UNIVERSITY I
THE OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE
STAT1000 Prinicipal of Statistics STAT1801 Probability & Statistics: Foundation of Actuarial Science STAT1007 Applied Mathematics - Statistics STAT0601 Statistical Methods in Physical Sciences
December 20, 2002
Time: 9:30 a.m. - 11:30 a.m.
Candidates taking examinations that permit the use of calculators m a y u s e any calculator which julfils thefollowingcriteria:(a) it should be self-contained,silent, battery-operated and pocket-sized and (b) it should have numeral-display facilities only and should be used only for the purposes of calculation. It is the candidate's responsibility to ensure that the calculator operates satisfactorily and the candidate must record the name and type of the calculator on the front page of t h e c m m i n a t i o n s c r i p t s . L i s t s of permitted/prohibited calculators will not be m a d e uvailable t o candidates for reference, and the onus will be on the candidate t o ensure that the culcdutor used will not be in violation of the criteria listed above.
Answer ANY FIVE questions. Marks are shown in square brackets.
1. (a) Consider three events A , B and C with P ( A ) = 0.50, P ( B ) = 0.80 and P ( G ) = 0.75. Suppose the probability that a.n outcome belongs t o C but n o t J A 01` 13 is 0.10 and the following pairs of events are independent:-
A and B,
B and C ,
(.4 n B ) a,nd ( B n C).
[4 marks]
( i ) Compute t,he probability t,hat an outcome belongso all three events. t
( i i ) A r e the events A a,nd C independent?
[ 2 marks]
[3 marks]
( i i i ) Show that. the three events are exhaustive.
I i n t : A V e n n d i a g r a m wl be useful for the above calculations. il
(1.)) Bolhy andPeter play with a fairdieandt,herule of thegame is as f o I I ours :If Bobby tosses a "6", then he wins. If not, then Peterwins if he tosses a "1" or "6" . If Peter does not win, then Bobby tosses and wins on getting a
%AS: STAT1000 Prin. of Stat. / 1801 P & S / 1007 AM-S / 0601 Phy. SC. 2
"1" or a "6". And so on.
(i) Show that this is a fair game to thetwo players.
[6 marks]
(ii) Show the conditional probability that Bobby wins eventually if he does not get a "6" on the first toss is 2/5. [5 marks]
Consider a discrete random variable whose probability generating funcX tion is given by ,
G(s) = E[sX] =
(i) Show that the mean and variance of X are equal to 2r.
(-)'s 2S
\
F
.
[5 marks]
(ii) Deduce the probability mass function of X and hence show that X is a negative binomial random variable.
Hint: You may use the result that
k-l+i
k-l
(CS)Z.
[5 marks]
Suppose that X has a binomial Bin(n,p ) distribution and Y has a Poisson Poi(X) distribution.
(i) Obtain the probability generating functions of X and
TC
[5 marks]
(ii) Using theresults in (i), show that thebinomialdistributionwith large n and small p with np being a constant can be approximated by the Poisson distribution with X = np.
[ 5 marks]
3. A manufacturer produces an electronic component whose lifetime, X , (in hours) has an exponential Esp(X) distribution. In (c) - (e), you may assume x = 0.001. Obtain the moment generating function of X . Hence, show that A-1 is the mean and also the standard deviation of X .
[ 5 marks]
Obtain an expression for the cumulative distribution function of X.
[2 marks]
Compute the probability that a randomly selected component has a lifetime between 500 hours and 1500 hours.
[2 marks]
L
%AS: STAT1000 Prin. of Stat. / 1801 P & S / 1007 AM-S / 0601 Phy. Sc. 3
(d) These electronic components are shipped to their customers in batches of 2000 items. Approximate the probability that a randomly selected batch contains at least 750 items whose lifetimes are betlween 500 hours and 1500 hours. [4 marks]
(e) A customer assembles this electronic componentt o a product and there is
a n operation cost associated with the lifetime this component. Suppose of thai the operation cost per item is given by
Find the mean and standard deviation of the operation cost per item. [7 marks]
4. Let X I ,X 2 ) . . . Xn form a random sample of size n from some proba.bility
)
distribution with mean p and variance 02. For (c) and (d), you may assume the probability distribution to be normal.
(a) Show that the sample mean and sample va.riance,
defined by
are unbiased for population mean p and population variance tively.
02, respec-
(11) If n is large enough, write down the distribution of State the theorem Vou use. Hence compute the probability that the sa.mp1e mean is within 1.5 standard devia.tions from the population mean. 15 marks]
x.
[6 marks]
(c) If o2 = 4 and t,he width t>he samplesize n.
of a 95% confidence interval for p is 0.98, find
[4 marks]
( d ) If we would like to reduce t,he width of the confidence interval in (c) by
at least 20%) how many extra observations do we need?
1 marks] 5
5. Supposetha,tthe coursework mark, X , andtheexaminationmark, Y , of a student in a large statistics class have the normal N ( p z ) ai) distribution and the N ( p v ,0:) distribution, respectively. A random sample of 1 2 students is talmn and t,heir- coursework marksandexaminationmarksarerecorded as follows:-
S&AS: STAT1000 Prin. of Stat. / 1801 P & S-/1007 AM-S / 0601 Phy. Sc. 4
X
Y
37.0
31.5
77.0
80.0
79.0 68.0 84.0 67.0 74.0 61.0
46.0
61.5 61.0 78.5 61.0 37.5 79.0 66.5 57.5 54.0
0 . ;
3
74.0 78.0 71.5 60.0
(a) Compute unbiased estimates for ps, py,: and 0
[3 marks] (b) Construct 95% confidenceintervals for the mean coursework mark and mean examination mark. [4 marks] (c) Is there any evidence that the mean coursework mark is higher than the mean examination markby at least 5 marks? Draw your conclusion based on the 0.05 level of significance. [6 marks]
(d) If the coursework mark contributes 25% to the final mark, W , of the course and the examination mark contributes 75%, construct a 95% confidence interval for the mean final mark. Without carrying out the formal procedures of hypothesis testing, test, at the 0.05 level of significance, whether the mean final mark is significantly different from 70 marks. [7 marks]
Note: In testing hypothesis, state clearly the hypotheses of interest, the distribution of the test statistic, the rule decision and your conclusion.
6. The ma.rlteting director of an insurance company would like to review the sales of an investment and life assurance product, the InvestLife. He found that out of 2340 policies sold, 585 of them terminated within one year of purchase. Let p1 be the population proportion of policies that terminated within one year. State the distribution of the estimator of the true proportion of policies that terminated within one year. Construct a 99% confidence interval for pl. Is this interval exact? Comment.
[2marks]
[4 marks]
A similar product, the SuperGrow,of one of its vivals had a sales of 4680 policies, of which 1053 policies terminated within one year. Is there any evidence that there is a significant difference in the termination proportions of the two products? Draw your conclusion based on the 0.05 level of significance.
The marketing director launched a campaign t o promote the product, InvestLife. After the promotion period, the company had sales of 5550 policies, of which 999 terminated within one year.
[6marks]
S&AS: STAT1000 Prin. of Stat.
/ 1801 P & S /
1007 AM-S / 0601 Phy. Sc. 5
(i ) Is the campaign successful to improve the termination proportion of the InvestLife? Draw yourconclusionbased on the 0.01 level of significance. [5 marks]
Note: In testing hypothesis, state clearly the hypotheses of interest,the distribution of the test statistic, the decision rule and your conclusion.
of ( i i ) If a confidence interval of the termination proportion the InvestLife afterthepromotionperiod is given by (0.168, 0.192), what is the corresponding level of confidence? [3 marks]
%AS: STAT1000 Prin. of Stat. / 1801 P & S / 1007 AM-S / 0601 Phy. Sc. 6
Table
Standard Normal Curve Areas
0
z
z 0.0 0.1 02 . 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.o
1.1 I .2
0.00
0.01
0.02
05080
0.03
0.55 17 0.59 10 0.6293
0.04
0.5 160 0.5557 05948 -0.6331 0.6700 0.7054 0.7389 0.7704 0.7995 0.8264
0.05
0.06 05239 0.5636 0,6026 0.4" 0.6772 0.7 123 0,7454 0.7764 0.805 1 0.83 15 0.8554 0.8770 0.8962 0.91 3 I 0.927 8 0.9406 0.95 15 0.9608 0.9686 0.9750 0.9803 0.9846 0.988 1 0.9909 0.993 1 0.9948 0.996 1 0.997 1 0.9979 0.9985 0.9989 0.9992 0.9994 0.9996 0.9997
2. I 2.2
2.0
1.8 1.9
1.3 1.4 1.5 1.6 1.7
2.3 2.4
,
2.9 3 .O 3.1 3.2 3.3 3.4
2.5 2.6 2.7 2.8
-
0.5398 0.5793 0.6179 Q.6554 0.69 15 0.7257 0.7580 0.7881 0.8 159 0.8413 0.8643 0.8849 0.9032 0.91 92 0.9332 0.9432 0.9554 0.964 1 0.97 13 0.9772 0.9821 0.9861 0.9893 0.99 18 0.9938 0.9953 0.9965 0.9974 0.998 1 0.9987 0.9990 0.9993 0.9995 0.9997
0.5000
0.5040 0.5438 0.5832 0.6217 0.659 1 0.6950 0.729 1 0.76 11. 0.79 10 0.81 86 0.8438 0.8665 0.8869 0.9049 0.9207 0.9345 0.9463 0.9564 0.9649 0.97 19 0.9778 0.9826 0.9864 0.9896 0.9920 0.9940 0.9955 0.9966 0.9975 0.9982 0.9987 0.999 1 0.9993 0.999s 0.9997
05978 0.5871 0.6255
05 120 .
0.6628
0.6985 0.7324 0.7642 0.7939 0.8212 0.8461 0.8686 0.8aa8 0.9066 0.9222 0.9357 0.9474 0.9J73 0.9656 0.9726 0.9783 0.9830 0.9868 0.9898 0.9922 0.994 1 0.9956 0.9967 0.9976 0.9982 0.9987 0.999 1 0.9994 0.9995 0.9997
0.8907 0.9082 0.9236 0.9370 0.9484 0.9582 0.9664 0,9732 0.9788 0.9834 0.9871 0.990 1 0.9925 0.9943 0.9957 0.9968 0.9977 0.9983 0.9988 0.999 1 0.9994 0.99% 0.9997
0.8708
066 .64 0.701 9 0,7357 0.7673 0.7967 0.8238 0.8485
0.8729 0.8m 0.9099 0.925 1 0.9382 0,9495 0.959 1 0.967 1 0.9738 0.9793 0.9838 0.9875 0.9904 0.9927 0,9945 0.9959 0.9969 0.9977 0.9984 0.9988 0.9992 0.9994 0,9996 0.9997
0.8508
05 199 . 0.5596 05987 0.6368 0.6736 0.7088 0.7422 0.7734 0.8023 0.8289 0.853 1 0.8749 0.8944 0.9115 0.9265 0.9394 0.9505 0.9599 0.9678 0.9744 0.9798 0.9842 0.9878 0.9906 0.9929 0.9946 0.9960 0.9970 0.9978 0.9984 0.9989 0.9992 0.9994 0.9996 0.9997
0.07 0.6064
0.5279 0.5675
0.08
0.09
0.6443 0.6808 0 7157 . 0.7486 0.7794 0.8078 0.8340 0.8 577 0.8790 0.8980 0.9 147 0.9292 0.94 18 0.9525 0.9616 0.9693 0.9756 0.9808 0.9850 0.9884 0.99 I i 0.9932 0.9949 0.9962 0.9972 0.9979 0.9985 0.9989 0.9992 0.9995 0.9996 0.9997
0.53 19 0.5714 0.6 103 0.6480 0.6844 0.7190 0.75 17 0.7823 0.8 106 0.8365 0.8599 0.88 10 0.8997 0.9 162 0.9306 0.9429 0.9535 0.9625 0.9699 0.976 1 0.98 I2 0.9854 0.9887 0.99 13 0.9934 0.995 I 0.9963 0.9973 0.9980 0.9986 0.9990 0.9993 0.9995 0.9996 0.9997
0.5359 05753 0.6141 0.65 17 0.6879 0.7224 0.7549 0.7852 0.8133. 0.8389 0,862 1 0.8830 0.901 5 0.9 177 0.93 19 0.9441 0.9545 0.9633 0.9706 0.9767 0.98 17 0.9857 0.9890 0.9916 0.9936 0.9952 0.9964 0.9974 0.998 1 0.9986 0.9990 0.9993 0.9995 0.9997 0.9998
S&AS: STAT1000 Prin. of Stat. / 1801 P & S / 1007 AM-S/0601 Phy. Sc. 7
Table
Crttiul V J u u f,,
for the t
Distribution
~-
a
V
.IO
1 2 3 3 -078 1.886 1.638 1.533 1.47 6 1.440 1.415 1.397
.05
.025
12.706 4.303 3.182 2.776 2.57 I 2.447 2.365 2.306 2.262 2.228 2.20 I 2.179 2.160 2.145
2.131 2.120 2.110 2.101 2.093
.01
.MIS
.#1
318.31
.0005
4
2.353
2.132
2.0 15 1.943 1.a95 1.860 1.833
6.3 14 2.920
3 I .821
3.747 3365
6.965 4.541
63.657
4.604
9.925 5.84 1
2.2 236
10.213 7,173
636.62
9
10 12 13
14
5 6 7 8
1.383
3.143 2.998 2.896 2.82 1 2.764
4.032 3.707 3.499
11
1.363
1.372
21 22
24
20
15 16 17 18 19
1.341 1.337 1.333 1.330 1.328 1.325
1.321
1.350 1 .345
1.356
1.812 1.796 1.782 1.77 1 1.76 1 1.753 1.746 1,740 1.734 1.729 1.725 1.72 1 1.717 1.714 1.711 1.708 1.706 I .703
1.323
1.319 1.318
1.3 16 1.313
23
i 5 26
28 29 30 40 60 I20
I
27
1.3 15 1.3 14 1.311 1.310 1.303 I .296 I .289 1.282
1.701 1.699
2.060 2.056 2.052 2.048 2.045
2.064
2.080 2.074 2.069
2.086
2.68 1 2.650 2.624 2.602 2.583 2.567 2.552 2.539
2.528 2.5 18 2.508 2.500 2.492 2.485 2,479 2.473 2.467 2.462
2.457 2.423 2.3 90 2.358 2.326
2 71a .
3.169 3.106 3.055 3.012 2.977
3.355 320 .5
4.785 4.501 4.297
5.893 5.208
3 1598 12.924 8.610 6.869 5.959 5.408 5.041
4.781
4.144 4.025 3.930 3.852 3.787
4.587 4.437 4.3 18 4.22 1 4.140 4.073 4.015 3.963 3.922 3.850 3.819 3.792
2.845
2.947 2.92 1 2.898 2.878 2.861
3.646 3.610 3.579
3.552. 3.527 3.505 3.485 3.467 3.450 3.435 3.42 1 3.408 3.396 3.385 3.307
3.686
3.733
3.883
2.83 1 2.81 9 2.807 2.797 2.787 2.779 2.77 1 2.763 2.756
2.750
3.745
3.767
3.674 3.659
3.690
3.725 3,707
1.697 1A84 1.671 1.658 1.645
2.012 2.021 2.000 1.980 1.960
2.660 2.617
2.704
3.646
3.232
3.090
3.160
3.551 3.460 3.291
2.576
3.373