MATH 355
Test 3 My. D ‘
Student Number: '. " .7 _
Show all your work. The total is 110 points.
1. (10 points) The annual amount Spent for maintenance and repairs in your house has an
approximately normal distribution, with a mean of $400 and a standard
FOUNDATIONS OF PROBABILITY: EXAM 1. NAME:
Ex 3 (a) State the definition of probability (3 axioms).
?r"\okhl\
usa.
k^o-t.:*-
a^,'w
t'
C&,t^*^tt a,w wa^+\ A.S ) scfw_.
r) iPC,ql > o
e\ FG) =t
a) iq A,.A., .,: ) ^i:*
pf m^fu*-lt,X z*e),."t'uuert'"^
s4y-a/\cx
2.27
2.28
The sample space is (SS, SR, SL, RS, RR. RL, LS, LR, LL}. Let A be the event of at least one vehi-
cle turning left and B be the event that at least one vehicle turns. Assuming equally likely out—
comes. we have
PMS) _ Pm) _ P(SLURLULSULRULL] 5/
3.2
3.3
a. P(X:0)= [ number of ways of no chosen household head]
choosing 0 out of 4 has incume ($18, 000
4
431014.
Rxal): [ number of ways of J [ only one given household head of J
choosing 1 out of 4 those chosen has income < $13,000
a
_ l
I
/'—"-\
—- 4
2.40 Let D =: event that the person has the disease. and T = event ofa test result positive for the disease.
P(D)P{71 D) _ {0.01) (0.90)
P(D)P{710)+P(5)P(n5) ' (0.01) (0.90) + (0.99) (0.10)
P (pl 73 2 — 0.0333 2.44 Assuming that each of the 365 days in
)"X 756' éofﬁv‘m‘, 7"”?
Warning: You must write/explain each step unless it is obvious or you CAN do it
mentally. If you don’t do so, you are risking point deduction unless you can
prove to me that you can do that step mentally in my office.
1. (5) Let Y
(4 points)
Each day, a woman will visit one politically oriented on line chat room among four that she likes.
She is equally likely to visit chat room 1 as chat room 2, and she is twice as likely to visit each of
these as she is to visit each of chat room
Warning: You must write/explain each step unless I think (not you think) it is
obvious or you CAN do it mentally. If you don’t do so, you are risking point(s)
deduction unless you can prove to me that you can do the missing step(s)
mentally in my office.
115* L
(15 pts) A survey classified a large number of adults according to whether they were diagnosed
as needing eyeglasses to correct their reading vision and whether they use eyeglasses when
reading. The proportions falling into four resulting categorie
Chapter 2: Foundations of probability
Oleksandra Beznosova
January 13, 2015
First denitions: Sample space
First denitions: Sample space
A sample space S is a set that includes all possible outcomes
of a random experiment listed in a mutually exclusive an
FOUNDATIONS OF PROBABILITY: HOMEWORK 3.
Ex 3.27 A prociency examination for certain skill was given to 100
employees of a rm. Forty of the employees were men. Sixty of the
employees passed the examination (by scoring above a preset level for
satisfactory
3.11 Let X = number on the ticket drawn. and G!- = net gain for box i, i=1. II. Then G; = X — l.
2
a. mag = 2(x—1)p,(x) = (ux)£%]+o[ g')+1[ l J = 0
x=0
3
2
E(Gf) = 2(x—l)2p,(x) = 1L ]+o[ 1 ]+1[ 1 J =
Ixﬂ 3 3
we» = Etc?) — [H0012 =
Hal'—
IAIN
h. E(G”)
LAID
the method of moment generating functions to ﬁnd the new distribuiien. we have
6.14 3. Using
k
Mm) = me“) = HELIX") =
E: |
k
a: = H (E (ex"f) (since Xi’s are independent)
= I
= (MK (r) ) " (since Xl’s are identically distributed) = ( i ~ p 4- pa) ”"
By t
1 1 1/2 I l/Z 33‘] 15 21
5.9 a. P[X1<§,X2> Z ] = In [1/4(x|+x2]dx2dxl : J‘o [T4, 5 dxl = a
2
_x x
b- = I (11+x2)dx2dxl = [x1x24—Ez]
f,(x1}f2(x2) = [xl+ l ][x2+ ¢x1+x2=f(xl,x2)
forOSx, $1,03xgsi.
Therefore, X 1 and X: are not independent. 5.11 Note that
5.26 Let Y. = number of family homes ﬁres, Y2 = number of apartment ﬁres, and Y3 = number of ﬁres in
other types of dwellings, out of 4 ﬁres. Then (Yl, Y2, Y3} has a muitinomial distribution with
parameters :1 = 4. pl = 0.73. p2 = 0.2, p3 = 0.07. Thus
H!
4.57 Let X = amount spent on maintenance and repairs. Then X has a normal distribution with parame-
ters ti = 400, (I = 20 and
_ X_400 450-400
P(X>450) — P[ 20 >_20 ]
z P(Z>2.5) = 0.54.4933 = 0.0062.
— set
4.58 Letxo = budgeted amount. Then P (X>.re) =