactly one person gets a correct order; (e) the probability that no one gets the pizza he or she ordered.


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S33. A standard deck of 52 cards has been reduced to 51 because an unknown card was lost. With no other information, we’d certainly say P(lost card is ♠) = P(lost card is a spade) = 13

52 = 14 . However, we take a random card from the deck, and we observe

that this card is ♦ (diamond). Find the conditional probability

P(lost card is ♠ | selected card is ♦)

Think of this as a two-step game. At the first step a random card is lost, and at the second step one of the remaining 51 cards is selected and observed to be ♦. S34. An ordinary die is to be rolled ten times. Find the probability

(a) of getting the side with six dots exactly once in ten rolls; (b) of getting the side with six dots exactly twice in ten rolls; (c) of getting the side with six dots at least once in ten rolls; (d) that the total of all ten rolls will be an even number.

S35. The following list shows the probability distribution for X, the number of renewals ordered by 28 current subscription holders to a magazine. The probability that any individual subscription holder will renew is 0.22.

x P[X = x] P[X ≤ x] 8 0.118514 0.85662 0 0.000952 0.00095 9 0.074282 0.93090 1 0.007519 0.00847 10 0.039808 0.97071 2 0.028628 0.03710 11 0.018373 0.98908 3 0.069980 0.10708 12 0.007341 0.99642 4 0.123363 0.23044 13 0.002548 0.99897 5 0.167015 0.39746 14 0.000770 0.99974 6 0.180576 0.57803 15 0.000203 0.99994 7 0.160071 0.73810 16 0.000046 0.99999

(a) Find the probability that the number of renewals will be less than 12. (b) Find the probability that the number of renewals will be at most 10. (c) Find the probability of fewer than 8 renewals. (d) Find the probability of at most 11 renewals. (e) In the column P[ X = x ] appears the number 0.180576. Exactly what

arithmetic does Minitab do to produce this number?


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S41. The random variable X has the probability facts listed below. Find the expected value of X and the standard deviation of X.

x 0 1 2 3 4 p(x) 0.05 0.05 0.10 0.50 0.30

S42. The random variable X has the binomial distribution with n = 20 and p = 0.20. The random variable Y has the Poisson distribution with parameter λ = 3.5. Indicate which random variable has the larger mean and which has the larger standard deviation. S43. The number of phone calls arriving at the front desk of MetaWell, Inc., follows a Poisson distribution with mean λ = 3.0/hour. Jim is the person in charge of answering the phone. If Jim sneaks out for a ten-minute doughnut break, what is the probability that he will not miss any calls? S44. Westfall Bank issued 54 home-equity loans in 2002. It turns out at 14 of these are completely inactive. (This means that the credit line was established in 2002 but never used.) If an auditor selects 8 accounts at random, what is the probability that the selection will include exactly 2 inactive accounts? S53. The Magi-Kleen air filter is made in three different locations.

The assembly plant in New Bedford makes 60% of all the filters, and 1% of its filters will be returned by the customers because of defects.

The assembly plant in Fall River makes 30% of all the filters, and 2% of its filters will be returned by the customers because of defects.

The assembly plant in North Adams makes 10% of all the filters, and 3% of its filters will be returned by the customers because of defects.

In the process of packing the filters for shipment to retail outlets, the filters from the three sources are intermixed. Suppose that a filter is returned for additional repairs. What is the probability that this filter was made in New Bedford?


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S54. An auditor who has investigated the travel and lodging expenses for the sales staff of a large corporation has determined that 88% of the claims showing an average daily expense of $150 or more have adequate documentation, while only 65% of the claims showing an average daily expense below $150 have adequate documentation. It happens that 70% of the claims made by the sales staff generate daily averages over $150. What percentage of the claims without adequate documentation are for claims with daily averages under $150? S55. A university alumni office is trying to get recent graduates to subscribe to a glossy new alumni magazine. If the offer is presented over the telephone, it is happens that 35% will subscribe immediately. You wish to know the probability that, among 25 persons called, at least 10 will subscribe immediately. You have access to Minitab. For each of the following, indicate which of these would enable you to get the desired solution. Your responses could be “good” or “no good.”

(a) Ask for the binomial with n = 25, p = 0.35, and find the cumulative probability for 10 successes.

(b) Ask for the binomial with n = 25, p = 0.65, and find the cumulative probability for 15 successes.

(c) Ask for the binomial with n = 25, p = 0.35, find the individual probabilities for 0, 1, 2, …, 25 successes and add up those probabilities corresponding to 10, 11, 12, 13, …, and 25 successes.

(d) Ask for the binomial with n = 25, p = 0.65, find the cumulative probability for 15 successes, and subtract this answer from 1.

S56. The probability of winning at a certain carnival game is 0.4. If you play this game ten times, what is the probability that you will win at least twice? Solutions begin on next page....


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SOLUTIONS S1. The only serious requirement is sorting the list into ascending order. (Descending order will also work.) 28.7 29.2 30.6 31.1 34.4 35.4 37.2 The median is the center value, 31.1. The quartiles could be found by any one of a number of rules.

The lower quartile could be given as the value in position n +=

+14

7 14

= 2; this

value would be 29.2. The upper quartile would be the second value from the high end, namely 35.4.

S2. There is no right-versus-wrong distinction here, but some answers are better than others. For example, 1.5 years is certainly a plausible answer. The value 0.8 year is too low, and 3.2 years is too high. S3. (a) 18 weeks; (b) 23 weeks; (c) 13 weeks. (These could be off by 1 or 2 because of difficulty in reading the picture exactly. S4. (a) List P has the smaller standard deviation. (b) List Q has the smaller standard deviation.

For list Q, you will have

Sxx = x xi −∑b g2 = 150 (18-20)2 + 200 (20-20)2 + 150 (22-20)2

= 150 [ 4 + 4 ] = 1,200

It follows that list P will have the smaller standard deviation.

(c) This is a bit trickier. For both lists, the average value is 20. For list P, you will have

Sxx = x xi −∑b g2 = 100 (18-20)2 + 100 (19-20)2 +100 (20-20)2 + 100 (21-20)2 +100 (22-20)2

= 100 [ 4 + 1 + 0 + 1 + 4 ] = 1,000


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S5. (a) The original list contained the sum 230 + 250 = 480. The revised list replaces this with 200 + 280 = 480. The total does not change, so the average is still 240.

(b) Since s = 25.88 = ( )2

101 1ix x−−

∑ = x xi −∑ b g2100

, you will find that

x xi − = ×∑b g2 2100 25 88. = 66,977.44. Originally this sum contained (230-240)2 + (250-240)2 = 200. After the corrections are made (and noting that the average remains at 240), these are replaced by (200-240)2 + (280-240)2 = 3,200. Thus, the correction causes the sum to grow by 3,000, up to 69,977.44.

Now find s = 69100,977.44 ≈ 26.45.

This is a difficult problem. S6. (a) The average home cost is higher in Denver. N : boxplots are not designed to

reveal averages. (b) The median home cost is larger in Denver. T : the median (represented by the

central line within the box) is slightly higher in Denver. (c) The IQR is larger in Denver. T : the IQR is the length of the box, and it is

slightly larger in Denver. (d) If you combined the data from the two cities, the median would exceed $110,000.

T : more than half the values from each city exceed $110,000, and this fact would necessarily be so in the combined sample.

(e) The midrange is higher in Indianapolis. F : the midrange is slightly higher in

Denver. (f) The lowest recorded home cost in this survey comes from Denver. True. (g) The largest home cost in this data base exceeds $200,000. T: this happens in

Denver. (h) More homes were surveyed in Denver than in Indianapolis. N: there is just no

way to know this.


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(i) In Indianapolis about half the home costs were above the local median, while in Denver about half the home costs were below the local median. T : the statement is correct but quite vacuous.

(j) The generally lower home costs in Indianapolis were a consequence of lower

construction costs. N : this statement is pure speculation. S7. (a) List A has 20 entries with a total of 20 × 510 = 10,200.

List B has 15 entries with a total of 15 × 580 = 8,700.

The combined list has an average 10 8 70020 15

1835

,200 , ,900++

= = 540.

(b) The individual lists have standard deviations around 80. Note that the combined

list will spread out a little more. It would be quite reasonable to assert that the standard deviation is in the interval (80, 120), probably at the lower end of the interval, since the means are not too far apart.

Here’s the exact result. In general, if the A-list has m items with average a and standard deviation sA , and if the B-list has n items with average b and standard deviation sB , you can show that the standard deviation of the combined list is

m s n s mnm n

a b

m n

A B− + − ++

−

+ −

1 1

1

2 2 2a f a f c h

In this case, the expression is

20 1 76 15 1 88 20 1520 15

510 580

20 15 1

2 2 2− × + − × +×+

−

+ −

a f a f a f

= 7 651 7647, . ≈ 87.45


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S9. (a) The average is certainly 135. (b) With the average unchanged, it must happen that Sxx is the same for both lists.

For the expanded list, Sxx = x xi −∑b g2 has one extra summand, but that

summand is (135 - 135)2 = 0. For the list of 20 values, s = 14 = S Sxx xx

20 -1 19= ,

so that Sxx = 19 × 142 = 3,724.

The value of Sxx is not changed by including this particular 21st value. Thus, for

the list of 21 values, we have s = 3 72421 1

3 72420

186 2, , .−

= = ≈ 13.65.

S10.

(a) The average would be 4,650 pounds × 1 kg2.2 pounds

≈ 2,114 kg. Actually, this is

the average in kilograms whether you convert the separate 25 values or not! (b) It’s 660

2 2.kg = 300 kg.

(c) The median of the combined data cannot be determined from the information

given. (d) This is obtained by noting the total amount

15 × $140,000 + 10 × $100,000 = $2,100,000 + $1,000,000 = $3,100,000

spread out over 25 firms. The average is $3,100,00025

= $124,000.

S8. (a) 22 (b) -$110 (c) The median is between the 11th and 12th values, which are -$20 and -$10, so that

we give the median as -$15. (d) The lower quartile (by one rule) is to be found at the 6th position; the 6th value

is -$60. (e) The order is actually ascending. For instance, the row with -0 ⏐321

represents -$30, -$20, -$10.


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S12. Here’s what it should look like: ┌───────────┬─────────┐ ─────┤ │ ├───────────────────────── └───────────┴─────────┘ ----+----+----+----+----+----+----+----+----+----+----+----+----+----+ 110 120 130 140 150 160 170 180 190 200 210 220 230 There’s no reason to worry about showing extreme points with dots and circles. First of all, the problem does not provide enough information to enable you to do this. Secondly, there can’t be any points beyond the so-called fences; note that IQR = interquartile range = 170 - 125 = 45 and “upper inner fence” = UQ + 1.5 IQR = 170 + 1.5 × 45 = 237.5, so that the maximum is within the “upper inner fence.” Similarly, the minimum is within the “lower fence.” S13. (a) is false. (b) is true; the upper quartile and the median can actually be equal if there are many

tied values in the set. (c) is false; the interquartile range is a measure of spread (or variability) and thus

does not have to have any particular relationship to the median. (d) is false; the standard deviation must always be ≥ 0. The statement is a

spectacular blast of meaningless jargon. (e) is false. (f) is true. (g) is false. While the statement is generally true, it need not be; examples in which

this happens are described as “Simpson’s Paradox.” (h) is true; it would be hard to deny that this is one of the major attractive features of

the mean.

S11. For (a) you could use 8 , 10 , 200. For (b) you could use 8 , 10 , 12. For (c) you could use 0 , 10 , 12.


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S14. There are only twenty values, so it seems best to put these in only a few stem positions. If you use the tens digit to set the stem (and the units digit for the leaves, ignoring the tenths digit), you get the following:

1 | 89 2 | 001 2 | 23 NOTE: 1 | 8 means 18 2 | 445 2 | 77 2 | 88889 3 | 01 3 | 3 | 5

This was made by Minitab (and then edited), but a hand-made version would be similar. You could use a “double-stem” to spread this out a bit:

1 | 89 2 | 0012344 NOTE: 1 | 8 means 18 2 | 57788889 3 | 01 3 | 5

The display above uses each stem position twice, with leaves 0 through 4 in the first position and 5 through 9 in the second. S17. Imagine that Anna is to be seated first, and Joe is to be seated second. No matter what seat Anna gets, exactly two of the remaining five seats will be next to her, so the probability is 2

5 . S18. (a) 175

620 ≈ 0.2823 (b) 30

210 = 1

7 ≈ 0.1429 (c) 30

175 ≈ 0.1714

(d) 175

620174619

× ≈ 0.0793

S19. The probability of losing in one turn is 0.72, and the probability of losing all ten times is 0.7210. Thus, the probability of winning at least once is 1 - 0.7210 ≈ 0.9626.


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S20. P( two ♥ | two ♠) = ( )( )

P two and two P two

♠♥♠

=

132

132

261

525

132

393

525

FHGIKJFHGIKJFHGIKJ

FHGIKJ

FHGIKJFHGIKJ

FHGIKJ

.

The numerator involves the 261FHGIKJ , as we must account for the one card which is neither ♠

nor ♥. The denominator involves the 393FHGIKJ , and as we must account for the three cards

which are not ♠. The calculation continues as

132

132

261

132

393

FHGIKJFHGIKJFHGIKJ

FHGIKJFHGIKJ

=

132

261

393

FHGIKJFHGIKJ

FHGIKJ

= 78 26×9,139

≈ 0.2219

S21. P(Urn 1 | yellow ) = P Urn 1 and yellow

P yellowa fa f =

P Urn 1 P yellow Urn 1P yellow

a f b ga f .

Certainly we have P(Urn 1) = 1

2 and P(yellow | Urn 1) = 1015 = 2

3 . We need P(yellow).

P(yellow) = P( (Urn 1 and yellow) or (Urn 2 and yellow) ) = P (Urn 1 and yellow) + P(Urn 2 and yellow) = P(Urn 1) P(yellow | Urn 1) + P(Urn 2) P(yellow | Urn 2) = 1

21015

× + 12

1030

× = 12

Thus we complete P Urn 1 P yellow Urn 1

P yellowa f b g

a f = 12

23

12

× = 23

.


n Page © gs2012 20

This can be done as well by the “hypothetical 100,000” method. For 100,000 plays there will be (expected counts) 50,000 from each urn.

Green Yellow TOTAL Urn 1 50,000 Urn 2 50,000 TOTAL 100,000

The picks from Urn 1 will result in 1

3 green balls and 23 yellow balls.

Similarly, picks from Urn 2 will result in 2

3 green balls and 13 yellow balls. This gets us

to the following:

Green Yellow TOTAL Urn 1 16,667 33,333 50,000 Urn 2 33,333 16,667 50,000 TOTAL 50,000 50,000 100,000

Given that a yellow ball has been selected, the conditional probability that this came from

Urn 1 is 33,33350,000

≈ 23

. If we had recorded exact values instead of rounded values, this

would have come out exactly as 23

.

Then P(3 or 4 or 5 heads) = P(0 or 1 or 2 heads). This symmetrically accounts for all the probability, and thus P(3 or 4 or 5 heads) = 1

2 . S26. The believable values are (a) 16 oz, (b) 35 cars, (c) 3 inches, (d) $200, (e) $1.20.

S22. The probability that there will be exactly k heads is 5

0 5 1 0 5 5

kk kF

HGIKJ − −. .a f =

50 55

kFHGIKJ . ,

and we could evaluate this for k = 3, 4, 5. However there is a much simpler way to get to the solution. Observe that

P(3 heads) = P(2 tails) = P(2 heads ) P(4 heads) = P(1 tail ) = P(1 head ) P(5 heads) = P(0 tails) = P(0 heads )


n Page © gs2012 21

S27. It looks like you will need to consider the top five studios (Buena Vista, Sony, Warner Brothers, Fox, Paramount). There is no dominant studio in this business, as the top 88% of the film titles are spread rather evenly among nine studios.

S28. The mean is easily found as x = 869.0462

≈ 14.0168. To get to the standard

deviation, we’ll find first

Sxx = ( )2

2 ii

xx

n− ∑∑ = 12,929 -

2869.0462

= 12,929 - 755,230.521662

≈ 12,929 - 12,181.1374 = 747.8626

Then find s = 1

xxSn −

= 747.862661

≈ 12.26004188 ≈ 3.5014.

S29. Here the sample size is n = 58.

To find the lower quartile, we look for position ( )14 1n + = 1

4 × 59 = 14.75, which we round up to 15. Thus LQ = Q1 = 8.6.

If you want to get overly precise, you’d note that we want the first of the 8.6 values in the stem-and-leaf. If you had the original list, you could identify the lower of the two values starting with 8.6. This extra effort is simply not worth the extra precision.

The median appears in position ( )1

2 1n + = 12 × 59 = 29.5. We’ll take the average

of the 29th and 30th values. Numerically this is 12 (11.0 + 11.3) = 11.15.

The upper quartile requires position ( )3

4 1n + = 34 × 59 = 44.25, which we round down to

44. Then we find UQ = Q3 = 12.9. Operationally, it’s easiest to get this value by counting 15 positions from the high end.

S30. The probability that a non-black number will come up on a single turn is 1 - 919

=

1019

≈ 0.5263. The probability that Carla gets a non-black number on every turn is then

0.52634 ≈ 0.0767. Thus the probability that Carla will see at least one black number is 1 - 0.0767 = 0.9233.


n Page © gs2012 22

S31. Use G1 and G2 to denote the garlic pizzas. Use P and V to denote the other pizzas. There are 4! = 24 equally likely patterns in which these can be given to the four diners in A-B-C-D order:

D I N E R D I N E R D I N E R D I N E R A B C D A B C D A B C D A B C D G1 G2 P V G2 G1 P V P G1 G2 V V G1 G2 P G1 G2 V P G2 G1 V P P G1 V G2 V G1 P G2

G1 P G2 V G2 P G1 V P G2 G1 V V G2 G1 P G1 P V G2 G2 P V G1 P G2 V G1 V G2 P G1

G1 V G2 P G2 V G1 P P V G1 G2 V P G1 G2

G1 V P G2 G2 V P G1 P V G2 G1 V P G2 G1

The correct matches are shown in boldface. You can see that

12 times in the 24 patterns have A getting a garlic pizza 12 times in the 24 patterns have B getting a garlic pizza 6 times in the 24 patterns have C getting a pepperoni pizza 6 times in the 24 patterns have D getting a vegetable pizza

If we tally up the patterns according to the number of correct matches, we get

2 patterns have exactly four matches 0 patterns have exactly three matches 10 patterns have exactly two matches 8 patterns have exactly one match 4 patterns have exactly zero matches

The corresponding probabilities are 224

, 024

, 1024

, 824

, and 424

. These should not be

reduced to lowest terms, but it would be reasonable to give them all with denominator 12. It would also be reasonable to give them as decimals. S33. This can be done by applying the conditional probability definition.

P(lost card is ♠ | selected card is ♦) = [ ][ ]

P lost card is selected card is P selected card is

♠ ∩ ♦♦

We’ll calculate the numerator as follows:

[ ]P lost card is selected card is ♠ ∩ ♦


n Page © gs2012 23

= [ ]P lost card is P selected card is lost card is ♠ ⎡ ♠⎤⎣ ⎦♦ = 1 134 51×

The denominator P[selected card is ♦] must be 1

4 . After all, one of the four suits must be selected, and there is no reason that these should not be equally likely. Our final answer then is 13

51 . This is slightly larger than 14 . Thus, once we see a

diamond, it become slightly more likely that a spade has been lost. S34. This deals routinely with a binomial with n = 10 and p = 1

6 .

(a) P[ exactly one six in ten rolls ] = 1 910 1 5

1 6 6⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠ =

9

10

5106

× ≈ 0.3230.

(b) P[ exactly two sixes in ten rolls ] = 2 810 1 5

2 6 6⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎝ ⎠ =

8

10

5456

× ≈ 0.2907.

(c) P[ at least one six in ten rolls ] = 1 - P[ no sixes in ten rolls ] =

0 1010 1 510 6 6

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

= 1051

6⎛ ⎞− ⎜ ⎟⎝ ⎠

≈ 0.8385.

(d) After nine rolls, your accumulated total will either be even or odd.

If the total after nine rolls is even, the probability that the tenth roll will come up 2, 4, or 6 (and keep the total at even) is 1

2 . If the total after nine rolls is odd, the probability that the tenth roll will

come up 1, 3, or 5 (and convert the total to even) is 12 .

In any event, the probability of an even total after ten rolls is 12 .

S35. (a) P[ X < 12 ] = P[ X ≤ 11 ] = 0.98908.

(b) P[ X ≤ 10 ] = 0.97071. (c) P[ X < 8 ] = P[ X ≤ 7 ] = 0.73810. (d) P[ X ≤ 11 ] = 0.98908.

(e) P[ X = 6 ] = 0.180576 refers to the arithmetic 6 22280.22 0.78

6⎛ ⎞⎜ ⎟⎝ ⎠

. You are

pleased that Minitab does the calculation.


n Page © gs2012 24

S41. First find

E X = 0 × 0.05 + 1 × 0.05 + 2 × 0.10 + 3 × 0.50 + 4 × 0.30 = 2.95

S42. Certainly E X = np = 20 × 0.20 = 4 and E Y = λ = 3.5. Clearly X has the larger mean. The standard deviation for X is np p1−a f = 20 0 2 0 8× ×. . ≈ 1.7889, while the standard deviation for Y is λ = 3 5. ≈ 1.8708. So Y has the larger standard deviation. S43. The X be the number of calls in this ten-minute period. If the Poisson rate is

3.0/hour, then the rate that applies to X will be 10 minutes 3.060 minutes

× = 0.50. Thus, P[ X = 0 ]

= 0

0.5 0.50!

e− = e- 0.5 ≈ 0.6065 = 61%.

S44. This is a straight hypergeometric problem with N = 54, D = 14, and n = 8. We ask the probability that X, the number of inactive in the sample, will be equal to 2. The calculation is this:

P[ X = 2 ] =

14 402 6

548

⎞ ⎞⎛ ⎛×⎟ ⎟⎜ ⎜

⎝ ⎝⎠ ⎠⎞⎛⎟⎜

⎝ ⎠

This calculation is easiest, of course, with software. If you need to do it by hand, you might evaluate all three binomial coefficients separately …. or … you could set up the whole calculation in one place to take advantage of cancellations. Let’s do the three parts separately:

142⎞⎛⎟⎜

⎝ ⎠ = 14 13

2 1××

= 91

There are several ways to get to SD(X), but certainly the easiest is to work through E X2.

E X2 = 02 × 0.05 + 12 × 0.05 + 22 × 0.10 + 32 × 0.50 + 42 × 0.30 = 9.75 Then Var X = 9.75 - 2.952 = 1.0475, and finally SD(X) = 1 0475. ≈ 1.0235.


n Page © gs2012 25

406⎞⎛⎟⎜

⎝ ⎠ = 40 39 38 37 36 35

6 5 4 3 2 1× × × × ×× × × × ×

= 3,838,380

548⎞⎛⎟⎜

⎝ ⎠ = 54 53 52 51 50 49 48 47

8 7 6 5 4 3 2 1× × × × × × ×× × × × × × ×

= 1,040,465,790

The probability is thus 91 3,838,3801,040,465,790

× ≈ 0.335708 ≈ 34%.

S53. We can answer this by tracing through a hypothetical group of 100,000 filters. Here’s how it starts:

Returned Not Returned TOTAL New Bedford 60,000 Fall River 30,000 North Adams 10,000 TOTAL 100,000

Using the indicated return rates, we can update the diagram thus:

Returned Not Returned TOTAL New Bedford 600 60,000 Fall River 600 30,000 North Adams 300 10,000 TOTAL 1,500 100,000

Finally, the detail can be completed as

Returned Not Returned TOTAL New Bedford 600 59,400 60,000 Fall River 600 29,400 30,000 North Adams 300 9,700 10,000 TOTAL 1,500 98,500 100,000

It should be emphasized that this display gives the expected counts. Any actual run of 100,000 filters would be subject to statistical fluctuation.


n Page © gs2012 26

We do see, however, that for every 1,500 filters which are returns, 600 came from New

Bedford. Thus, we have P(NB⏐returned) = 6001,500

= 0.4

S54. The easiest way to handle this is probably the “hypothetical 100,000” method. Imagine a (fictitious) set of 100,000 claims. The starting information is this:

Documentation adequate? YES

Documentation adequate? NO

TOTAL

Claim less than $150 30,000 Claim more than $150 70,000

TOTAL 100,000 This display shows simply that 70% of the claims have daily averages of $150 or more. Now note that there should be 65% × 30,000 = 19,500 claims of less than $150 with adequate documentation; this number goes in the upper left corner. Similarly, 88% × 70,000 = 61,600 goes in the lower left corner. This gets us to the following:



TOTAL

Claim less than $150 19,500 30,000 Claim more than $150 61,600 70,000

TOTAL 100,000 The remaining information can be completed easily, giving



TOTAL

Claim less than $150 19,500 10,500 30,000 Claim more than $150 61,600 8,400 70,000

TOTAL 81,100 18,900 100,000 Now, to answer the question, we have 18,900 claims without adequate documentation, and 10,500 of these have daily averages below $150. The requested conditional

probability is 10,50018,900

≈ 0.5555 ≈ 56%.


n Page © gs2012 27

In conditional probability terms:

P(less than $150 | inadequate ) = P less than $150 inadequate

P inadequate∩

= 10,500

10,500100,00018,900 18,900

100,000=

This problem is not really difficult, but it gets complicated because you have to sort through heavy verbiage. S55. (a) Ask for the binomial with n = 25, p = 0.35, and find the cumulative probability

for 10 successes. (b) Ask for the binomial with n = 25, p = 0.65, and find the cumulative probability

for 15 successes. (c) Ask for the binomial with n = 25, p = 0.35, find the individual probabilities for

0, 1, 2, …, 25 successes and add up those probabilities corresponding to 10, 11, 12, 13, …, and 25 successes.

(d) Ask for the binomial with n = 25, p = 0.65, find the cumulative probability for 15 successes, and subtract this answer from 1.

Methods (b) and (c) will work. If X represents the random number of people who will subscribe, then X ~ Bin(n = 25, p = 0.35). We want to find P(X ≥ 10). Method (a) will obtain P(X ≤ 10), which is not what we want. Method (b) works with p = 0.65, so it is apparently talking about Y, the number who would not subscribe. The event equivalent to (X ≥ 10) is (Y ≤ 15). Since (b) computes P(Y ≤ 15), it gives what we want. Method (c) will obtain P(X ≥ 10), albeit painfully, and this is our objective. Method (d) is also concerned with Y. However, it obtains 1 - P(Y ≤ 15) = P(Y ≥ 16), which equals P(X ≤ 9). This is not our objective.


n Page © gs2012 28

S56. Let X be the random number of times you’ll win. You want P[X ≥ 2], and it’s easiest to calculate this as 1 – ( P[X = 0] + P[X = 1] ). Now

P[X = 0] = 0 10100.4 0.6

0⎛ ⎞⎜ ⎟⎝ ⎠

≈ 0.006047

P[X = 1] = 1 9100.4 0.6

1⎛ ⎞⎜ ⎟⎝ ⎠

≈ 0.040311

Thus, P[X ≥ 2] = 1 – (0.006047 + 0.040311) = 1 – 0.046358 = 0.953642.

INFORMATION ABOUT EXAM A n - New York Universitypeople.stern.nyu.edu/.../Exams/E1infoSpring2012.pdf · 2012. 2. 9. · INFORMATION ABOUT EXAM A OF 2012.FEB23 ... The horizontal axis

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