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Unformatted text preview: MAT ‘221 Final Exam Dec. 16, 2004 ver. A Signature: Instructions: 0 You should have 10 pages (5 sheets) and 9 problems. 0 Write the answers and Show the main steps of your work on
this test sheet. 0 Your ﬁnal answers must include the appropriate units (eg.
dollars, dollars per week, miles per hour, etc.) o If you use a table, state the table used: for example, 1.887 (from Table W).
o if you use a function on the T183 (or T189) write out the com mand you entered as well as thve‘result:
for example, 0.0668 (normalcdf(10,1.5,0,1)). DO NOT WRITE ON THE REST OF THIS COVER SHEET! (Your instructor will use this sheet for recording your scores.) Problem 1(12) Problem 4(12) Problem 7(1’2) Problem 200) Problem 5(9) Problem 8(9) Problem 3(1‘3) Problem 602) Problem 902)
‘ I):\RT 1(34) PART 2(33) PART 3(33) Tolflltmo) PART 1: Chapters 1 & 2 Problem 1. (12 points) Consider the following 17 test scores: 58, 43,
53, 31, 62, 58, 52. 65, 52, 33, 37, 67, 48, 59, 48, I4, 67 (i) Make a stem plot of these scores (ii) For this data set: (a) Give the ﬁvenumber summary of these scores
0 o (b) List all suspected outliers using the 1.5 x [QR rule. If you
think: that there are none, say so. ’ (iii) These data have a mean of 49.82 and a standard deviation of 14.46 Suppose these data are transformed by the linear trans
formation )" = 2X — 10. (a) The mean of the transformed data set is: (b) The standard deviation of the transformed data set is: (c) The median of the transformed data set is: MAT 221 Final Exam Yer. A 3 Problem 2. (10 points) The random variable X of temperatures in a
certain refrigeration facility is normally distributed with a mean, ,ux, of 14.4 degrees Fahrenheit and a standard deviation, ox of 4.2 degrees
Fahrenheit. ‘ (i) What is the probability that temperature will exceed 20 degrees? (ii) Compute the probability P(—4 S X < 0). (iii) Compute Q1 for this distribution. (iv) Roughly what percent of the time do you expect the temperatures
to fall between 6 degrees Fahrenheit (6 = 14.4 — 2 X 4.2) and
20.8 degrees Fahrenheit (22.8 = 14.4 + 2 x 4.2). Justify your
answer. 4 MAT 221 Final Exam Va. A Problem 3. (12 points) Consider the following data on heights (2n
inches) of dating couples:
Observation # 1 2 3 4 5 6 7 l/Vomen (I—data) 60 66 64 65 70 65 66
Men (y—data) 64 72 68 68 71 65 70 (i) Make a scatter plot of the data:
yaxis
75 70
65
60 55 xaxis 50 55 60 65 70 75
(ii) Given the following information,
T = 65.14, 51 = 2.97, 25131111: 31179. g = 68.29, 3y : 2.98,
compute the correlation coefﬁcient. Circle your answer below.
(If you cannot compute the correlation coefﬁcient, circle the
right—most option and use that value to complete this problem.) .77 .27 O —.‘27 .77 (iii) Circle the equation of the linear regression line for predicting
y from .1‘: g: 35.72 + 51:, y = 6829‘ g1)=18.11+.771‘,
y) : 51.35 + .271. g = 85.22 ~ 3271. (iv) Using this regression line, estimate the height. of a man that
would date a woman of height 63 (inches). \ MAT 221 Final Exam Ver. A 0! PART 2: Chapters 3 & 4 Problem 4. (12 points) (i) A research team is interested in the possibility that drug A or
drug. B will decrease the number of colds in adult males. They
plan a year long study involving 31,596 male physicians. Each
will take a drug A tablet or placebo each day and a drug B
tablet or a placebo each day. (a) Who are the subjects? (b) List the treatment(s): (c) What is(are) the response uariable(s)‘.2 (ii) We wish to take a sample of NAT 221 students. For each of
the following sampling options, circle the correct description of
the sampling technique
(a) Randomly select 4 sections and randomly select 8 students from each of these sections. Circle one: Simple Random Sample Multistage sample
Stratiﬁed random sample None of these (1)) Write the names of all student from all sections on slips
of paper and randomly select 82. Circle one: Simple Random Sample Multistage sample
Stratiﬁed random sample None of these (c) Select all of the students from. section 1. Circle one: Simple Random Sample ‘Multistage sample Stratiﬁed random sample None of these 6 MAT 221 Final Exam Ver. A Problem 5. (9 points) Let .4, B, C, D, E and F denote events in a
probability space. (i) Given that PM) = .45, P(B) = .6 and P(A and B) = .15,
compute (a) P(A or B) =
(b) P(AIB) =
(C) P(AC) = (ii) Given that P(C) = .45, P(D) = .6 and that events C and D
are independent and compute (a) P(C and D) =
(b) P(DlC) = .
(c) P(C or D) 2
(iii) Consider the Venn diagram below.
(a) Shade in the area that rcprcaents P(E and FC). (1)) Given P(E) = .45, P(F) = .75 and P(E and FC) = .15,
ﬁll the probabilities in each of the remaining three areas of
the Venn diagram. MAT 221 Final Exam Yet. A ‘1 Problem 6. (12 points) ~ (i) A jar contains 8 $1 bills and 2 $5 bills. You take two bills from
the jar, one at a time without returning them.
(a) Construct a tree diagram for this process and label the
branches of the tree with appropriate probabilities. (b) Let X be the random variable that takes on the dollar
amounts that you can win and ﬁll in the following table: X 82 $6 $10
p. (ii) Consider the random rariable (a) Compute the mean of this random z'ariablc: (1)) Compute the standard deviation of this random variable: 8 MAT 22] Final Exam Ver. A PART 3: Chapters 5 & 6 Problem 7. (12 points) When tossed, an unfair coin comes up heads
with a probability of .4. a ‘V (i) What is the probability that, in ﬂipping this coin 30 times, you
will have exactly 20 headsf2 (ii) Compute the mean and standard deviation for the binomial
distribution of the number of heads in 30 ﬂips of this coin: ll Ill, (iii) Use the normal approrimation to approximate the probability
that in 30 ﬂips there will be 10 or fewer heads. MAT 221 Final Exmn Ver. A 9 Problem 8. (.9 points) A random sample of 275 Syracuse University
undergraduate students was selected and they were asked for the cost
of their text books. The average cost for the students in the sample
was $295. Suppose the textbook costs are normally distributed with a
standard deviation of $69. (1) Find a 95% conﬁdence interval estimate for the true mean cost
for the whole population. (ii) How large of a sample would be needed to estimate the true
mean cost within i810 with 95% conﬁdence? 10 MAT 221 Final Exam Ver. A Problem 9. (12 points) The mean and standard deviation of the ACT
scores of high school seniors is 20 and 6, respectively and the test scores
are normally distributed. The mean test score for a random sample of
45 seniors who have had a calculus course is E = 22.3. 13 this suﬁlcient
evidence to claim that students taking calculus do better on the ACT
test? We denote the mean of the ACT scores of all seniors who have
had a calculus course by u and assume that the standard deviation for that group would be the same as that for the entire population.
(i) State the hypotheses: H05
Ha: (ii) Find the value of the zstatisticfot‘ testing the hypotheses in
part a. ' (iii) Compute the Pl’lllUC of the :statistic you found in part b. (iv) Is this statistically signiﬁcant (circle your answers) at the 5% level? Yes No at‘the 1% level? Yes No Tables and Formulas for MAT 221 .
Chapter 1 : Looking at DataDistributions , . 1 _ 1: +1 + +1 _ 1 .
0 Mean. J. ~—l——2—n_;§):zz
7 V   3_ —« _(Ii5)2+lr25)2+~+(1.,—i)2
. o Vanance. s n_1 ECU1 _ _—'_‘—nT*‘“ __ 0 Standard deviation: 3 = ‘/,L_1 (I1 — 5:)2 = 5—1:“: 1:12 — Ill—‘2) o zscore: : = 1—;E (x = ”+2”
Chapter 2 : Looking at DataRelationships
7172(asfi) (m) 0rr= (4)212w sy n—l szsy o Leastsquares regression line : g = a + bx, where b = ”r3: and
a = g — bi:
Chapter 4 : Probability: The Study of Randomness 0 Probability Rules
— For any event A: P(.4C) = 1 — P(A). — For disjoint events A and B: P(A or B) = P(A) + P(B). — For anyeventsA and B: P(A or B) = P(A)+P(B)——P(A and B).
— For independent events A and B: P(A and B) = P(A)  P(B). — For any events A and B: P(A and B) = P(A)  P(BA).  When PM) > 0, P(B}.4) = We — Bayes's Rule: PHIB) = W
0 Finite Random Variable X
— Mean: ll.\' = 11121 +1'~2P2 + ' ‘ ' + ﬁlm = 213131"
— Variance:
0\=('Ii/1X2')Pi+ +(1k(—/l.\'2)PA=Z(Ti— Ml?!) — Standard Cl€\1dt10n 0\ — V: —;I,')x 2]) — ”(2117192) #13} 0 Random \ariables X and l
— If a and b are ﬁxed numbers, then
#(a+bX) = a + bllxi U(a+b.\') = bUX
— For any random variables X and Y:
MA’H') = #X + My
— For independent random variables X and Y: 0(,\+y) = Mai +0% and 0(X_y): ”0% +03“ ‘5 Chapter 5 : From Probability to Inference
o Binomial distribution: X ~ B(n,p) Binomial coefﬁcient: (2) : #4)“
wheren!=nx(n—l)x(n—2)><~><3><2>< 1.
— Binomial probability: my = k) : (Z)pk(l — [mu—k). fork=0,l,...,n — For the actual count, X, ﬁx 2 np ox = np(1 ~ p) — For the sample preportion, 15, #13 2 p, 17,, = w 0 Let [51" be the mean of an SRS of size n from a population having
mean u and standard deviation 0. Then #2 = u, 0'5 = 0/\/T_L
Chapter 6 : Introduction to Inference o A level C conﬁdence interval for p ( a known, SRS from a normal
population): 90% 95% 99% — #L * '
miz «5' Z fromMO’l) .1645 1.96 2.576 0 Sample size for desired margin of error m : 2*0 2 .
'n = .
m 0 Test statistic for H0 : it = #0 (assuming that a known and the
sample is an SRS from a normal population): =l—l'7—‘y0 o/ﬁ in Table entry for z is
the area under the standard normal curve to the left of z. .4 .0003
—33 00%
3.2 .0007
—3.1 .0010
~3.0 .0013
2.9 .0019
—2.8 .0026
—2.7 .0035 .
—2.6 .0047
—2.5 .0062
—2.4 .0082
—2.3 .0107
—2.2 .0139
—2.1 .0179
—2.0 .0228
—1.9 .0287
—1.8 .0359
—1.7 .0446
—1.6 .0548
—1.5 .0668
~1.4 .0808
—1.3 .0968
—1.2 .1151
—1.1 .1357
1.0 .1587
—0.9 .1841
—0.8 .2119
—0.7 .2420
——0.6 .2743
0.5 .3085
~0.4 .3446
>113 .3821
112 .4307
~414 .4002
1).11 .5000 . .0003
.0005
.0007
.0009
.0013
.0018
.0025
.0034
.0045
.0060
.0080
.0104
.0136
.0174 .0281
.0351
.0436
.0537
.0655
.0793
.0951
.1131
.1335
.1562
.1814
.2090
.2389
.2709
.3050
.3409
.3783
.4168
.4562 077‘) 4960 . .0003
.0004
.0006
.0009
.0012
.0017
.0023
.0032
.0043
.0057
.0075
.0099
.0129
.0166
.0212
.0268
.0336
.0418
.0516 .
.0630
.0764
.0918
.1093
.1292
.1515
.1762
.2033
.2327
.2643
.2981
.3336
.3707
.4090
.4483 4880 . Probability TABLE 1A Standard normal probabilities .0003
.0004
.0006
.0008
.0012
.0016
.0023
.0031
.0041
.0055
.0073
.0096
.0125
.0162
.0207
.0262
.0329
.0409
.0505
.0618
.0749
.0901
.1075
.1271
.1492
.1736
.2005
.2296
.2611
.2946
.3300
.3669
.4052
.4443 4840 .0003
.0004
.0006
.0008
.0011
.0015
.0021
.0029
.0039
.0052
.0069
.0091
.0119
.0154
.0197
.0250
.0314
.0392
.0485
.0594
.0721
.0869
.1038
.1230
.1446
.1685
.1949
.2236
.2546
.2877
.3228
.3594
.3974
.4304 4761 .0003
.0004 ‘.0005 .0008
.0011 .0015 .0021
.0028
.0038
.0051
.0068
.0089
.0116
.0150
.0192
.0244
.0307
.0384
.0475
.0582
.0708
.0853
.1020
.1210
.1423
.1660
.1922
.2206
.2514
.2843
.3192
.3557
.3936
.4325
.4721 .0003
.0004
.0005
.0007
.0010
.0014
.0020
.0027
.0037
.0049
.0066
.0087
.0113
.0146
.0188
.0239
.0301
.0375
.0465
.0571
.0694
.0838
.1003
.1190
.1401
.1635
.1894
.2177
.2483
.2810
.3156
.3520
.3897
.4286
— . . .4681 .0002
.0003
.0005
.0007
.0010
.0014
.0019
.0026
.0036
.0048
.0064
.0084
.0110
.0143
.0183
.0233
.0294
.0367
.0455
.0559
.0681
.0823
.0985
.1170
.1379 .1611 .1867
.2148
'.2451 .2776
.3121
.3483
.3859
.4247
.4041 . ........‘.d Tables Table entry for z is
theareaunderthe
standard normal curve
totheleftofz. Probability TABLE 'A .. Standard upwa‘lo‘pzjobabilitie’s (continued) 99999
hww—o 99¢
\JO\U1 pp
\000 hiwru 4 C>k>d>kzb~01Lixiul—23 NluIululqy—oay—HH__.___ .5398
.5793
.6179
.6554
.6915
.7257
.7580
.7881
.8159
.8413
.8643
.8849
.9032
.9192
.9332
.9452
.9554
.9641
.9713
.9772
.9821
.9861
.9893
.9918
.9938
.9953
.9965
.9974
.9981
.9987
.9990
.9993
_900:
.9997 .5438
.5832
.6217
.6591
.6950
.7291
.7611
.7910
.8186
.8438
.8665
.8869
.9049
.9207
.9345
.9463
.9564
.9649
.9719
.9778
.9826
.9864
.9896
.9920
.9940
.9955
.9966
.9975
.9982
.9987
.9991
.9993
.9995
.9997 .5478
.5871
.6255
.6628
.6985
.7324
.7642
.7939
.8212
.8461
.8686
.8888
.9066
.9222
.9357
.9474
.9573
.9656
.9726
.9783
.9830
.9868
.9898
.9922
.9941
.9956
.9967
.9976
.9982
.9987
.9991
.9994
.9995
.9997 .5160 .5557
.5948
.6331
.6700
.7054
.7389
.7704
.7995
.8264
.8508
.8729
.8925
.9099
.9251
.9382
.9495
.9591
.9671
.9738
.9793
.9838 .5199
.5596
.5987
.6368
.6736
.7088
.7422
.7734
.8023
.8289
.853]
.8749
.8944
.9115
.9265
.9394
.9505
.9599
.9678
.9744
.9798
.9842
.9878
.9906
.9929
.9946
.9960
.9970
.9978
.9984
.9989
.9992
.9994
.9996
.9997 .5636
.6026
.6406
.6772
.7123
.7454
.7764
.8051
.8315
.8554
.8770
.8962
.9131
.9279
.9406
.9515
.9608
.9686
.9750
.9803
.9846
.9881
.9909
.9931
.9948
.9961;
.9971
.9979
.9985
.9989
.9992
.9994
.9996
.9997 .5675
.6064
.6443
.6808
.7157
.7486
.7794
.8078
.8340 _...... .85//
.8790
.8980
.9147
.9292
.9418
.9525
.9616
.9693
.9756
.9808
.9850
.9884
.9911
.9932
.9949
.9962
.9972
.9979
.9985
.9989
.9992
.9995
.9996
.9997 .5753
.6141
.6517
.6879
.7224
.7549
.7852
.8133
.8389
.8621
.8830
.9015
.9177
.9319
.9441
.9545
.9633
.9706
.9767
.9817
.9857
.9890
.9916
.9936
.9952
.9964
.9974
.9981
.9986
.9990
.9993
.9995
.9997
.9998 C2.) ...
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