JOSEPH NAUS
STATISTICS DEPARTMENT
RUTGERS UNIVERSITY
FINAL EXAM
IS ON TUESDAY DECEMBER 16 AT 8 AM . ROOM :
FRELINGHUYSEN B5
(Room B5 is directly below our regular room)
Bring working calculator, student ID.
960:379:01 Basic Probability &
Statistics
Fall 2003
Dr. Naus
Frelinghuysen A5 CAC W 11:30-12:50, Fri
1:10-2:30
571 Hill Center Office Hours
Wed 4-5 pm
Ph: (732)445-2695
e-mail: naus@stat.rutgers.edu
Teaching Assistants : Hailing
Jiang Office hours Tuesday 4:30 to 5:30
; Thursday 4:30 to 5:30
Office Room 557 Hill Center
You can also go to any
of the following Teaching assistants for help: (all in Hill Center)
Jiangang Fang, Room 553
Thursday 12:50-2:50
Qiankun Sun Room 553
Friday 3:30 to 5:30
Jue Wang Room 555
Tuesday 12:30 to 2:30
Zhaoyu Luo Room 457
Tuesday 4:30 to 5:30 , and Thursday 1 to
2
answers to stapled practice exam:
1a) 0.2938 b) 0.0956
2)a) 0.2584 b) 0.6981 3)a)77 b) 30 c) 91
4) 112 5) 4290
6a) 0.4 b) 0.3 7a) 1/80 b) 0.34375 c) 6.04167
d) 7.2482 8) 0.11 9a) 540 b) 1215 c) 3375
d) 810
10) 0.06040 11a) 330 b)
60 12a)0.2029 b) 212.20 c) 0.0853 13a)
12.5 b) 12.5 c)0 d) 25
14) a) 0.3056 b) 0.05556 c) 90,000
answers to long sheet practice exam
1) a) 0.646 b) 0.3830 c) 5.44 d) 0.4173 e) 0.4013
2a) 2.25 b) 0.3375 c) 0.03704 d) 0.2593
3) a) 5 b) 0.01011 c) 0.1755 (you
can show that if X and Y are independent Poisson random variables, that
X+Y is a Poisson random variable. Given this, you can easily
solve parts c and d)
d) 5 e) 66 4) a) 0.00243 b) 0.04261
c) 8.5 d) 5.25 e) 0.85 f) no
5) a) 0.38095 b) 0.625 6a) 0.001389 b)
0.36806
(note that 6a) is just 1/6! = 0.001389, which is straightforward. To do 6b) is much more challenging. Let Ei denote
the event that Box i is labeled correctly. The probability that no
box is labeled correctly = 1 - P(UEi) . Now apply
the principal of exclusion and inclusion: P(UEi) = Sum P(Ei) - Sum P(Ei
and Ej) + SumP(Ei and Ej and Ek) - ....(and so on)
By symmetry P(E1) = P(E2)= ...=P(E6) = 5!/6!, so that Sum P(Ei) = 6 (5!/6!)=
1
Similarly P(E1 and E2) = P(E1 and E3) =... = 4!/6!. There are 15
possible pairs (combinations of 2 from 6), so
that SUM P(Ei and Ej) = 15 (4!/6!), and so on.).
Text: Anthony J. Hayter Probability
and Statistics for Engineers & Scientists,
Second Edition., Duxbury Publishers 2002.
Week
Reading Problems
(turn in even*, problems)
Due
Sept 3-5
1-20 p. 8/5, 6*, 7, 9 ;
p. 14/3, 10*, 11, 12*
Sept 10
Sept 10-12 21-43
p. 36/2*, 3, 5, 6*, 11, 12*, p.43/1, 6*,8*, 9
Sept 17
Sept 17-19 44-68
p.54/1,2*,8*,9; p. 60/1,4*; p.
69/1,3,4*,8*,11,18* Sept 24
Sept 24-26 73-96 p.
84/3, 5, 8*; p. 96/1, 3, 4*, 5, 6*
Oct 1
Exam 1: Oct 3rd : Material thru page
70.
Oct 1-3
98-122 p. 108/2*, 7, 11 ; p. 122/1, 6*,
10*
Oct
8
Oct 8-10 123-152 p.
139/3, 4*, 6* ; p. 152/1, 3, 4*, 6*
Oct 15
Oct 15-17 156-179 p. 170/1,2*,
4*, 9; p. 179/1, 2*,3, 6*
Oct 22
EXAM 2: October 24th (Cover material thru p.
152)
Oct 22-24 180-197 p. 184/1, 3,
4* ; p. 192/4*, 6*, 7 ; p. 197/1, 4*
Oct 29
Oct 29-31 199-211, 232-246 p. 202/1,
2* ; p. 211/2*, 3, 4* ; p. 247/1,3,4*,6* Nov 7
Nov 5-7 248-270
p. 258/1, 2*, 3 ; p. 271/1, 2*, 3, 5
Nov 14
Nov 12-14
Exam 3: November 19th (covering material thru
p. 246)
Nov 19-21 271-283
p. 284/5,7
Dec 3
Nov 26 (Note that Wednesday meets on Fridays
scheduled time: 1:10-2:30)
Dec 3-5 286-313
p. 313/1 ; p. 291/1, 4*
Dec 10
Dec 10 Last day
of class. Review
Tuesday Dec 16
8am-11am FINAL EXAM
Exams 1,2,3 are each worth 20 points. Final Exam
worth 40 points.
Homework: The answers to the odd numbered problems are in the
back of the book.
Only hand in the even (*) numbered problems on date due.
You do not have to hand in homework, but if you do, it should be done
independently, and neatly, and should show work as well as answers.
It should be done by due date (no late homeworks accepted). Students
who have turned in acceptably at least 8 of the 11 homeworks, will
be given the opportunity to answer bonus questions (worth a total of 20
points) on the final exam.
.Answers to assignment 1:
p. 8/6 Sample Space: { (red, dull) ;
(red, shiny) ; (blue, dull); (blue, shiny)
p. 14/10 a) 0.42 b) 0.18 c)
0.58 = 1 – 0.42 d) 0.66 = 0.24 +
0.39 + 0.03
p. 14/1212 the event two heads in succession occurs in 3 out of
8 possible outcomes: (hht ; thh ; hhh)
3/8
Answers to assignment 2:
p. 36/2 a) 0.46 b) 0.18 c) 0.61 d) 0.07
e) 0.88 f) 0.38 g) 0.72 h) 0.43 i) 0.30
j) 0.13
p.36/6 P(B) is between 0.3 and 0.9
p. 36/12 P(red U Shiny) = 0.575 ; P(Dull AND
Blue) = 1 - 0.575 = 0.425
p. 43/6 P(on time AND satisfactory) = P(on time)P(satisfactory
| on time) = 0.77(0.85) = 0.6545
p. 43/8 in a 4-year period there are 1461 days = (365+365+365+366).
There are 4x12 = 48 days that fall on first of month . P(falls
on 1st of month) = 48/1461 = 0.0329
There are 4(31) = 124 days in March in the four year period, and
of these 4 are on the first of the month.
P(falls on 1st of month | it falls in March) = 4/124 = 1/31 = 0.0323
There are (28+28+28+29) = 113 days in February in the four year
period, and of these 4 are on the first of the month
P(falls on 1st of month | it falls in February) = 4/113 = 0.0354
Answers to assignment 3:
p. 54/2 a) (12/52)(12/52) (with replacement) ; without
replacement (problem 1) (12/52)(11/51). So probability is greater
with replacement)
b) (26/52)(26/52) = 0.25 (with replacement) > answer with out
replacement (which is (26/52)(25/51)
c) P(one care is red, the other black) = P(lst is red AND 2nd Black)
+ P(lst is Black AND 2nd is red)
= (26/52)(26/52) + (26/52)(26/52) = 0.5. This is
less than answer without replacement = 2(26/52)(26/51)
p. 54/8 we did this in class: P(at least 2 people out of n share
a birthday) = 1 - p(no 2 people share a birthday)
= 1 - {(364/365)(363/365)...(366-n)/365).
p. 60/4 P(tagged) = {0.1 x 0.45) + (0.15 x 0.38) + (0.5 x
0.17) = 0.187
P(specieis 1 | tagged) = {0.1 x 0.45)/0.187 = 0.2406
P(species 2 | tagged) = (0.15x0.38)/0.187 = 0.3048 ;
P(species 3 | tagged ) = 0.4545
p. 69/4 numbe of full meals = 5 x 3 x 7 x 6 x8 = 5040
bumber of meals with just soup or appetizer = (5 x 7 x 6 x8) +
(3 x 7 x6 x8) = 2688
p. 69/8 100 ! / 97 ! 3! = 100 (99)
(98)/6 = 161,700
b) 91,881 c) 91881 / 161700 = 0.568 d) 17 {83
x 82)/2 = 57851
e) number of samples with 1 or 0 broken lightbulbs = 91881 + 57851
= 149732
P(sample contains no more than 1 broken bulb) = P( 0 or 1 broken
bulb) = 149732/161700 = 0.926
p. 69/18 Let C(r,N) = N!/r!(N-r)!
a) C(12, 43) / C(12,60) = 0.0110
b) C(3,25) x C(3,12) x C(3,5) / C(12,60) = 0.00295
p. 84/8 x -1 0
1 3
4 5
p(x) 1/6
1/6 1/6 1/6 1/6
1/6
F(x) 1/6
2/6 3/6 4/6 5/6
6/6
p.96/4 b) P(X LE 2) = F(2) = 1/4 c)P(1 LE X
LE 3) = F(3) - F(1) = (9/16)-(1/16) = 8/16
d) f(x) = dF(x)/dx = x/8 for 0 LE x LE 4
p. 96/6 a) A = 5.5054 (found by setting integral
of f(x) to 1, and solving for A)
b) F(x) = 5.5054{0.5x - 0.33333((x-0.25)^3) - 0.06315}
for 0.125 LE x LE 0.5.
c)F(2) = 0.203
p. 108/2) E(X) = {1x1x(1/36)} +{2(1x2)x1/36} + {2(1x3)x1/36}
+ {[(1x4)+(4x1)+(2x2)]x1/36}+.....
= 1(1/36) + 2(2/36) + 2(3/36)+ 4(3/36)
+ 5(2/36) + 6(4/36) + 8(2/36) + 9(1/36)+.....+ 36(1/36) = 12.25
p. 122/6 if c.d.f. is F(x) = x^2/16 for 0 <
x < 4 , then probability density function f(x) = dF(x) dx
f(x) = x/8 , for 0 < x < 4. To find E(X)
integrate x f(x) ; to find E(X^2) integrate (x^2) f(x)
E(X) = Integral (between 0 and 4) x (x/8) dx = (x^3)/24 evaluated
at x=4 less (x^3)/24 evaluated at x=0
= (4^3)/24 = 64/24 = 8/3
E(X^2) = Integral (between 0 and 4) x^2 (x/8) dx = (x^4)/32 evaluated
at x=4 less (x^4)/32 evaluated at x=0
= (4^4)/32 = 256/32 = 8
Var(X) = E(X^2) - {E(X)}^2 = 8 - {(8/3)(8/3) = 8 - (64/9) =
8/9
b) standard deviation = square-root of (8/9) = 0.94
c) to find the lower quartile, find the value x, for which F(x) =
0.25
F(x) = (x^2)/16 = 0.25 , gives (x^2)
= (16)(0.25) = 4 ; or x = 2
since F(2) = 0.25 , 2 is the lower quartile (or
25th percentile).
To find the upper quartile, find the value x, for which F(x) = 0.75
(x^2)/16 = 0.75 , gives (x^2) = 12 , or
x = square-root of 12 = 3.46
since F(3.46) = 0.75, 3.46 is the 75th percentile (or
uper quartile).
d) The interquartile range is the upper quartile - lower quartile
= 3.46 - 2 = 1.46
p. 122/10)By Chebyshev's inequality, 75% of the times should be between
the mean less 2 standard deviations to the
mean + 2 sd = 75.0 - 2(7.3) to 75.0 + 2(7.3) or:
60.4 < X < 89.6
89 % of times should fall between the mean - 3 sd to mean + 3 sd
, or 75.0 - 3(7.3) to 75.0 + 3(7.3),
or 53.1 < x < 96.9
p. 139/4a,b) Y
0 1
2
3 marginal of X
X
0 1/16
1/16 0
0 2/16
1 1/16
3/16 2/16 0
6/16
2 0
2/16 3/16
1/16 6/16
3 0
0 1/16
1/16 2/16
marg Y
2/16 6/16
6/16 2/16
c) P(X=0 , Y= 0) = 1/16 ; P(X=0) = 2/16
, P(Y=0) = 2/16 : P(X=0, Y=0) does not = P(X=0)P(Y=0)
so X and Y are not independent
d) E(X) = 1.5 ; E(X^2) = 3 ; Var(X)
= 3 - (1.5 ^2) = 0.75
by symmetry, Y has E(Y) = 1.5 ; Var(Y) = 0.75
e) E(XY) = 44/16 ; COV(X,Y) = E(XY) - E(X)E(Y) =
(44/16) - (1.5)(1.5) = 0.5
f) P(X=0 | Y =1) = P(X=0, Y=1)/P(Y=1) = (1/16)/(6/16) = 1/6
P(X=1 | Y=1) = (3/16)/ (6/16) = 3/6 = 0.5; P(X=2 | Y=1) = 1/3;
P(X=3|Y=1) = 0
E(X|Y=1) = (0 x 1/6) + (1 x 0.5) + (2 x 1/3) + (3x0) = 7/6
E(X^2 | Y=1) = (0 x 1/6) + (1 x 0.5) + (4 x 1/3) + (9x0) = 11/6
Var(X | Y=1) = (11/6) - {(7/6)(7/6)} = 17/36
p. 139/6 a,b)
Y
0 1
2
marginal of
X
X
0 25/102
26/102 6/102
57/102
1 26/102
13/102 0
39/102
2 6/102
0
0
6/102
marg Y
57/102 39/102
6/102
c) X, Y not independent: e.g., P(X=2, Y= 2) =
0 , which does not = P(X=2)P(Y=2) = (6/102)(6/102)
d) E(X) = 0.5 ; E(X^2) = 21/34 ; Var(X) =
25/68 ; E(Y) = 0.5 ; E(Y^2) = 21/34 ; Var(Y)
= 25/68
e) E(XY) = 13/102 ; COV(XY) = (13/102) -
(0.5)(0.5) = -25/204 ; f)Correlation(X,Y) = -1/3
g) P(Y=0|X=0) = 25/57 ; P(Y=1|X=0) = 26/57 ; P(Y=2|
X=0) = 6/57
P(Y=0 | X=1) = 2/3 ; P(Y=1 | X=1) = 1/3 ; P(Y=2
|X=1) = 0
p. 152/4 length = A1 + A2 + B ; E(length) = E(A1)
+E(A2) + E(B) = 37+37+24 = 98
Var(length) = (by independ) Var(A!) + Var(A2)+Var(B) = (0.7^2)+(0.7^2)+(0.3^2)
= 1.07
p. 152/6 E(average weight) = 1.12 kilograms
Variance(average weight = (0.03^2)/25 = 0.000036
; standard deviaition = 0.0012 kilograms
b) for 0.03/sqrt(n) LE 0.005, n GE (0.03/0.005)^2
= 36. so n = 36.
p. 170/2 a) 0.1147 b) 0.0004
c) 0.2785+0.3798+0.2330 = 0.8913 d) 0.00003085 e)
1.2 f)1.056
p. 170/4 n = 9, p = 0.09 : a) 0.1507 b) 0.1912
E(X) = 0.81
p. 179/2 a) P(X=5) = 0.2074 b) P(X=8) = 0.0464 c)P(X
LE 7) = 0.9037 d) 0.1792
p. 179/6 a) 2.703 b) 8.108 c) P(X LE 10) =
0.7794 (negative binomial p = 0.37, r=3 d)P(X=10)=0.0718
p. 184/4 P(5 men , 7 women|16
men , 18 women) = Combin(5 from 16)Comb(7 from 18)/Comb(12 from 34)=0.2535
P(5 men , 7 women|1600 men , 1800 women) approx = Prob (exactly 7 "successes"
in 12 trials | p = 1800/3400)
= P(X = 7 | n = 12, p = 18/34) = 0.2131 , where X is Binomially
distributed, with n=12, p = 18/34
p. 192/4 P(X=0) = e^(-2.4) = 0.0907 ;
P(X GE 4) = 1 - P(X LE 3) = 1 - P(X=0)-P(X=1)-P(X=2)-P(X=3)= 1
- {e^-(2.4)}{1+2.4+((2.4^2)/2)+((2.4^3)/6)}
= 0.2213
p. 192/6 use Poisson with Mean = 4. a)P(X=0) = e^-4 = 0.0183
b(P(X GE 6) = 1 - P(X LE 5) = 0.2149
p. 197/4 use multinomial distribution: E(number dead) = 22(0.08)
= 1.76 ; E(number slow) = 22(0.19) = 4.18
E(number medium) = 22(.42) = 9.24 ; E(number strong) = 22(0.31)
= 6.82
a) P(3,4,6,9) = {22!/3!4!6!9!}(0.08^3)(0.19^4)(0.42^6)(0.31^9)= 0.0029
b) (P(5,5,5,7) = {22!/5!5!5!7!}(0.08^5)(0.19^5)(0.42^5)(0.31^7)
= 0.00038
c) Number died is Binomial random variable with n = 22, p = 0.08. P(number
died LE 2) = P(X=0)+P(X=1)+P(X=2)
where X is Bin(n=22, p = 0.08), P(X LE 2) = 0.7442
p. 202/2 a) 1.515 b) 0.0491 c) F(x) = (x-1.43)/0.17 for
1.43 LE x LE 1.60 d) F(1.48)=0.2941 e) F(1.5)=0.412
p. 211/2 a) 10 b) 0.3679 c) 0.3935 d) , for exponential:
P(total wait time > 15| waited 10) = P(addit time>10)=0.3679
e) E(X) = 10, conditional distribution of additional w.t. U(0,15)
p. 211/4 a) F(5) = 0.7878 b) Binomial (n= 12, p = 0.7878) has E(X)
= np = 9.45, Var(X) = 2.01
c) P(B(12, .7878) LE 9) = 0.4845
p. 247/4 a) 0.9693 b) 0.7887 c) 0.9293 d) 0.2884 e)
0.0326 f) x = -4.4763 g) -4.7071 h) 2.18064
p. 247/6 solve P(X LE 10) = P((X-u)/sigma LE (10-u)/sigma) = P(N(0,1)
LE (10-u)/sigma) = 0.55
and P(X LE 0) = P(N(0,1) LE (0-u)/sigma) = 0.4 gives u =
6.6845 and sigma = 26.3845
p. 258/2 a) P(N(-1.9-3.3, 2.2+1.7) GE - 3) = 0.1326 b) 0.5533
c) 0.3900 d) 0.6842 e) 0.4781 f) 0.0648
p. 271/2 a)exact propbablity=0.0106, normal approx. 0.0079
b) exact prob = .6160, normal approx =0.6172
c) exact prob is .9667, normal approx = .9764 d) Exact prob
= 0.3410, Normal approx 0.3233
.
