# 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

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.

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

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

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
.