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Nhu vfy, qua kh6u lya chgn biiSn ta dgOc
(*)
Y = -0,304+0,083x1 -0,031x3 +0,004x] +e.
Chuong
3
2:
G
BA lraP
1
0
-1
'
-2
$6.1.8Ar TAP CHTI0NG r
Hinh 5.9. PhAn du chuin ho6 theo quan sat c0a s6 tieu
dQ tan
Ki6m tra phan du cta m6 hinh ndy: Ching h4n, theo chi s5 i ta th6y c6
2 gi|triphAn du chu6n ho6 (tmg vdi quan s6t thri 6 vi thrl l0) vuqt qu6 2; vi
ph4m thir hai ld d; khA nhd t4i c6c quan s6t ll *24. Dir sao hai vi ph4m niy
!
(*)
ldm m6 hinh cu5i cirng.
c6 vi ph4m tl6ng kC. Ta lga chqn
cirng kh6ng a6n ndi nio. Pfr6n du chuAn ho6 x6p theo x1, x2 hay
ddu kh6ng
#
l.l.
The lifetime of a pC is defined
the tirne
it
as
takes for the pC to
break down.
a) Define a random experiment that
involves the lifetime ofthe pC.
b) What is the sample space?
c) Define two events that are disjoint.
d) Define two
events that have a
nonempty intersection.
1.1. Tu6i thg cria mQt chi6c pC
tfnh tir hic n6
bit
tlu-o.c
ttAu ho4t riQng
ttrin khi h6ng.
a) Xdc dinh ttri nghigm ngiu ntridn
gin v6i tu6ithg cria pC.
b) Kh6ng gian miu 6 d6y td gi?
c) X6c ttlnh 2 bi6n c6 xung khfc.
d) X6c tlinh 2 birin c6 c6 giao khric
trting.
D&
b)S = IR* = (0;
-);
c) (0;1000) vi (>2000);
d) (0;1000) and (900;2000).
1.2*. Customers access an Automated 1.2*. Circ khrich
hing lui t6i mqt chiiic
Teller Machine (ATM). They want
m6y rrit tidn tu rlQng. Hq muiin rrit
to withdraw random amounts of
mQt lugng ti6n ngSu nhi6n 50 ngdn
money in multiples of 50 thousand.
tl6ng mQt. Hiiy chi 16 kh6ng gian
VND. Specify the sample space. Is
m6u. Edy phrii chnng ld kh6ng
it the discrete one? Specift three
gian mdu roi r4c? Chi ra 3 bi6n ci5
events of interest.
quan
tim.
D,S. S = {50,1 00,...,1 04} ;
Dfrng; (<
t03), (t03; 5.103),
15.103+t04).
232
233
L3**. xdt thi nghiQm ngiu nhi6n tung
con xirc xic don I litr vi di5m s6
diu ch6m hi€n tr6n mdt. Gie sfr
courrting the number of dots facing
P({6})=0,3 vd t6t ci c6c m[t
tup. Assume that P({6})=0.3 and
1.3**. Consider the random experiment
of tossing a single die once and
all
other faces are equiprobable.
kh6c
,-
,.A
A
suat cua cac Dlen co
A={2,4,6}, B={1,5},
6 = {1. 2. 3, 4} ancl D= Au(BnC)
A= ?) 4,6\., a = {1, s} ,
g = {1, 2,3, 4},. D = A
Let P(A; = 0.9; P(B) = 0.8 . Show
that P(A n e) > o.z .
1.5*. Given that P(A) = 0.9:
a)
b)
0,58; 0,28; 0,56vd0,72.
P(A):0,9; P(B):0,8. Chitng
t6r[ng e(anB)>0,7.
1.5*. Gia
st
n(nn
P(A) = 0,9;
P(B):0,8;
B) = 0,75, tim
DS; a) 0,95; b) 0,15; c) 0,05:
1.6. Prove the Boole's inequality
\
i=t
1.7**. Consider tlre switching network.
It is equally likely that a switch
will or will not work. Find the
all
alternatives are possible;
khi ndng d6u c6 thii
h) B-E (Bose-Einstein) the
b) B-E (Bose-Einstein)
particles carrnot be distinguished,
all alternatives are possible;
thr3 phAn biQt
c) F-D (Ferrni-Dirac) - the particles
cannot be distinguished, at most
c) F-D (Fermi-Dirac)
one particle is allowed in a box.
chrla nhi6u
c6c
prob. that a closed path will exist
between terminals A and B.
-
kh6ng
phdn biQt
-
kh6ng th6
c c6c hgt, mQt hQp
ntr6t t tr4t.
duo.
n! n!(m-l)!
a) "' : b)
.
,n,-,
(m+n_l)1,
n!(m
n)!
-
.
1.9*. MOt thf nghiQm ngiu nhi€n c6
space S={a,b,c}. Suppose that
kh6ng gian miu S={a,b,c}. Gi6
P{a,c} = 0.75 and P{b,c}=0.6.
srlr P{a,c} =0,'15, P{b,c}=0,6.
1.9*. A random experiment has sample
Find the probabilities
of
the
elementary events.
1.7**. Xdt mOt mach tliQn nhu hinh v€.
Cdc c6ng tic it6ng hoic md v6i
Tim xic suSt cria c6c bitin cti so
cap.
oS. e(a) = O,+; P(b) =0,25;
P(c) = 9,35'
l.l0**. lf m
students born on
independent days in
adterrding a lecture.
D& 0,688.
'
lg93 are
Find the
probabilitiy that at least two of
them slrare a birthday arid show
that
p>l
,2
*n"nm=23.
ft0**.
Gi6 sri c6 m sinh vi€n sinh nrm
t993 ttang tham dU gio giAng. Tim
x6c su6t
itnhit2
sinh vi6n cirng c6
ngiy sinh vd chftng t6 ring
p
>1
khi m=23.
ESi
234
duo.'c;
ttucr. c cric het, t6t cd
khi ndng ddu c6 thiS dugc;
--]-l'm!
'[!^,)=r(o,)
khi nlng nhu nhau. Tim x6c su6t
el6 c6 it ra mQt dudng d5n gita 2
diu n5i A vi B.
tru6c (m5i hpt chi d trong
c6c trudng hgp sau:
a) M-B (Maxwell-Boltzmann) c6c hat coi ld kh6c nhau, t6t ci cdc
(Maxwell-Boltzmann)
g)
1.6. Chring minh b6t tling thric Boole
i
h4t
a) M-B
DS.
n
Ue, l=I(n,)
\i=r )
hQp chqn
cl e(nne).
c) P(AnB).
ta d4t ng6u nhi6n n
(ph6n t&) viro m > n hQp. Tirn x6c
su6t p cl€ c6c h4t du-o. c tim th6y d n
I h$p). X6t
r(nuB);
b) r(e - n);
P(A-B);
Pl
each
al
P(AuB):
(n
in
1.8. Chtng
box). Consider the following cases:
the particles are distinct,
u(Bn.C).
1.4. Cho
0.8: P(A n B) = 0.75, find
P(B) =
rS;
preselected boxes (one
ddng khn ndng. Tim x6c
Find the probability of the events
.
1.4.
li
1.8. We place at random n particles in
m > n boxes. Find the probability p
that the particles will be found in n
l-
36s!
(365
-
m-)!365*
235
l.l1**. A train
and a bus arrive at the
station at random between 9 A.M.
and l0 A.M. The train stops for l0
rninutes and the bus for a minutes.
,1.._
1.ll**. Tiu ho6 vd xe bus tdi ga t4i m6t
thoi di,3m ngSu nhi6n tir 9 tl6n l0
gid. Tdu dirng trong l0 phrit cdn xe
bus dirng a phrit. Tim a tl6 xric su6t
xe kh6ch vi tdu ho6 g{p nhau
bing 0,5.
l.ind a so that the probability that
the bus and the train will
meet
\
'r
1.17**. Consider the experiment of 1.17**. Xet thi nghiQm tung 2 con xirc
throwing the two fair dice. You are
xic c6n aOi. Si6t ring tting c6c n5t
then informed that the sum is
l6n hon 3.
greaterthan 3.
a) Tim x6c su6t biiin c6 2 mdt
gi6ng nhau khi kh6ng biiit th6ng
a) Find the probability of the event
equals 0.5.
D.s.
1.12. A fair die is rolled two tirnes. Firrd
oS:
We have two coins; the first
l.
is
1.13*. C6 2 tt6ng tiBn, mQt cdn d6i, mQt
c6 2 mf;t s6p. Rtit ngiu nhi6n I
d6ng tidn, tung n6 2 lin vd d6u
hign m4t s6p. Tim xdc su6t ddng
it nvice and heads shows
both times. Find the probability
vv€ toSS
that the coin picked is fair.
tirSn
rtt
PS:
l.
b) Find the probability of the same
v6i th6ng tin dl cho.
duo.
manufacturing plants
produce sirnilar parts. Plant I
produces 1000 parts, 100 of which
are defective. Plant 2 produces
2000 parts, 150 of which are
defective. A part is selected at
random and found to be defective.
c ld ddng ti6n cdn d6i.
5
plant
that n(nln) defined by 1.I4. Chring t6 ring e(ele) theo
Eq. (1.2.1) satisfies the three
(1.2.1) thod mdn 3 ti€n d6 cta xdc
lwo
b) Tim x6c su6t cfra bii5n c6 tr€n
ps,
1,1.
6'33
\
xu6t.
I?
D& 0,4.
1.14. Show
axioms of a probability, that is:
dE n6u.
1.18**. Hai nhi m6y s6n xudt nhirng
linh kiQn giting nhau. Nhd m6y I
sin xu6t 1000 linh kiQn, trong d6
c6 100 li h6ng. Nhd m6y 2 san
xu6t 2ooo linh ki€n, trong d6 c6
150 la h6ng. Chgn ng5u nhi6n I
' linh kiQn vi th6y ring n6 bi h6ng.
Tim x6c suSt n6 do nhi m6y I sin
What is the prob. that it came from
1.18*".
6
two-headed.
We pick one of the coins at random,
tin
event with tlre information given.
xic cdn A5i Z tin.
Tim x6c sudt aii t6ng s5 n6t bing 7.
is 7.
fair and the second
phirt.
1.12. Tung con xric
the probability that the sum of dots
l.l3*.
60-J100
that two faces are the same without
the information given.
.V
1.19**.
su6t, d6 ld:
A lot of
100 semiconductor chips
contains 20 that are defective. Two
1.19**. L6 hdng 100 chip bdn din c6
chira 20 clrip b! h6ng. Chgn ngiu
ay
n(nle)>o;
ay
e(nln)>o;
clrip.s are selected at random,
uy
e(sla)=
uy
n(sle)= r ;
replacement, frorn the lot.
a) X5c su6t chitic thri nhSt b! h6ng
a) phat is the probability that the
li
first one,selected is defective?
b) X6c su5t chii5c thti 2 b! h6ng li
bao nhi6u, bii5t ring chii5c thfr nh6t
bi h6ng?
r
;
c) n(n,t.lA2lB)=
c; P(A,r-,rerln)=
r(e, ls)+ r(e,In)
if
A'1n Az=A
1.15*. Show that
if
P(AlB)>P(A)
then
P(BlA)>P(B).
if
P(A), P(B)
rhen e(ele)> r(sle).
1.16. Show that
e(n,la)+ n(nrle)
;r,
niSu A1
b) What is the probability that the
r\A2=A.
1.15*. Chirng minh ring
second one selected
niSu
e(ele)>n(e)
thi P(B|A)>P(B).
ring n6u
P(A) > P(B)
1.16. Chung minh
thi P(AIB)> e(nla).
nhi6n 2 chi6c kh6ng
without
'
is
defective
bao nhi6u?
giverr that the lth on" was defective?
c) X6c suSt d6
c) What is the probability that both
bao nhi6u?
are defective?
l{p lai.
ci
2 chiiSc bi h6ng ld
D,S: a) 0,2; b) 0,192; c) 0,0384.
v--
I contains 1000 bulbs of 1.20**. HQp I g6m 1000 b6ng ddn,
trong d6 lO% bi hdng. H$p 2 g6m
which l0 percent are defective.
2000 b6ng, trong d6 5% bi h6ng.
Box 2 contains 2000 bulbs of
1.20*,'. Box
which 5 percent are defective. Two
bulbs are picked from a randomly
'.
du-o.
duo.
c rirt ra t& mQt h6p
c chgn ng6u nhi6n.
selected box.
a) Tim x6c su6t
a) Find the probability both bulbs
h6ng.
are defective.
b) Gie st} ring cd 2 b6ng ddu bi
h6ng, tim *6c su6t d,6 chirng ttu-oc
b) Assuming that both are defective,
find the probability that they came
fi'om box
l:
1.21**. SLrppose that laboratory test to
detect a certain disease has the
:
,
n(nla)
B*
B
= o,ee;
n(n[)
{krit qui ki6m tra duorg tinh}
%o
ddn si5 bi bQnh niy.
bi6t rang kiSt
is the probability that a
bQnh,
qui ki6m tra li duong
7, respectively. Catch one mouse
frorn the first cage and one mouqe
from the second cage then 6rass
them to the third cage. Finally
catch one mouse from the third
cage. Find the probability that this
mouse is male.
238
binary
communication channel. The
channel input symbol X may
:rssume the state 0 or the state l.
Because of the channel noise, an
1.23*. Xet k€nh th6ng tin nh! ph6n. DAu
viro X cfia k6nh
trang th6i
du-o.
c xem nhu 6 2
0 ho4c l. Do c6 nhi6u
diu ra 0 c6 thii ring
k6nh truydn,
and vice versa.
v6i dAu viro I vi ngucr.c l4i. K€nh
dugc tl{c truorg bdi x6c su6t
truydn k€nh po,Qo,pr, Ql, x6c
The channel is characterized by the
tlinh theo
channel transition probability
Po,Qo,Pl, and ql, definedbY
o, = R(vrl*o),p' =n(volxr),
irrput 0 may convert to an output I
oo =
oo
e(lr l*o),
p, = R(volxr),
=e(volxs), andql =R(vrlxr),
here xs and x1 denote the events
=0)
and (X
= l),
respectively,
and y6 and y1 denote the events
qo = P(vo
lxo),q, = P(y, l*, )
trong d6 xs
vi
x1
kf
hiQu biiSn ci5
(X = 0) vi (X = l), tuong ring; yo
vi y1 kf hi€u bii5n c5 1v=0) vi
(Y = l) tuong ring. Chir f ring,
p0+q0=l:pr+ql.DAt
vi pt = 0,2.
(Y = 0) and (Y = l), respectively.
P(xo) = 0,5, po = 0,1
Note that po + qo = I = Pl * q1. Let
P(xo) 0.5, po 0.l, and pt = 0.2.
a)
a) Find P(ye) and P(y1).
(de) la tr4ng thrii cfia dAu vdo?
:
:
b)
If a 0 was observed at the
Tim P(ys) vn P01).
b) Ni5u thdy 0 d diu ra, xac suit tI6 0
c) NiSu th5y 1 d diu ra, x6c su6t d6
tinh.
output, what is the probability that
I (de) Ia tr4ng th6i cta tliu
D,S..0,165.
a 0 was the input state?
d) Tfnh x6c su6t sai
c) lf a I
DS. a) 0,55;0,45; b) 0,818;
ry
1,22**. We have two cages of 1.22t't'. C6 2 l6ng chuQt thi nghiQm,
experimental mice. In the first
I6ng thrl nh6t c6 l0 con chuQt dgc
there are l0 male and 15 female
ones; in the second there are 8 and
1.23*. Consider the
(X
= o,oo5
Tinh x6c su6t m6t ngudi bi
and 0.1 percent the population
person has the disease given that
the test restrlt is positive?
:
vit 0,1
= o.oo5
actually has the disease.
What
I.
vdi A = {ngudi kiiim tra c6 bQnh},
{event that the test result is
positive). It is known that
r(ale)
a) 0,0061 9; b) 0,80; c) 0,08
thu dugc kiSt qui sau ddy:
iras the disease)
= o.ee;
D$
1.21**. Gid st ring, bing x€t nghiQm tt6
phet hiQn mQt lo4i bQnh nguoi ta
leveni that the tested person
e(ela)
hai b6ng ttdu bi
b6ng h6ng.
fol lowin g statisiicq. Let
4
ci
rittirh$p 1;
c) GiA sri ring ci 2 b6ng ddu bi
h6ng, tim x6c suSt d6 chiiSc b6ng
tiiSp theo rfit ttr hQp tE chgn li
c) Than find the probability that
the next bulb picked from the
selected box will be defective.
'
Hai b6ng
vi
15 con chuQt c6i; l6ng thti
II c6
was observed at the
output. what is the probability that
a
I
d)
c) 0,889; d)
lim
vdo?
P.
.
0,1 5.
was the input state?
Calculate
error P..
the probability of
8 con chuQt tllrc vd 7 con chuQt cdi.
Bit I
con
tu l6ng I,
mQt con tir
II rdi dua sang ldng III; sau
d6 bit I con tir l6ng III. Tinh x6c
l6ng
su6t d.l con
niy
ld chugt
DS:0,467.
239
1.24**., The incidence of an illness in
the general population is q. A new
medical procedure has been shown
to be effective in the early
detection of tlre illness. The
probability that the test corlcctly
identifies someone without ihe
illness as negative is 0.95.
a) Find the
probability that
someone with the illness will get
positive result.
higu qui d6 ph6t hipn s6m benh
ndy. X6c su6t x6t nglriQm chi th!
if it functions when at least
of the components functions.
Assume that the components fail
independently and that the
ihing m6t nguoi khrSng bi bQnh c6
probability of failure of component
t.24**. Tf
lQ mic mQt losi bQnh cria
cQng d6ng ld q. Quy trinh mdi t6 ra
x6c sudt it6 h€ hoat dQng.
a) Tim x6c suAt phin ring duong
tinh c0a ngudi kh6ng c6
functions.
os
bQnh.
b) Bitit ring q=0,0001
vi
bao nhi€u?
c) Biiit ring q = 0,2 vi x6c sudt dd
mgt ngudi khi c6 k6t qui duong
tinh s€ bi b€nh ld 0,8. Tim x6c sudt
phin trng duong tinh cria ngudi c6
tlrat you have the illness?
c) Now it's known that q = Q.2
and someone with positive result
will be ill with probability 0.8.
Find the probability that someone
with the illness actually
1.28. Let S be the sample space of 1.28. Gia sri S li kh6ng gian m6u c6c
thi nghiQm vd S={A,B,C},
an experiment and S={A,B,C} ,
P(A)=p, P(B)=q, and P(C)=r,
where p,q,r>0. The experiment
is repeated infinitely, and it is
b€nh.
assumed
experiments are independent.
Find the probability that the event
A occurs at least once after the nth
c) 0,80; 0,92.
correct.
in order to establish
occurs before B.
1.25. Bao nhi€u phuong trinh ban cAn
a6 ttrii5t l$p tinh rtQc t6p cria 5 biiSn
c6z
D& 65.
1.26. Let g = [0;
l] x [0; t]. Assume that
P(A) is equal to the area of A. Find
t\Yo independent events
A, B that
rlo not have a rectangular form.
1.27**. A systcnr consisting ofseparah:
componcnts is said to be a parallel
240
1.26. Gia sfr S=[0; I] x [0;
l].
Cho rang
P(A) bing diQn tich A. Tim 2 biiin
lgp A, B mi kh6ng c6 d4ng
cO aQc
_o chB nhflt.
R(B)=e,
P(C)=r,
vdi p,q,rr9. Lap lai thf nghiQut
vd h4n lAn vi gii sri ring c6c thi
nghiQm thinh c6ng ln dQc lflp.
Tim x6c su6t tt€ bii5n cti A x6y ra it
nh6t I IAn sat thi nghiQm thri n r6i
sau d6 tim x6c su6t cira bi6n c6 A
xiy
experiment and than firrd the
probability of the event tlrat A
Find the probability that the test is
independence of 5 events?
P(A)=p,
successive
ES. a) 0,05; b) 0,165;
positive result.
' to check
that the
Tim x6c su6t xet nghigm tfiing.
has
1.25. How marry equations do you need
i=t
su6t xdt nghiQm chi th! tfiing mgt
ngudi b! bQnh c6 k6t qui duo:rg
the illness as positive is 0.99. You
test
the
prpbability that
It
l-flni.
xdc
take tlre test. and the result is
positive. What is the probability
the
thinh phin hoat tlQng. Gi6 sir c6c
thinh phin h6ng h6c m$t c6ch tlQc
lfp vi xic su6t h6ng cira thinh
phin thri' i ld p;, i=1,2,...,n.'fim
pi, i- 1,2....,n. Flnd the
probability that the system
correctly identifies someone with
and
n6u n6 ho4t tlQng khi
i is
k6t qud 6m tinh ld 0,95.
tinh ld P=0,99. B4n thgc hiQn
x6t nghiQm cho k6t quA duong
tfnh. Khe ndng b4n b! mic b€nh ln
b) It's given that's q =0.0001
it nh6t rnQt
system
one
D,si
ra tru6c bitin c6 B.
I
; P(A)/[P(A)+ n(e)].
n:
1.291'. Manufacturer sells 20000 1.29*. Nha sin xu6t b6n ra 20000 sin
products, 300 of which are
phAm, trong tt6 c6 300 phi5 phim.
defective. Distributor tests at
Nhi phin ptrOi t
,
random 100 produces, ifthere is at
rrost one fault, the products will be
accepted. Find the probability that
100 bO, n6u c6 kh6ng qu6 mQt b0
t6i ttri ctr6p nh6n 16 hing. Tinh x6c
su6t 16 hnng bi tir chi5i.
the batch ofgood is refused.
DS:0,443.
1.27*J'. MQt he th6ng c6c thdnh phin
ri6ng rE xem nhu mQt h€ song song
241
$6.2.8Ar rAP CHTIONG rr
2.1**. Two basketballers one by one 2.1**. Hai cdu thri lin luqt n6m b6ng
throw a ball into a basket until
vAo 16 cho d6n khi ndo b6ng tr0ng
there is one ball thrown into the
ri5 thi dirng n6m. Bi6t ring x6c su6t
basket. The probability of success
ndm trring cria m6i nguli tuong
in every throwing is 0.8 and 0.6 for
0ng li 0,8 vi 0,6 trong mdi tAn
,t
the fist anfthe second basketballer,
nem. Tim x6c su6t cia:
respectively. Find the probability
table of:
56lin
bi 56 lin
a) The number of throwing of the
first basketbaler;
D,S.
b) The number of throwing of both
b) P{Y = 2n
a)
basketbalers.
.
ndm cta cAu thri thri nh6t;
n6m cria c6 2 ciu
tht.
a) P{X = n} = 10,08)n-r. 0,92;
-
1} = 0, 08n-1. 0,8;
P{Y = 2n} = 0,08n-1. 0,12.
ages
2.2*.Mqlt nh6m tr6 gdm 4 em tu6ittr
from 0 to 3. Two children are chosen
at random then their ages are added
0
ngiu nhi6n hai em r6i
. :.
cQng hai tu6i cria chtng l4i v6i
nhau. Ggi X ln kiit qui. Tim ph6n
b6 x6c su6t cia X.
tti5n 3. Chgn
together. We call X the result. Find
the probability table of X.
2.3*. A personnel officer has a number 2.3*. Cin bQ phdng nhdn sg c6 mQt lo4t
of serious candidates'to fill four
ring vi6n tldng tlon vdo 4 ch6
positions. The chance for every
tilSng. Khi nlng thdnh c6ng cria
candidate is 0.6. Find the
m5i ung vi6n li 0,6. Tim xic suSt
probability that he has to consider
d,5 c6n b0 d6 phni xem x6t ding
7 persons correctly.
7 ngudi.
2.4**. An information source generates
symbols at random from
a fourletter alphabet {a, b, c, d} with
1, P(b)=i,
probabilities P(a)=
I
P(c)= P(d)=r.
A
D,S..0,166.
2.4**. Ngudn th6ng tin sinh ngiu nhi6n
cilc
kli
{a, b, c,
:
hi€u g6m
d} vdi
4
ch&
xric su6t P(")
II
P(b)=4, P(c)= P(d)=-.
coding
scheme encodes these symbols into
binary codes as follows:
a-0, b-10, c-110, d-lll
242
z_
of the code,
that is, the number of binary
denoting the length
/
2.2*. A group 4 children of the
Gqi X le BNN kf hiQu ttQ dii cria
m6, tt6 ln s5 4i hi€u nhi ph6n (s6
Let X be the RV (random variable)
lugc tl6 m6 mi ho6 c6c
thinh
a
-0,
kf
c1i
=;,
c6c x6c su6t R1X=l), P(X=2),
P(X =3) vi P(X >3).
of X?
Assuming that the generations of
symbols are independent, find the
prob. P(X = l), P(X =2),
P(X = 3) and P(X > 3).
D,s..
mE nhi
ll.
{t,z,sl; },
l,
l
o.
2.5*. Consider the experiment of
throwing a dart onto a circular
2.5*. Xdt thi nghiQm n6m phi ti6u vio
plate with unit radius. Let X be the
vi. Gqi X ln birin ngiu nhi€n chi
RV representing the distance of the
point where the dart lands from the
origin of the plate. Assume that the
dart always lands on the plate and
khoing c6ch tir tli6m phi ti6u cham
vio dia t0i tam cria dTa. Gii sri phi
ti6u lu6n roi vio dia vi ch4m vio
that the dart is equally likely to
nhau. Tim
land anywhere on the plate. Find
P(X < a), P(a
P(X
(a
mQt c6i tlTa hinh trdn b6n kinh tlsn
-.t cria tlia v6i.khi nlng nhu
mgi' tli6m
2.6*. a) Verif, that the function p(x)
DS: a2i b2
:a2
.
2.6*. a) Chung t6 rAng hAm p(x) x6c
defined by
ttinh bdi
x=o't'Z'"'
p(*)={;(;)- x=0,r,2, '
otherwise
L0
is a probability mass function
li
(pmf) of a discrete RV X.
cta BNN rdirac X nio tl6.
u) rina: (i) P(x = 2),
b)Tim: (D
't.l={i(*)cdn t4i
[o
I
(ii) P(x<2), (iii) P(x>l).
hdm khlii'luqng
D^s:
2.7**. Consider
a function
r(x) =1fe-1x2+x-a), -@ < x < @.
xlc
suSt
(pnifl
P(x=2),
(ii) P(x <2), (iii) P(x
1461
hiQu nny
phin nhu sau:
b-10, c-l 10, d-l
bit). Tap gi6 tri cia X li gi? Gi6 sr!
viQc sinh kj hi$u li ttQc lflp, tfnh
symbols (bits). What is the range
>
t).
""
'64' ' 6i;"')tr'
' il1:ii)
63
b)
I
v
2.7**.Xethim sti
r(x)=
uf
e-("2***u)' -@ <.x
.
243
.
Find the value of a such that (x) is
a probability density function (pdf)
of a continuous RV X.
Assuming that X is a normal RV
witlr mean 1000 and variance
Tim gi6 tri cria a sao cho f(x) ln
him rnflt d0 (pdfl cria BNN li6n
tgc X.
2500. Find the probability that
resistor picked at random
D& a=1.
A RV X is called a Rayleigh RV if 2.8. BNN X dugc gqi ld c6 phdn b6
its pdf is giverr by
Rayleigh niiu him m6t d0 cria n6
fy (x) =
a)
f
e-*2
rx (x)
Determine
the cumulative
distribution function (cdf)
Fx
(x)
every nine good chips. Let
=f
b) Ve Fy(x) vd
b) Sketch Fx(x) and fx(x) for
o= l.
2.10*. The number
of
telephone calls
arriving at a switchboard during
any l0-minute period is known to
be a Poisson RV X with l. = 2.
n
c) Calculate E[X],
VXl, Mod(X).
A
production line manufactures
1000-ohrn (C)) resistors that have
2.11'*.
l0
percerrt tolerance. Let
X denote
of a
resistor.
the
244
resistance
b) One year of use.
2.13. The radiat miss distance [in meters
(m)l of the landing point of a
parachuting sky diver from the
of the target area is known
to be a Rayleigh
RV X with
'l'
parameter o- = 100.
center
a) Find the probability that the sky
diver will land within a radius of
I 0 m frorn the center of the target
b) Tim x6c su6t kh6ng c6 cugc ggi
nio trong vdng l0 phfit.
ttiSn
b) Find the probability that no calls
arrive during any lO-minute
period.
c) Tinh
with a good chip, Find the
a) Six months of use;
a) Tim x6c su6t c6 qu6 3 cugc ggi
d6n trong vdng 10 ph6t.
will
^",
/"=-
probability that a chip purchased
randomly will fail before:
?"= 2.
any l0-minute period.
L=;
RV with parameter
for a defective chip and
I
2.10*. S(i cugc ggi diSn I t6ng ddi trong
l0 phtt le BNN Poisson X vdi
a) Find the probability that more
than three calls will arrive during
be
exponential
fy(x) khi o = l
2.9**. Consider a gaussian RV X with 2.9**. X6t BNN chuin X v6i cric tham
s6 p=-1, 62 =4. Vi6t him m{t
Parameters p=-1, o2 =4. Write
the pdf of X and calculate the
ttQ cfra X vd tinh c5c x6c su6t
following probabilities P(X>0),
P(X>0), P(X<-0,5), e(Xl<21.
P(X < -0.s), P(xl< 2).
ES: 0,3085; 0,5987; 0,6247.
X
to failure (in months) of
chips. It is known that X is an
time
e-*2rtzo2)u1*;.
a) Tim him ph6n b6 n*1x1.
.
lo4ib6.
be
computer memory chips, company
A produces one defective chip for
cho bdi
rtz"2)r,*r.
a
In thq manufacturing of
2.12**,
ElXl, VlXl, Mod(X).
D& a) 0,143; b) 0,135.
Y
2.11**. MQt d6y chuy6n sin xuit diQn
trd 1000 6m (O) ttusc phdp x6
dich l0%. Kf hiQu X la tr! sti cria
ttiQn trd. Gi6 sri X c6 phdn b6
'
vd
phuong sai 2500. Tim x6c sudt m$t
chi6c di€n trd chgn ng6u nhi6n bi
rejected.
4
2.8.
will
chuin v6i trung binh 1000
)_
ES..0,045.
2.12*r,. Trong viQc s6n xu6t chip nh6
mdy tinh, c6ng ry- A sin su6t I
chi6c h6ng v6i cd 9 chiric t6t. Gie
sir X li tu6i thg (theo th6ng) cria
c6c chip. Bi6t ring X ti BNN mfi
v6i tham
,o i,=]
Z
h6ng vd ,t =
"
+
l0
aoi voi ctrip
v6i chi6c chip t6t.
Tim x6c su6t di5 I chi6c duoc chgn
ngdu nhi6n sE bi h6ng:
a) Sau sdu th6ng sri dUng;
b) MQt nIm srl dgng.
DS; a) 0,501; b) 0,729.
2.13. D0 lQch (theo m6t) cria di6m tiiSp
d6t cta vfn rlgng vi€n nh6y dir t6i
tdm virng mgc ti6u li BNN X c6
ph6n b6 Rayleigh X v6i tham s6
o-=
100.
a) Tim x6c su6t tlii vQn clQng vidn
nhdy dir ti€p d6t trong vdng b6n
kfnh
r:
lOm tri tdm virng mgc
ti€u.
afea,
b) Tim b6n kinh r sao cho x5c su6t
b) Find the radius r such that the
probability that X > r is e-l .
deX>rbinge.l.
2.14r"'.lt is known that the floppy disks
produced by company A will be
defective with probability 0.01.
The company sells the disks in
u. ,S
a) 0,393; b) 14,142 (rn).
2.14"*. ei6t ring c6c dTa nh4c sin su6t
bdi c6ng ty A sE b! h6ng vdi x6c
su6t 0,01 . C6ng ty b6n dTa thdnh 16
I0 chii5c mQt vdi
lli
eldm bdo
li
s6
245
packages
of l0
guarantee
of
and offers
a
'.
replacement that at
I
of the l0 disks is defective.
Find the probability that a package
most
pr.rrchased
will
thay
ci
c6 qu6
16 n6u
I
ifia
bi
c) Find
rtt
and the variance
ra b! thay thti.
2.16*. Let X be an exponential RV with
-i'7"'
s6
I
ri6m tra ring,
a
sequence of 2.17t"'. X6t dny c6c phdp tht Bemoulli
vdi x6c su6t thinh c6ng p. Ddy ndy
Bernoulli trials with probability p
duo. c quan sit tt6n ldn thri thinh
is
of success. This sequence
c6ng tliu ti€n. Gii sir BNN X kf
observed until the first' success
hiQu sii lin th& cho tltin ph6p tht
denote
the
Let
the
RV
X
occurs.
thinh c6ng ttiu ti€n. Khi tl6, him
trial number on which this first
kh6i luqng x6c su6t (pmf) cta X
success occurs. Then the
of
is given by
px (k) = P(X =
k)
P(X
>c).
-
of RV X be given by 2.19. Hdm mit ttQ cta BNN X cho bdi
fy(x) = kxe-xu(x).
fy(x) = kxe-xu(x).
2.19. Let the pdf
a) Find the constant k, Mod[X].
E[x]. Elx2l,
a) Tim hang s6 k, Mod[X].
vtxl.
c) Find the pdf ofthe RV
b)
lE.
the first success occurs on trial X).
The RV X is called a geometric
ttugc ggi le BNN c6 ph6n tO trintr
hqc v6i tham sti p.
RV with parameter p.
a) Chung t6
a) Show that py(k) satisfies the
@
=t
-
ring py(k) thoi mdn
q
phuongtrinh
fRx(k)=I.
k=l
.
b) Tim him phdn bO
Tirn E[x], ElX2l, VlXl.
c) Tim him m6t tl$ cria
DS. a) k =1,
b) 2,
r*1x;.
1&.
Mod[X]=l;
6,2;
c) fu7(x)=2x3e-*2u(x).
2.20. Find the mean and variance of a
Rayleigh RV (see Prob.2.8).
2.20. Tim k! vgng vi phuong sai cria
BNN Rayleigh (xem bditap 2.8).
D& E(x)
(There must be k - 1 failures before
(Phni c6 k
c,d>0,
P[X>c+dlx>d]=P(X>c).
k)
I th6t bai trudc lin thri
thinh c6ng X iliu ti6n). BNN X
k=l
>c+dlx >d]=
\.
=(t-p)u-'p, k=1,2,...
p)o-' p, k = 1,2,...
BNN mii X v6i tham si5 l.
Chi ra reng BNN X c6 tinh ch6t
kh6ng c6 tri nhq chinh ld: V6i mgi
z.t{.*.xdt
the
cho bdi
px (k) = P(X =
lp*(L)
P[X
b) Finct
x
probability mass function (pmf)
2.18**. Consider an exponential RV X
with parameter 1.. Show that the
RV X
Tinh Med(X), so s6nh v6iE[X].
larger,
E[X] or Med(x)?
2.17t,,'. Consider
L3.
pp2
haS
memoryless
property, that is: For all c, d > 0,
I
r- I
ElXl =; vi V[X tL
L,'
..
Find Med(X). Which is
equation
DS..
c) n = I,2,...; 1,
2.16*. Gqi X le BNN ph6n b6 mfr tham
Verifr that,
r-_l
E[X]=: and V[X]=
-
V[X].
b)xe [n;n + l) : F(x) = I -(1 -p)n,
I.
DS:3-5:
"12
= (t
value E[X]
the
have to be replaced.
sai cria X.
X
c) Tim k! vgng E[X] vd phuong
sai V[X].
DS..0,004.
2.15. Let a RV X denote the outcome of 2.15. Gqi X le BNN chi kiSt qui khi rrlt
mQt con xirc xic cdn
tri trung binh (ki vgng) vi phuong
(expected value) and variance ofX.
parameter i'..
b) Find the cdf Fy(x) of X.
'expected
.
h6ng. Tim x6c su6t tt€ mQt 16 tluo. c
V[X]= (Z-ntZ)oz o0,429a2.
2.21**. Given that X is a Poisson RV 2.21*t', Cho
and Px (0) = 0.0498 , compute E[X]
and P(X > 3).
=GIi o;
X ld BNN Poisson vi
Px(0) = 0,0498 '
Tinh E[X] vn
Pfi
> 3).
DS. 3; 0,5767.
A RV X is the Pareto random
variable with parameter a, b
(a, b > 0) if its pdf is given by
2,22.
fx(x)
=
(a/b) (b/x)a*l, x € [b;
*)
X ld BNN Pareto vdi c6c
s6 a, b (a,b>O) n6u him
2.22. BNN
tham
mAt dg cria n6 cho bdi
fx(x) =(a / b) (b/x)a+l, x e[b; o)
247
'
a) Show that E[Xn] exists
if
and
'.
onlyifn
a) Chi ra rang E[Xn] t6n tai
vdchiniiu n
b) Find E[X] and
Elx2l (a> 2).
2.23*. Show that for a Cauchy RV with
the parameters a. b with pdf
tham s6 a, b v6i mQt <10
,.,
(b > 0) tlre mean does not exist.
(b > 0), k) vgng kh6ng tdn t4i.
x (x_a),
N(pr, o2 ) , evatuate E[X3 ]
2.24. Giasti X
- N(p,o2), tinh
DS: 3o2p[l + p3]
2.25. Suppose.that Z
-
N(0,1)
2.25.Gie sir ring Z
.
a) Evaluate ElZl.
a)
-
Ax--r0
xelR
+b.
J
xfxlxlep;dx,
,-,,
Ax
P({x
E[X3].
n
fi=t qxlen)p(Bn).
Elxl=f e1xln"in1e"1
i=l
2.28. Consider an integer, nonnegative
RV X. Show that
.
2.28,
X& BNN nguy6n, kh6rrg 6rn X.
Chring t6 ring
6
9o
Elxl=)P(x>k).
rinh Elzl .
Elxl=fe1x>r<1.
k=0
DS:
J2/tr
b) Chi ra
b) Show that
E[22"7= 1.3...(2n
- l)
/
2.29. Let
E[22")= 1.3...(2n -1).
th6y s6 qui cam tr€n mQt cdy tu6n
than 20 orallges and 30 ones with
Nguli ta d6m
thfi 600 cAy thi th6y 15 c6y c6 it
hsn 20 qui, 30 cdy c6 it hon 25
less than 25 oranges.
que.
with
less
a) Hey udc tugng
one tree.
binh tr6n mgt cdy.
with more than 60 oranges.
trees
b) U'6c lusng ty
si5
s6
li BNN Poison vdi tham
i..tim Mod[X] khi ].>l vd
khi l.<1.
0<1"<l : Mod[X]=0; t.> l:
uoarxt={lhl,
Lf-l
IQ
5l,l;b)29
2.30**. The cycle of the traffic light is 2
minutes of green followed by 3
nrimutes of red. What is expected
delay in the journey if you arrive at
qui cam trung
cdy c6 tir
60
qua trd l6n.
DS. a)
%.
2.27**. (Total Probability Theorem for 2.27* (Dinh lf x6c sulit toin phAn vdi
k) vgng). Xdt BNN X tr€n kh6ng
Expected Value). Consider a RV X
gian m6u S. Xdt ph6p phdn ho4ch
on a sample space S. Consider a
partitiorr {81,...,8n} of S. Define
{B1,...,Bn} cta S. Dat
248
when
2.2g. Giesri X
theo ph6n bl5 chuAn.
a) Estimate the average oranges in
b) Estimate the ratio of the
with
D,S..
of oranges in a tree follows normal
trees there are 15 ones
k=0
a Poisson RV
Mod[X]
?'. > I and then 1, < l.
2,.26**. O mQt vtng trdng cam, ngudita
ln the 600 counted
be
parameter tr". Find
ring
2.26)'*.ln the orange-region the number
distribution.
X
Ax
^i]o
Chi ra ring
l1
Etxl=
k = 1.2,...,n
,r"]e ao fy(x I B) =
Show that
.
N(0,1)
@
k = 1,2,...,n
*n"r" ,11* I ,y =
P({x<.X
2.23*. Chi ra ring a6i vOi BNN Cauchy
f*1x1=1-i-,
-
xfxtxl\)clx,
J
b)Tim E[X] vd Elx2l (a>2).
O- -. xeR
n(x-a)'+b'
E(xlnnl=
co
fy(x1=1
2.24. l-et X
E(XlB*1=
n6u
gL
the junction at a random time
tuniforrnly distributed over the
k! c0a tldn hiQu giao th6ng
g6m 2 phfrt xanh r6i d6n 3 ph0t rt6.
Tinh thoi gian cho trung binh cria
di n6u ban tti5n ngd tu t4i
m$t thdi ili€m ngiu nhi6n ph0n b6
chuy6n
tl6u trong khodng thdi gian 5 phrit.
X-
HD:
Hint: X - the delay, T
dirlm b4n d6n ngd tu.
\ 0
X=0;
2
tr6i lai
2.30**. Chu
whole 5:minute cycle?
- the time
when you arrive at the junction.
}'"e{2,3,"'l
Thdi gian chd, T
- thoi
0<'[<2:X=0;
2
.
D,S..0,9.
l0 shots 2.31. Gie sft mQt chi6c tdu chi6n bin l0
ph6t vio mpc ti6u vi cAn ft nh6t +
at a target, and it takes at least 4
2.31. Suppose a warship takes
249
hits to sink it. If the warship has a
record of hitting with 20% of its
'.
ph6t trring ai5 Aanir chim n6. Trong
a) Determine the value of k.
thli
gian ddi, nguli ta ghi
nhgn ring c6 20% lAn tiu bin
mQt
shots in the long run, what is the
clrance of sinking the target?
trring dich, co hQi
ti€u Ii bao nhi€u?
K
di5
a) X6c ttinh gi6 tri cria k.
b) Find the probability that
the
distance from the origin of the point
selected is not greater than a < R.
b6n chim mpc
c) Find the marginalpdf of X,
,s..0,1209.
b) Tim x6c su6t mA khoing c6ch tir
g5c d6n di6m chgn kh6ng vugt qu6
a
y.
vd Y.
2.32**. Wearever tires have a truct 2,32"*. Nhiing chi6c l6p cria hdng
record of lasting 56000 miles on
Wearever de ghi nhfn mQt k! lgc
everege, with a standard derivation
khring khiiip h tli tlugc trung binh
of 8000 miles, and a normal
56000 d{m v6i rtQ l6ch chuin 8000
distribution.
d[rn vi c6 ph6n b6 chuin.
.
a) What is tlre chance that a given
tire will last 50000 miles?
a) Tfnh xric su6t a,i
t cni6c l6p de
cho di dugc it ra 50000 dam.
b) Wlrat is the chance that all four
Wearever tire on my car will last
b) Tinh x6c su6t tlti cA 4 chitic liip
50000 rniles?
dugc ft ra 50000 d4m.
$6.3.
[0,
3.1*. X6t hdm
os
x,y < oo,
otherwise
Can this function be a cumulative
distribution function (cdf) of a
random vector (X, Y)?
3.2. Suppose we select one 'point at 3.2.
random from within the circle with
radius R. If we let the center of the
circle denote the origin and define
X and Y to be the coordinates of
the point chosen, then (X, y) is a
uniform random vector with the
prob. density function (pdf) given by
r*" 1*,y;
=
{k'
lo,
xe cria t6i s6 tli
8Ar TaP CHUONG rrr
3.1*. Consider a function
F(x.v)
.J t _ Jt -"-(**r),
Wearever tr6n
x2 + Y2
'R2
*2+y2rR2
F(*,y)
=
{t
-
"-(-*')'o
[0,
( x'Y < oo,
nguo, c
lai
Hdm ndy c6 thrS ld him ph6n b6
(cdf) ctia VTNN (X, Y) hay k}r6ng?
Gie sri ta chgn I di6m ng5u nhidn
trong hinh trdn b6n kinh R. N6u kf
hiQu tdm vdng trdn ld gi5c to4 116 vi
X vd Y li to4 tlQ c0a diiSm chgn,
khi d6 (X,D Ie VTNN vdi
mft
ttQ
hdm
x6c su6t (pdf) cho b&i
r*"
"' = {k'
'rr\1*,y1
*2 + Y2 < R2
[o' *2+Y2'R2
DS; a)
c)
1=];; Ul4;
rR' Rr'
Ir.fi'_l
fy(x)=1-r;r-'
l*l
Lo,
,
l*1, R
3.3**. A manufacturer has been using 3.3**. Nhe sin su6t
dirng 2 quy trinh
two different manufacturing
s6n xu6t kh6c nhau ae ian xudt
processes to make computer
chip nhd m6y tinh. Ci6 srl (X,y) te
memory chips. Let (X,y) be a
VTNN,. trong d6 X kf hiQu thdi
random vector, where X denotes
gian tti5n h6ng cria chip sin su6t
the time to failure of chips made
bdi quy trinh A vi y ld thli gian
by process A and y denotes the
trinh B. Gid st hdm mflt d0 cria
' piocess B. Assuming
that the pdf
(x,Y) ta
of (X,Y) is
f (*,y) = {ub"-(u**u'), x > o,y > o
f (*,v) = {uu"-(**u'), x > o,y > o
nguoc lai
[0,
[0,
otherwise
where a=10-4 and
determine P(X >
y).
b=l.2xl0-4,
trong d6 a=10-4, b=1,2.10-4,
tinh F(X > Y).
D,S:b/(a+b)=9,545.
3.4*. Let (X, Y) be a random vector, 3.4*. Gia srl (X,
n le VTNN, trong d6
where X is a uniform RV over
X ld BNN phdn b6
vi Y li BNN mfr v6i tham s6 5,
RV with parameter 5, and X and y
X vi Y tlgc lfp.
are independent.
a) Tim mft
b) Tim P(Y < X).
,
b) Find P(Y < X).
D,S..
a)fxy(x,y)
--[zs"-tt,
lo,
b) e-l = 0,368.
0
l4r
251
35. Cia sft m{t ttQ cira (x, Y) cho bdi
3.5. Let the pdf (X, Y) be given by
fxv (*,y)
{*t-*(vnr)'
=
[0,
x > o'Y > o
>
fxv (*,y) = {*t-*(v*t)' x o'Y > o
[0,
otherwise.
a) Show that fay (x,y)
a) Chring t6
satisfies
I
J Ifr"t*,y)dxdy=1
ring fly(x,y)
;;"
Ti---a, d0 bi6n cria X vi
D& a) e-*, 1x>0;;
Ol
marginat pdf of X and Y.
3.6**. The
iaf of a random vector
"-"*
xt -
(*,y)= j x >0, y)0,
"-g*
cr,
),
0>0;
lo,
t'
otherwise.
b) Tim hnm kh6i lugng x6c
su6t
mass
b) Find the marginal pmf of X
bi6n cria X vd Y.
and Y.
c) Phni ching X
c) Are X and Y independent?
D,S:
and
Y
are
t6p.
c) Find
c) Tim
a)pyy(0,0) = 0,45; pay(0,1) = 0,05;
pxv (1,0) = 0, l; p1y(1, l) = 0,4.
b)px(0) =px(l) = 0,5;
Pv(o) = 0,55;py(l) = 0,45.
3.8**. The pdf of a random
(X, Y)
10,
c)
3.7. Consider the birrary communication
Prob. 1.22. Let
(X, Y) be a random vector, where
X is the input to the channel and Y
is the output ofthe channel. Let
1t
o 1x'Y 12'
a) Find the value of k.
b) Tim mdt tiQ bi6n cia X, Y
lo,
b) Find the
.otherwise
marginal pdf
of
ft
X
c) X
vi Y ln tlQc l6p?
DS: a)k =
I
r;
0
-'
'
>0.{l -e-o}, v ) o
y<0
x<0 [0,
10,
nguo.clai
c) Kh6ng.
3.9*. Let (X,Y) be a random vector.
-e-ax)(l -e-oY1.
1.22. Gia sri
H hing s5).
and Y.
x
(X, Y)
lA
VTNN trong d6 X ln diu vio k6nh
vd Y li dAu ra cria k6nh. Gii sfr
r(x=0)=9,5t
o
\ "'
ngugc l4i
[0,
a) Tim k.
- e-a Xl - .-0 );
tfp
f
wherekisaconstant.
3.7. Xdt k6nh th6ng tin nh! ph6n nhu &
bni
3.8**. Hdm phdn b6 cfia (X,Y) cho bdi
is given by
c) Are X and Y independent?
{l -"-"*,
b) (l
vector
c) Dfine.
vi Y ld tlQc
D.9l
u;
\a
f(x-v)=lo(**Y)'
\ "'
P(x<l,Ysl), P(x, *,Y>y).
l,Y < t), P(x > x,Y > y).
vi Y tl6c l6p?
vi Y.
Tim him ph6n h5 bi6n cua X
independent.
in
function (pmf) of (X, Y).
probability
nguo.c l4i.
b) Cht?ng t6 ring X
as
su6t
a) Find the
Y.
e-o* Xl ),
[,, "-o*
Fxv (*,y) = j * >0, y 20, g, F>0;
b) Show that X
channel
a) Tim hnm kh5i lugng x6c
cta (X, Y).
3.6**. Him ph6n b6 cria VTNN (X.Y)
a)
<
o)= o.l,
l/(y+l;2,1y>0;.
a) Find the marginal cdf of X and Y.
P(x
=
,YI
I
10,
r;x
cho bdi
is giverr by
[(t
t' -
=
r(v = tlx = o)= o,l
va r(Y=0lX=l)=0,2.
and P(Y=olX=t)=0.2.
tho&
If*"(*,v)dxdy=1.
b)
Fxv
P(v
c l?i.
co@
@@
(X, Y)
ngu-d.
min phuong trinh
the eqution
,,
R(x=o)=6.5,
,
Show that
[e(xv)]2 e(xr)e(vr).
This is known as the Cauchys
Schwarz inequality.
3.9*. Cho
(X,U HVTNN. Chungt6rdng
[E1xv1]2 s e(x'?)e(v'?). r.i,
tting thfc (BDT) niy c6 t6n li
BDT Cauchy-Schwarz.
i/D: E[(X-aY)2]> 0, Vae R.
253
N
3.10**. The pdf of a random vector 3J0**. Hdm
cho bdi
(X, Y) is given by
, [k*y, 0
otherwise
to,
t
mQt <10 cria
VTNN (X, Y)
. [k*y,
" 0
fxy (*,y)
^ (*,y)
fxv
wherekisaconstant.
trong d6 k ld hing s6.
a) Find the value of k.
a) Tim gi6
3.11**. Consider the random
x
vector
(X, Y) of Prob. 3.8.
a) Find the conditional pdf
r1a (vlx) and raly (.lv)
=
D^S. a)
(
\
X vi Y
tlQc
lQp?
b)rrnh
r
lo,
a) Show that f1y(x,y)
the equation
O@
I If*"(*,y)dxdy=1.
t;; e(x >tlv =y).
satisfies
D^Si
a)f;(x)=2x, (0
4,to
< x,y < 2;
bdi
[1"-*'r.-Y,x)o,y>o
f (x,y)=.{ra) Chung t6
nguo.c
ring fyy(x,Y)
3.16. The pdf of a random vector (X, Y)
is given by
l3i.
fxy (*,y)
thon
@@
I If*"(",v)dxdy=1.
P(x,rlv=11.
b) Are X arrd Y correlated?
254
D^S: a)
Kh6ng; b) Kh6ng.
=
{t-(**')'
|.0,
x > o,Y > o
otherwise.
Y) le VTNN chuAn. Tinh
E(Ylx).
px + p(x - py)oy / ox
3.16. Him mat dO crha VTNN (X, Y)
cho bdi
fxv (*,y) = {"-(**')' x > o'Y > o
10,
nguo. c
l4i
a) X
b) Find the conditional pdf of X.
b) Tim m$t iIQ
vi Y
le dQc lfp?
DS; a) Dring; b)
,
.
a) Are X and Y independent?
3.17. Let (X1,...,Xn). be an n-variate
D& b) e-l/r.
3.13. Suppose that a random vector 3.13. Gie srl VTNN (X, Y) ph6n bli
(X,Y) is uniformly distributed over
(Prob.
3.2).
a unit circle
' a) X vi Y li tlQc lflp?
a) Are X and Y indepenA"i,tt
b) X vi Y Ii tuong quan?
3.15. Cho (X,
D,S.
m6n phuong trinh
;f,
y.
a) Tim mdt tlQ bi6n ctia X vd
e(vlx) va e (xlv).
3.15. Let (X,Y) be a normal random
vector. Determine E(Ylx).
[0,
c lai.
b) Tinh trung binh ttiAu ki6n
3.12. Him m6t tlO cta VTNN (X,Y) cho
otherwise.
nguo.
uli,to
a.
'32
. [1.-*rr"-Y,x)o,y>o
v6i him
b) Compute the conditional means
.
b)
f(x,Y) = i
X
DSi
jffi,o
te VTNN
and Y.
e(0.v.Il*=,).
(rl-)=
y)
|.U,
a) Tim mat do tli€u kien
r1,1(vlx) va r*1y(xlY)
3.14. Gia st (X,
mflt ttg
r*r(*,y)={1' o
E(ylx) ano e(xlv).
3.11**. xet VTNN (x,Y) d BT 3.8.
a) rvlx
3.12. The iaf of a random vector (X,Y)
is given by
('> ocy
otherwise.
t;,
a) Find the marginal pdf of
8; b) Kh6ng.
.
b)FindPl 0
f*" (*'Y)
nguo.c l4r
to,
tri k.
b) Phai ch[ng
b) Are X and Y independent?
3.14. Let (X, Y) be the random vector
with the pdf
normal random vector with its pdf
given by Eq. (3.6.a). Show that if
the covariance of X; and X; is
zero for i *
j, that is,
'. (t
then X1,...,Xn are independent.
(x > 0).
sri (X1,...,X1) ld
J,
VTNN
chuin n thinh phAn v6i him mft
tlQ chi ra d c6ng thric (3.6.4).
Chung t6 ring, n6u X; vi X; ti
li
cov(x;,x.;)=E,j={"i i=i
[o i*i,
kh6ng tuong quan v6i i *
cov(x;,x.;)=t,.,=lli :=l
,*
LU
3.17. Gia
e-x'
thi X1,...,X1 ErtQc l{p,
j,
tr?c
3.18. The pdf of (X,
Y) is given by
(-u
fxv(*.y)=]:-"
' 10,
rtxy(x'YJ=1
/-.-.\ ["-',
^
' 10,
o
0
correlation coefficient of X and
nguo.c l4i.
a) Tim mat dO tti€u kien crha
b) Find the conditional cdf of Y,
b) Tim hdm ph6n U6 al6u kiQn cria
ring X = x.
Y be two independent
identically normally distributed
RVs with zero mean and variance
4. Find the probabilities that the
random pirint (X,Y) belongs to the
circle witlr center at (0,0) and
3.19. Let X and
0
vi
3.23.
If X - U(0;l), derermine
of Y=aX+b, a,belR.
DS.
ngiu nhi6n (X,Y) thuQc vio
hinh trdn tfrm t4i (0, 0)
3 (m).
Ds. flraglx)
vi
= e*-Y, (y > x);
Fy1(rlx)={'^-t*-''
'
b6n kinh
l''
x
x>Y'
l.9,
the RV Y = -'ln X.
If X is N(0;2),y=3X2,
BNN Y
find
=-lnX.
3.26*. Niiu x - rrr(o;z), y =3X2,, tim
ElYl, VlYl, fy(y) vd ev(v).
^-y/12
D^S.'6;72;:-r;
zllty
.
I
3.21*. Which following
matrices are
covariance ones:
,,(; l)' (? i),
"
.,(; i), ,, (;' l),
3.21*. Circ ma trQn nio sau
(z
o\
(z
z+o6[(y -t4)teQ))'
9.27),t
a)
rf
and
256
r=(; l), *" the RVs X
Y independent?
Gii st (X, Y).ld VTNN
v6i ma trfln tuong quan f, .
a) N6u
vi Y li
3.28. Let Y = tan X. Find the pdf of
if X - U(-r l2; n /2) .
y
chuin
r=[; l), "" BNN x
tlOc l6p?
. Let y = x2 . Find the pdf of y
N(0,1).
if x -
D,S.'a, b.
3.22*r'. Let (X, Y) be a normal random
vector with covariance matrix I.
I
r'\
")[; t),D[, tl
,,[; i), ,, (;, i),
v'
3.22r'*.
tr6i lai.
3,24t', The RV X is N(5,2) and 3.24*. Gia srl BNN X-N(5;2) vi
y =2X+4. Find E[y], V[yJ
Y =2X+4. Tim E[y], V[y] vn
fv (Y)'
fv(Y).
325. The RV X is uniform in the 3.25. BNN X c6 ph6n bii tt6u tr6n
interval (0, l). Find the density of
khoing (0, l). Tim m{t rtQ cria
ElYl, VIYI, fy(y) and Fv(y).
x-N(0,4), Y-N(0,5).
x-N(0,4), Y-N(0,5).
Vdi a>0:
b<*
I\ /
3.26*.
3.20. L{p lai Bnitflp 3.19 v6i
vi Y.
the pdf 3.23. N6u X - U(0;l), tim hhm m6t d0
cria Y=aX+b, a,be IR.
phuong sai 4. Tim x6c su6t d6
di,5m
\r
tuong quan gita X
3.19. Gia sfr X vi Y ldz BNN ttQc l4p
cirng phdn b5 chu6n vdi trung binh
radius 3 (rn).
3.20. Repeat the Prob 3.19 for
y.
l\ titn
z=(o 9J'
he t'
-6I.
Y,bi6tring X=x.
X=x.
b) N6u
Ans: a')Yes: b)
Y bi6t
a) Find the conditional pdf of Y,
giventhat X=x.
giventhat
b) rf z=(o I \ find the
\ r s)'
3.18. Hdm mflt ttQ cta (X, Y) cho bdi
.td
3.27*t. Gi6 sri Y=X2. Tim him m6t
ttQ cria Yn6u X - N(0,1).
3.28. Gia
st Y=tanX.
cta Y n6u X
-
Tim him mdt
U(-rc / 2; x I Z)
D,Sj BNN Cauchy vdi tham si5
S.Z'q. Show
that,
if the RV X has a
Cauchy density
with a=0; b=l
(see Prob. 2.19) and Y= arctanX,
then'Y is uniform in the interval
(-n12,-xl2).
l.
3.29. Chung t6 ring n6u BNN X c6 mft
ilQ Cauchy v6i a=0;
(xem
b=l
vi Y= arctanx, khi
d6 Y phin b6 ddu tr6n
bni tflp 2.19)
(-n12,-nl2).
257
3.30. Let X be a continuous RV with the
pdr ry
(.)
|
=
^'
^-x
t;,
pdfofY
[0,
otherwise.
l;Tt'
fv(v)=
is
0
li
rv
DS:*exnr${r;$;0.
BNN li6n tpc vdi mft
x> o
c<0.
Tim ph6p bi€n dOi Y
cho mdt tlQ cria Y ld
Y = g(X)
It
Cia sfr X
["-*,
dQ rx(x)=t;,
::l
Find the transformation
such that the
S.SO.
-
3.34. Consider
g(X)
l+,
ocy
[0,
nguo.c lai.
(v)=l z./v'
Y=eX.
Find E[Y] by using fy (y)
then by usirrg f1 (x).
3.31.
and
and
3.32. The a-centered clipper is described
by the following transformation
'
[*-u,
g(x) =
sao
I
lx+a,
x
e(*)
=,
=.1
<-a.
a) Plot this fuiction for
o,
x
vii
Y
a
2u2.
are
X
and
c) What type of the RV Y when X
v
g(X)
.
c) Dgng cria BNN Y
BNN li6n tuc.
li
gi ntiu X
li
3.33**. Suppose that (X,Y,Z) is a j.sg**. Gii sri ring (x,Y,z) le VTNN
chuln voi vecto trung binh
normal random vector with mean
p = (0,1,2)T
vi
rira
t{n
tuctng quan
( t 0,5 0,5)
tt
t=l 0,s 2 0 l. Tim
[o,s o 4)
mat d0 cria BNN T
vA
tinh pry, pyz.
tlQc
3.35. Gia st rEng, X vi Y li nhtrng
BNN chuin tirc ttQc l{p. Tim m$t
tlQcfiaZ:X+Y.
DS: Z - N(0,2).
Y be independent
of Z=XY.
3.36. Gia srl X
vi Y ld niing BNN phdn
U5 adu tren
(0; l). Tim mit tlg cta
Z=XY.
. [-m*, 0
3.37. Gie srl X vd Y li nhiing BNN
DS. f"(x) = {
'
X
and
Y be independent
tic
standard normal RVs. Find the pdf
chu6n
of Z=XIY
Z=XlY
.
aQc gp. Tim
m{t tlg.cua
1
a.
tlQ cria Y =
X
DS.'BNN Cauchy.
b) Tim hdm phin bi5 vi hAm mdt
RV T = X-2Y +32
pxy, pyz.
and
+
gi6 tri
b) Find the cdf and pdf of
the pdf of the
and calculate
X
also
1.1
uniform RVs over (0; l). Find the
<-a.
cta
(
o.s\
tt t o.s
matrix E=|0.5 2 0 l. Find
[0., o 4)
Z is
vi tlugc
ofa.
vector p = (0,1,2)T and covariance
3.36. Let
3.37. Let
a) Ve dd thi hnm ndy v6i
various
y=g(X).
'
then
independent standard normal RVs.
l*l = u
[x+a,
is continuous.
respectively,.
3.35. Suppose that
x>a
Z = X + Y. Chung t6 ring,
vi Y li nhtng BNN Poisson
l{p vdi tham s6 X.,, f,2 tuong
ung thi Z cf,rng le BNN Poison v6i
tham s6 lq + ?,,2.
m6taboi 6nhxgsaucl6y
[x-",
3.34. Xdt
ntSu
FindthepdfofZ=X+Y.
3.32. B0 x6n, hu6ng tdm a tlon
I
l*l
if
X and Y are
Poisson RV with parameter
cia sr! X ld BNN c6 ph6n bi5 ddu
tr6n(0,1)vi Y=eX.
Tim E[Y] bing c6ch dirng fy (Y)
rtii sau tt6 dtrng fx (*).
x>a
j0,
X + Y. Show that
independent Poisson
RVs with parameters ?y and ?,2,
ES; y=11_"-*)2.
3.31. Let X be an uniform RV over (0,1)
Z:
3.38**. Let X and Y be two RVs with
joint pdf
'
(*,y)
joint cdf
fxy (*,y). Let Z= max(X,v) .
a) Find the cdf of Z.
(x,y)
vd hem m$t dq d6ng thli
f*" (*,y) . DAt z =max(x,v) .
b) Find the pdf of Z if X and Y are
a) Tim him phin b6
independent.
b) Tim him m4t tlO cria Z niSu X
Y tlQc lflp.
Fxv
and
hdm ph6n b6 tt6ng thdi'Fxy
ciaZ.
vi
DS; a) Fz(x) = F1v(x,x);
him
=X-2Y +32
f7(x)=--:--=-, x€lR.
zr(l + x')
Y.
3.38**.
Gii sti X vi Y lA 2 BNN vdi
.a
b) f2(x) = fx(x)Fv(x) + Fy (x)fy(x)
3.39. A voltage
V
is a function of time
and is given by
t
3.39. MOt hi€u tli€n thii V ld hi.rn cria
'
thdi giair t vi cho bdi
'. v(t) = Xcosort + Ysinrot
v (t) = X cos ort + Y sin
in which
a) Show that X
li tAn sii g6c kh6ng d6i,
X = Y - N(o;"'z) vd chong tlQc
is a constant angular
frequency and' X=V-N(O;o2)
trong d6
and they are independent.
t0p.
a) Show that V(t) may be written
a) Chrng t6 rdng c6 th6 vitit V1t)
as
co
V(t)=Rcos(o:t-@).
c,:
that R and @ are independent.
-O).
b) Tim him ph6n bti cria BNN R
vd chring t6 ring R vd @ ttQc lap.
b) fe(r)
E^Si
=;|e
'2
tt2o2)
,(r
> o)
fs6(r,0) = fn(r[o(0)
.
,...,Xn be n independent 3.40*. Ci6 srl X1,...,X1 li n BNN tt$c
RVs each with the identical pdf
lfp ctrng ph6n b6 v6i hdm mat ttQ
f(x). Find the pdf of
(x). Ci6 sri W = Min(X,, ...,Xn).
3.40*. Let
b) X
b) Show that X and Y
are
D,Si
vi Y kh6ng tlQc lap.
a) E[XY]=ElXl ElYl;
b) Elx2y2l + E1x2l Ely2l.
3.44. a) The function g(x) is monotone
increasing and Y=g(X). Show
3.44. a) Hdm g(x) tlon rliQu tIng
Y = g(X) . Chring t6 ring
that
(*), if v > g(x);
F(x.v)=la*
,J '
lP" (v), if y c g(x).
b) Find Fxv (*,y) if g(x)
vi
(x), ndu v >e(x);
Lr" (v), ndu y < g(x).
.r*.rr)=fFx
,r
'
b) Tim Fx" (*,y) ntiu g(x) ttcn
is
tliQu gi6m.
monotone decreasing.
X1
W=Min(X,,...,Xn).
Tim hdm m4t
D&
3.41. Let
X1,
X2 and
X3
be
independent standard normal RVs.
=Xr +X2 +X3,
YZ=Xr-X2,Y3=XZ-X:
Yz = Xr
the joint pdf of
-X2,
Y3 = Xz
-Xr
vi
cria
Esi
(l
\
)
^-[rri+1vi+1v5+1vztt
w"'
I
3.42. Write down the pdf of the random
vector variable
(x,Y) - N(0,
6, 4,9, -0.1)
.
" 2t 2t 2
3.42. Vi6t ra him mat dO cria vecto nglu
nhi€n chuin
(X,Y):
N(0, 6,4,9, -0,1)
.
3.43. Let X and Y be defined by 3.43. Gia st X vi Y xlc tllnh bdi
X = cos@, Y = sin @, trong d6 @
f, = cos@, Y = sin @, where @ is
la BNN phdn bti tt6u tr6n (0;2n).
a' random variable uniformly
distributed over (0; 2zr).
vi Y dQc l{p v6i
mflt
d0 mfi
fx
(*)=o"-"*u(*);
fy (y) = pe-Pru (v).
Find the densities of the following
Tim him m{t tlQ cta cic BNN:
RVs:
a)2X+Y; b) X-Y; c) X/Y;
d) Max(X,Y); e) Min(X,Y).
X-Y; c) I;
Y'
d) Max(X,Y); e) Min(X,Y).
3.46. Chring t6 ring:
3.46. Show that:
Y3.
3.45. Cdc BNN X
.
f" (y) = p"-pru (y).
a) ZX+Y; b)
Tim hdm mflt tlQ cldng thdi
Y1,Y2
3.45. The RVs X and Y are independent
with exponential densities
fy (x) = cre-"*u (x);
cia W.
- F(*)]n-r .
3.41. Gie s[r X1, X2 vi X3 li nhtng
BNN chuAn tic tt0c lap. Cho
Yr
Y;,Y2 and Y3.
tIO
fyy(x) = nftx)[l
Let Y,=Xl *X2+X3,
Determine
"260
a) X vA Y khdng tuong quan,'
uncorrelated.
independent.
du6i dangV(t) = Rcos(ot
b) Find the pdf of RVs R and show
and y
cia 2 mdt tlQ chuAn ld
a) The convolution of two normal
a) T(ch chfp
densities is a normal density;
mat dO chuin;
b) The convolution of two Cauchy
densities is a Cauchy density.
b) Tfch ch$p cta 2 mdt d0 Cauchy
X is of discrete type
taking
tlre
values xn with
,
P{X = xn} = pn and the RV y is of
3.47. The RV
continuous type and independent
of X. Show that
W=
XY,
if Z=X+ Y
ld mdt ttQ Cauchy.
3.47. BNN rdi rqc X nhgn gi6
tri xn v6i
P{X=xn}=pn vA BNN li6n tgc
Y tlQc l{p v6i X. Ch&ng t6 ring
nlu Z=X+Yvd W=XYthi
and
then
Chung t6 ring:
261
f7(z)=lr*
(z
-
s6.4.BAr raP CHTIONG rV
' fr(r)=I& (z-xn)pn;
x" )R";
4.1*. HSng san xu6t khi de nghi€n ctu
automobile engine is being
thdi gian d6nh hla kh0i
cta dQng co 610. Khi thti vdi mQt
. .!.
manufacturer. The following times
chi6c xe tdi, ngu&i ta thu duo. c c5c
(in seconds).were obtained for a test
s6 tieu sau (tlon v!: gi6y): 1,70;
vehicle: 1.70, 1.92, 2.62, 2.45, 3.09,
1,92; 2,62; 2,45; 3,09:, 3,15; 2,53;
3.15,2.53,and l.19.
1,19.
4.1*. The "cold start ignition time" of
f*(*)=+**[*),,
fwr*)=T**[*),",
f1(x) .
Let Y =lXl. find the pdf of Y in
terms of f* (*).
3.48. Let
X
be a RV with pdf
3.48. Gia s&
X
li
f*(*).Gii
BNN v6i him mSt
ttQ
Y=lxl. Tim him
mflt ttQ cria Y qua f* (*).
srl
,,S:
ry (x) =
Y = sin X, where X is
uniforrnly distributed over (0; 2z).
Find the pdf of Y.
3.49*.
Let
.
t.,*,'J:ol
{l:,,.,
3.49*. Cho Y = sin X, trong tl6 X c6
phdn b6 ddu tr6n (0;2n). Tim him
mat d0 cta Y.
D^Si
f"(y)=
lr",[4),
10,
-l
3.50. Consider an experiment of tossing
3.50. Xet thi nghiQm tung d6ng ti6n c6n
a fair coin 1000 times. Find the
probability of obtaining more that
d5i looo dn. Tim x6c su6t nh6n
dugc qui 520 m{t s6p:
520 heads:
a) By using formula (3.7.10);
b) By formul a (3.7 .12).
a) Dtng cdng thric (3.7.6);
P(Y < x)
^v
DS: a) 2,1 89h; b) 2,195h.
4.2. A second formulation of the gasoline
was tested in the same vehicle, with
the followirrg tirnes (in seconds):
I.80, 1.99, 3.13, 3.29, 2.75, 2.87,
3.40,2.46, 1.89 and 3.35. Use this
new data along with the cold start
times reported in Exercise 4.1 to
4.2. Hinh thric
tht
hai cria su tl6nh hia
tlugc thri vdi cirng chilic xe tai vi
thu dugc sl5 tieu thdi gian nhu
sau: 1,80; 1,99 3,13; 3,29;2,75;
2,87;3,40;2,46; 1,89 vi 3,35. Srl
.
,(..^
1...^
dung so ll9u nay cung vor so lr9u
d bdi t$p tr6n vd thbi gian xu6t
xiy dung dd thi so
s6nh d4ng hinh h$p. ViiSt mO ta vd
th6ng tin mi bpn th6y
duo. c
ttr c6c
tni nay.
4.3. Suppose that we have a sample 43. Gia sri ta c6 m6u x1,...,xn vi ta
x1,...,x,1 and we have calculated
dE tinh duo. c Xn vi 5l cho miu
Xn and Sfr for the sample. Now an
niy. B6y gid ta c6 tiiip quan s6t
(n + l)u' observation becomes
th& n + l. Gi6 sri Xnal vi Sl*, U
available. Let fn*1 and Sl*1 be
trung binh miu vi phuong sai
the sample mean and sample
m6u hiQu chinh cho m6u s& dgng
variance for the sample using all
tit ci n + I quan s6t. '
' n*lobservations.
computed using
fn
Xnal
can
be
and xn*1.
- t)Sl +(n(n
+ l))(xn*1
a) Xn+l c6 th6 duo. c tinh nhu th6
ndo khi sri dung Xn vd xn*1?
b) Chi ra
b) Show that nSl*, =
(n
262
l$ch
DS: a) 2,3313; b) 0,6817.
a) Show how
-?i t Jt')
ttQ
cta d& liQu.
aO
b) Dirng c6ng thirc
.
b) Xay dpng d6 th! dang hinh h$p
b) Dirng c6ng thr?c (3.7.12).
a) By using formula (3.7 .6);
+ttZ-7i/J^)
b) Construct a box plot of the data.
plots.
x6c su6t 0,90:
P(Y < x) = tD((x
chuAn m6u.
a) Dirng c6ng thirc (3.7.10);
than 200 cars to lrave entered the
parking lot with probability 0.90:
By using formula
a) Tinh trung binh miu
sample standard deviation.
ph6t l4nh tt6
DS. a) 0,1038; b) 0,0974.
vi
a) Calculate the sample mean and
construct comparative box plots.
Write an interpretation of the
inforrnation that you see in these
3.51. The number of cars entering a 3.5f. 35 xe con tt6n mOt bEi il{u xe c6
parking lot is Poisson distributed
ph6n b5 Poisson v6i vdn t6c 100
with a rate of 100 cars per hour.
xe tr6n gid. Tim thoi gian cAn thiiit
Find the time required for more
d6 c6 qui 200 xe vio bdi tt{u v6i
.b)
an
-Xr)2.
(n
ring
nSfr*, =
- r)Sl +(n(n + l)[xn*1 -Xn)2.
a random
sample of size 2n from a
population denoted by X, and
4.4*. Suppose we have
E(X)=p
and
V(x) =o2 .L"t
E(x)=lr ve V(X)=02.Dit
x,=*It
x,=*I*, and X2 =*fl*,
a
random sample from a population
p
and variance o2.
vi
cria
a _ Xr + X2 +...+
a
estimator
population variance
'
is
_ 4Xl -2X3 + X5
best? In
b) U'6c lugng nio
variance
X2 and
4.6*. Data on pull-off force (pounds) for
connectors used in an automobile
engine application are as follows:
79.3, 75.1, 78.2, 74.1, 73.9, 75.0,
77.6, 77.3, 73.9, 74.6, 75.5, 74.0,
75.9, 72.9, 73.9, 74.2, 79.1, 75.4,
6.3, 7 5.3, 76.2, 7 4.9, 79.0, 7 5.1.
a) Calculate a point estimate of the
7
of
all
connectors in the population. State
which estimator you used and why.
b) Calculate a point estimate of the
separates
the
the
a) Ering; b)
"!
.
Similarly,
Sl ur" the sample mean
and sample adj. variance from
u6c
a
second independent population
with mean p2 and variance ol.
The sample sizes are n,and n2,
respectively.
.
a)
Show that X, - X2 is an
urrbiased estimator of lrt - pz .
b) Find the standard error of
X, - X, . How could you estimate
trit nh6t, theo
nghia nio?
D&
pull-off force value that
li
the standard error?
X be a Bernoulli random
variable. The probability mass
4.8. Let
b) Tinh udc lugng
r(*;p)
cila
mQt
tQch
.DS. a)
i=75,60;
b) fD = 74,28;
.
4.7. Xnva Sfl ta trung binh m6u vd
phuong sai m6u hi€u chinh tir
mQt t6ng th.3 vdi trung binh p
vd phuong sai
X2 va
of.
Tuong tg,
Sl ld trung binh m6u vd
phuong sai m6u higu chinh tir t6ng
th6 d$c l4p thri hai v6i trung binh
p2 vd phuon g sai oj. Kich thu6c
mdu tucrng ring Id n1, n2 ..
a) Chi ra ring X, -
Xz
lugng kh6ng chQch cria pt
-
Id udc
ltz
.
'cfia
b) Tim dO l€ch chu6n
X,-X, . Bgn u6c luqng d9 lQch
chuAn ndy th6 nio?
4.8. Cia sti X ld BNN Bernoulli. Hdrn
kh6i luqng x6c sudt ld:
(
func. is
74,9;78,0;75,1.
a) Tinh udc luqng ctidm ctia lgc
k6o trung binh c0a t6t ci cdc bQ
li6n k5t trong tiing th6. N6i 16 u6c
lugng nio tli sft dgng vi t4i sao.
phuong sai t6ng th6 vn dQ
DS: b)
6r.
4.6*. Lgc k6o cta bQ truydn trong dQng
co 6t6 nhu sau: 79,3; 75,1; 78,2;
74,1; 73,9;75,0; 77,6; 77,3; 73,8;
74,6;75,5; 74,0; 75,9; 72,9; 73,8;
74,2; 78,1; 75,4; 76,3;75,3;76,2;
c) Tinh u6c lugng di€m cta
c) S=1,6967.
a) Phdi chdng c6 hai u6c luqng d€u
li kh6ng chQch?
what sense is best?
mean pull-off force
of
and
population standard deviation.
X6
"z____- 3
a) Is either estimator unbiased?
b) Wrich
the
chuin t6ng thii.
c) Calculate point estimates
"|__-____E--'
a,6
_ Xr + X2 +...+ Xo
264
o2. X6t c6c
p:
phuong sai
ofp:
q
J
.
4.5**. Gia s0'X1, X2,...,X61i miu nglu
nhi€n tir t6ng th6 v6i trung binh p
Consider the following estimators
A
-z
from
4.7. Xn and Sf are the sample mean
and adj. sampte variance from a
population with mean p and
nio ld t6t hon? Giii thich.
lu-o-ng sau
_ 4Xr -2X3 + X5
population
strongest 5002.
vd X, =;lt*,
your choice.
having mean
in tlre
li 2 u6c luo. ng cria p . U'dc lugng
be two estimators of p. Which is
the better estimator of p? Explain
4.5**. Let Xt, X2,..., X6 denote
the weakest'50% of the connectors
4.4*. Gia sft ta c6 mdu ngSu nhi6n kich
thu6c 2n tir t6ng th6 kli hiQu ti X,
=
{0.
('
[0,
-P)'-*'
x = o'l
otlrerwise
p is the parameter to be
estimated. Find the likelihood
where
function and the loglikelihood one
of a randorn sample of size n.
f(x;p) =] n- (r -P)'-*'
[0,
x=
o'l
nguo. c
lai
trong d6 p ld tham sti cAn u6c
lugng. Tim hAm hSp ly vi loga
him hqp lj cira miu ngiu nhi€n
kich thu6c n.
nfra s6 lpc k6o th6p nh6t.
265
DS. pxl+"'+xn(1-p)n, (xi >0);
(xr +... + x, )lnp + nln(l
-p)
.
4.9. In the normal distribution case, the 4.9. Trong trudng hqp phdn bii chuin,
maximum likelihood estimators of
udc lugng hSp ly cyc tlpi cfia p vd
o2 ld
p and o2 were
[ = X and 6.' =
i
i=l
(*,-X)' l^
Find the maximum
likelihood
estimator of the function
r''(u,"2)=
.
lr=X va 6''
=Ii=l
(x,-x)'1 f".
Tim udc lugng hqp lli cgc tl4i cfra
ham
h(r,o2)=,',F=o.
.,f' = o.
I(*,-* f h
4.10**. Transistors have a life that is 4.10**. fu6i thg cria c6c b6ng tldn rtiQn
exponentially distributed with
tri c6 ph6n bii mfi vdi tham sii )'.
parameter 1,. A random sample of
MQt mdu nglu nhi6n kich thuo. c n
' n transistors is taken. What is the
tlugc rtt ra. Tim hdm mflt tlQ
thdi cfia miu.
pg. 1n"_l(x1+...+xn), (xi >0).
4.11*. ASTM Standard E23 defines 4.11*. Ky thu6t CM,l tlo nlng lugng
ndn vi thuong tlugc dirng ili x6c
standard test methods for notched
ttlnh mQt v{t liQu c6 thay aOi ttr
bar impact testing of metallic
d6o sang ddn hay kh6ng khi nhiQt
materials. The Charpy V-notch
(CVN) technique measures impact
tlQ giim ddn. Ph6p do nlng luqng
n6n (J) tr6n nhi1ng miu th6p A238
energy and is often used to
dem cit d 60oC nhu sau: 64,0;
determine whether or not a
material experiences a ductile-to65,0; 64,5; 64,6; 64,5;64,0; 64,6;
brittle transition with decreasing
64,8;vd64,3.
': ndng lugng chlu n6n
temperature. Ten measurements of
Cii sri' idng,
at
.
I
irnpact energy (J) on specimens of
c6 phin b6 chu6n v6i o = lJ. Tim
impact energy (J) on specimens of
khoing tin c$y 95Yo cho p- ning
4238 steel cut near 60oC are as
lugng ch!u n6n trung binh.
follows: 64.0; 65.0; 64.5; 64.6;
64.5: 64.0; 64,6; 64,8; and 64.3.
266
impact energy.
DS: (64,478+0,653).
''n
A confidence interval estimate
is desired for the gain in a circuit
4.12x*.
on a
semiconductor device.
Assume
that gain
4.12**, Ngudi ta mu6n u6c lugng
khoing tin c{y cho clQ suy giim
distributed with standard deviation
o =20.
trong mQt m4ch cria thii5t bi ben
d6n. Gii st} tls suy giim c6 ph6n
bi5 chuAn vbi o :20.
a) Tim khodng tin cQy 90%o cho 1t
a) Find a 90% CI for p
khi n=20,f =1000
is
normally
when
n=20andX=1000.
D^t.
of the sample?
Assume that impact energy is
normally distributed with o = lJ.
Find a 95% Cl for p, the mean
b) Find a
n =10 and
99oh
b) Tim khoing tin cfy 99Yo cho
CI for p
pt
khi n=10,f =1000.
DS: a) (1000 t 7,4);
when
x=1000.
b) (1000+r6,3).
\
4.I3**. n = 50 random samples of water 4.13**. Nguli ta l6y ngiu nhi6n 50 m6u
from a fresh water lake were taken
nu6c tir mOt h6 nu6c s4ch vd dem
and the calcium concentration
ki,3m tra n6ng dQ canxi (mdl).
(milligrams per liter) measured. A
Khoing tin c{y 90% cho n6ng ttQ
90% Cl on the mean calcium
canxitrung binh ld 0,49 < p < 0,82.
concentration is 0.49 < p < 0.82.
a) Khoing tin cfy 99% tfnh tluoc
a) Would a
tt mdu nu6c d6 ld rQng ra hay hgp
99Yo
CI
calculated
from the same sample data been
lei?
longer or shorter?
b) X6t ph6t bi€u sau d6y: CA 90%
. ^. -l
:,
co hQi
tl€ p ndm giiia 0,49 vd 0,82.
b)
Consider
the
following
statement: There is a 90Yo chance
Phdt bi6u
niy c6 iling khdng? Giii
that p is between 0.49 and 0.82. Is
thich cdu
tri ldi cria b4n.
this statement correct? Explain.
c) Xdt ph6t bi6u sau ddy: NCu 6y
ngdu nhi6n 100 miu nu6c tir h6 vd
tinh ra khoing tin cfly 90%o cho 1t,
c)
the following
, statement: If n = 100 random
Consider
sarnples of water from the lake
were taken and the 90% CI on pr
computed, and this process was
repeated 1000 times, 900 of the
CIs will contain the true value of
vd qu6 trinh niy lap lai 1000 lAn
thi c6 cO 900 khoing tin cgy chfa
p. Phrit bi6u ndy c6 dfrng kh6ng?
Giiithich c6u tri loi c[ra ban.
267
p. Is this
statement correct?
.
Explain.
It is knowrr that ring diameter
is normally distributed with
o' = 0.001 rnillirneters. A random
sample of 15 rings has a mean
diarneter of 7 = 74.036 millimeters.
Construct a 99Yo two-sided
corrfidence interval on the mean
a)
piston ring diameter.
Constiuct
a
confidence bound
95%o lower-
on the
mean
piston ring diameter.
life fbr a new
4.14x*. Mgt hlng sdn xu6t vdng gdng
cho tlQng co 6t6. Bi6t ring tluong
kinh vdng c6 ph6n bii chuAn v6i
o:0,001 mm. M6u ngSu nhi6n l5
vdng c6 dudng kinh trung binh
rr.rbber compound
and has built 15 tires and tested
thern to end-of-life in a road test.
The sample mean and standard
deviation are 60139 and 3645
kilometers. Find a 95oZ confidence
interval on mean tire life.
a) Tim khoAng tin c4y hai phia
99Yo cho tlubng kinh vdng gdng
trung binh.
duong kinh vdng ging trung binh.
a95%o confidence interval on mean
current required. State
any
ne-cessary assumptions about the
tunderlyirrg distribution of the data.
268
distributed.
n = l0
ddi h6i tt6 thu dugc mfic sSng ndo
tt6. MQt m6u l0 b6ng cho ta
f =317 vh S=15,5. Tirn khoing
tin cfy 950/o cho ddng trung binh
ddi h6i (mcA). Ph6t bitiu nhtng gid
thi6t cAn thiiSt ve ph6n b6 cta s5
IiQu.
E& (317+lt,t).
A
is
normally
random sample of
4.19**. A company that manufactures
light bulbs has advertised that its
75 watt bulbs burn an average of
800 hours before failing. The
consumer organization must decide
tivi c6 thii
c udc Iuong bdi viQc do dong
n6i
assumption of
cans yields a sample
standard deviation of 5 : 4.8
milligrams. Find a 95% two-sided
confidence interval fsr o.
cao su m6i, 6ng chiS t4o 15 chi6c
du-o.
the
canned peaches
k! su vd liip nghidn cri'u tutii
tho cia lOp d6i v6i mQt h5n hqp
vd 3645 (km). Tim khoing tin c{y
t6ng th6?
4.18. The sugar content of the syrup in
.
vir dem chtng thri nghigm tr6n
dulng cho tli5n h6ng. Trung binh
mdu vi tIQ l€ch chuin m6u h 60139
phAi
interval on mean natural
frbquency. Is there evidence to
d.5n
gin thi6t vd tinh chuin cria
ES: (231,67 +1,46).
nornrality in the population?
4.15. Mgt
4.16. D9 s6ng cfia ddn hinh
90% hai
tp nhi€n trung binh. C6
ld binh thulng hay kh6ng khi
crja tAn s6
Find a 90% two-sided confidence
support
b) Xay dsng bi6n tin c{y du6i
(khoing tin c4y phdi) 95% cho
95Yo cho tu6ithg l6p trung binh.
DSj (60139t2018).
4.16. The brightness of a television
picture tube can be evaluated by
measuring the amount of current
required to achieve a particular
brightness level. A sarnple of I0
tubes results in X=317
and 3 =15.5. Find (in microamps)
natural
229.48,232.58
mm.
DS: (74,0361 0,0007); 7 4,0356
A
researclr engineer for a tire
rnanufacturer is irrvestigating tire
4.15.
E
f=74,036
on the
delamination
c) Dirng.
4.14**. A manufacturer produces piston
rings for an automobile engine.
b)
of 4.17*. MQt tac gih m6 ti hiQu (mg c0a
sg phAn lnp l6n tin s6 tg nhi6n cta
frequency of beams made from
chim tia sinh ra tt nhirng bin
composite laminates. Five such
composit. NIm chilm nhu th6 du'o. c
delaminated beams were subjected
thu thdp vi tin s6 thu dugc tuong
to loads. and the resulting
ring nhu sau (Hz): 230,66;233,05;
frequencies were as follows (in
232,5 8 ; 229,48 ; 232,5 8.
hertz): 230.66, 233.05, 232.58,
Tim khodng tin cfy
phfa
4.17*. An article describes the effect
DS. a) RQngra; b) Kh6ngi
whether or not to allocate some
of
its financial resources to
. countering the company's
'
advertising campaign. So tlrat it
can make an informed decision, it
begins by purchasing and testing
100 of the disputed light bulbs. In
this experiment, the l00light bulbs
burned an average of X=745.1
hours before failing, with a sample
4.18. Lugng tlucrng cria sir6 trong
nhtng chi6c can c6 ph6n b5 chu6n.
Miu nglu nhi6n n : l0 can cho ra
d0 lQch chuAn 5 . = 4,8 mg. Tim
khoing tin cay hai phia 95%o
cho
o.
E&
(3,30;8,76).
ts
4.19**. MQt c6ng ty sin xu6t b6ng
tldn tld quing crio rdng b6ng tti€n
75W cria h9 tl6t sring trung binh
800 gio trudc khi h6ng. T6 chric
nhitng nguli ti€u dirng cAn phii
quyiit Ainn xem c6 ph4t tiBn li€n
quan ittin chi6n dlch quing c6o cria
cdng ty hay kh6ng. Vi thi5 h9 quy6t
ttinh rrit ng6u nhi6n vi ki6m tra
100 b6ng tldn khiiSu kiQn. V6i thi
nghipm niy, 100 b6ng rldn s6ng
trung binh *.=745,1 gid tru6c khi
ch6y v6i dO lgch chuAn m6u
s = 238,0 gio. Ph6t bi6u gi6 thuy6t
vn d6i thuy6t phir hqp cho tinh
269
of s = 238.0
hours. Formulate null and
. hu5ng ndy. Tinh x6c su6t f nghTa.
Phdi chlng k€t qui ndy tl6m bio
b6c b6 gii thuyiSt tai mirc f nghia
a = 0,05?
DS:2,13=zo,olll
standard deviation
alternative hypotheses that are
appropriate for this situation.
Calculate a
probability.
significance
Do
these results
warrant rejecting the null
hypothesis at a significance level
of
cr
P- gi6
'{
DS:28022.
suspension helmets used by
motorcycle riders and automobile
riders and automobile race-car
. drivers was subjected to an impact
test, and on 18 of these helmets
m6y ttinh chi lutr hdnh ttuqc dA
ngh! ki6m tra ch6t luqng vi th6y
some damage was observed.
a) Find a 95Yo two-sided CI
(confidence interval) on the true
proportion of helmets of this type
that would show damage from this.
b) Using the point estimate of p
obtained from the preliminary
sample of 50 hemets, how many
helmets must be tested to be 95%
confident that the eror in
estimating the true value of p is
c6 l8 chiiSc c6 nhiing hu hai.
a) Tim khoang tin cfy hai phia
95Yo ciuatj l€ thuc cta s6 mfi d4ng
niy s6 b! hu h4i sau cuQc kirSm tra.
b) Sri dpng u6c tugng di6m cta p
thu duo. c tir mdu 50 mii n6i tr6n,
bao nhi€u mfi cin phii tluqc kiiim
tra d3 vdi tlQ tin cfly 95%, sai sii
trong udc luqng gi6 tri thg"c cfia p
=I!--f)'
if it
is required to
days, and the average temperature
zo12=2213.
.
p vd tlQ lQch chuin I m6t. CAn phii
tiiln hdnh bao nhi6u ph6p tto niSu
mu6n x6y dyng mQt khoing tin
cQy mfc 90Yo cho p v6i dq r$ng
20cm?
DS:270.
is found to be 98oF.
a) Nhi€t tlQ nudc sC duo. c chip
nhfn vdi mrlc cr = 0,05 li bao
a) Should the water temperature be
nhi6u?
judged acceptable with o = 0.05?
b) P-gi6 tri cho ki6m dinh niy
b) What is the P-value for this test?
bao nhi6u?
c)
G0
Repeat
b) for the average
temperature l02oF.
4.22. Ngu&i ta nghi6n criu t! lQ c6c
integrated circuits produced in a
mech tich phin h6ng trong qu6
: trinh in inh lit6. Ki6m tra mdu 300
photolithography process in being
m4ch thi th6y c6 14 mach b! h6ng.
studied. A random sample of 300
circuits is tested, revealing 14
Tim khoing tin cfy hai phia 95%
Nguli ta quy6t dlnh ding m6y do
4.24*, The mean water temperature 4.24*. Nhiet tlQ nu6c trung binh t& 6ng
downstream from a power plant
xi th6p l4nh nhd m6y nhiQr di$n
cooling tower discharge pipe
kh6ng n6n vugt qu5 l00oF. Tru6c
should be no more than l00oF.
kia nguli ta th6y ring tlQ lEch
Past experience has indicated that
chuAn cria nhict d0 nu6c ld 2oF.
the standard deviation of temperature
NhiQt dO nudc dugc tlo nglu nhi6n
in 2oF. The water temperature is
9 lin 6 m$t ngiy chsn sin vi th6y
measured on nine randomly chosen
ring nhiQt dQ trung binh li 98oF.
less than 0.02?
defective
bi h6ng bdi
logi ndy.
dg cao laze d6 tto chi€u cao p cria
nggn nfi cao nh6t trong virng. Bi6t
J,
ring c6c phdp do dilng m6y tlo tlQ
cao laze c6 gi6 tr! trung binh bing
laser
20 centirneters?
a) (0,36t0,133);
u1 n
by the
construct a 0.90-level confidence
interval for p that has a length of
nh6 hon 0,02.
D^Si
4.23*.
of the highest mountain in the
country. It is known that
altimeter have an expected value
equal to p and a standard deviation
of I meter. How many measurements
19 c6c mach
c1r
D.Sj (141300 t0,024).
measure the height p
should be made
4.il**, Miu ng6u nhi€n 50 chiiSc mfi xe
270
to
measurements made
A' random sample of
4;22. The fraction of
ft is decided to use a laser
altimeter
: 0.05?
50
.
c0ng
4.23r'.
4.20. Consider the tire life data in 4.20.Xet sii tiou tu6i thq l6p trong Bii
Exercise 4.15. Find a 95%o lower
tfp 4.15. Tim gi6i han tin c{y.dudi
confidence bound for o2 .
95%ocho o2.
4.21**.
cia
CI on the fraction of
defective
circuits produced by this particular
tool.
tri = 0,01 l; dring.
tj
defectives. Find a 95% two-sided
4.23. A
"
li
c) L{p lei b) khi nhiQt tlQ trung
'binh
li
120T.
manufacturer produces
crankshafts for an automobile
engine. The wear ofthe crankshaft
lds:u}t nhi miy sin xu6t trpc truy6n
.) tlQng cho tl$ng co 6t6. Ngudi ta tlii
Lr
t-dil itQ mdn (0,0001 inch) cia
after 100 000 miles (0.0001 inch)
is of interest because it is likely to
tryc khuj,u sau 100 000 d{m, bdi vi
tl6 tlugc xem
li
tli€u quy6t dinh
271
have
an impact on warranty
'.
claims. A random sample of n = l5
shafts is tested and f = 2.78 . It is
knowrr that o = 0.9 and that wear
is normally distributed.
mdn c6 phdn bi5 chudn.
a**) Hly
a**) Test
Hs
:
lt=3.25?
4.26. The rainfall in acre-feet from 20
clouds that were selected at
random and seeded with silver
nitrate follows: I8.0, 30.7, 19.8,
27.1, 22.3, l g.g, 3 l.g, 23.4, 21.2,
27 .9, 3 I .g , 27 .1 , 25 .0, 24.7 , 26.9,
2l .9, 29.2, 34.9, 26.7, 3 1.6.
a**) Can you support a claim that
mean rainfall from seeded clouds
exceeds 25 acre-feet? Use cr = 0.05.
b*) In there evidence that rainfall
is normally distributed?
:p;r 3 (cr= 0,05). .
b) Tim sric mpnh cta ki6m ttinh
n6u P =3,25.
if
V
HD: a) lZl=O,gql <2o12
.=1,96;
b) P=0,35, F=l-0,35
4.1,6. Luqng mua (don
vi miu -
bu6c,
kho6ng t233m3; ctta20 tl6m m6y
dugc chgn ngiu nhi6n
'
vi
c6 sri
dsng ch6t k6t hat nitrat b4c Ii
18,0; 30,7; 19,8; 27,1; 22,3; 18,8;
31,8; 23,4; 21,2; 27,9; 31,9; 27,1;
25,0; 24.7; 26,9; 21,8; 29,2; 34,8;
26,7;31,6.
@
c6 th6 chdP nhfln khing
2u"
dlnh rdng, lugng mua trung binh tir
cdc d6m mdy c6 srl dgng c6c ch6t
if
the true mean rainfall is 27 acre-feet.
manufacturer
of
272
not
exceed 2%.
a)
Formulate
and test
qualified. Use
cr:
an
,&
a) Hs:p=0,02 / H,:p >0,02;
b) Find the P - value for the test in
b) Z = 0,452 =
parr (a).
used in a jet-turbine aircraft engine
is a random variable with
x6c dlnh
du_o.
b) Tim P - gi6 tri.
0.05.
4.28. The effective life of a component
c tl6nh gi6 ld tlat ch6t
Iugng hay kh6ng (l6y o:0,05).
m6y c6
z03z6
*
p = 0,326
.
4.28. Tu6i thg hiQu dsng cira mQt b0
phfln s& dpng trong tlQng co m6y
5000 hours and standard deviation
bay turbin phAn lgc ld BNN vdi
trung binh 5000 gicr vd tlQ lQch
40 hours. The distribution of
chuAn 40 gid. Phdn b5 cria tu6i thq
mean
life is fairly close to a
nonnal distribution. The engine
effective
hi€u dung i6t gAn v6i ph6n bri
manufacturer introduces
chuin. Nhd s6n su6t giOi thiQu mgt
sg cii ti6n tron! qu6 trinh sin su6t
improvement
into
an
thiiSt
the
b! niy mi d6 n6ng tu6i thq
vi giim d0
manufacturing process
for this
component that increases the mean
trung binh I6n 5050 gid
lQch chu6n xudng cdn 30 gio. Gi6
coi lugng mua tir c6c tl6m miy
frdrn the "old" process and a
random sample of nz =25
Tim xdc sudt d6 sg kh6c biiit gita
hai t4rng binh m5u it nhdt ld 25
dirng chdt kiit hat c6 phin h5 chu6n?
components
hay
!b)'C6 li
binh thuong khdng khi ta
li
27.
vugt qu6 2%.Mdu ng6u nhi€n 250
kinh chria 6 kinh h6ng.
nr
=16
b0
tt
l6
Yi2r**. MQt hiing sin xu6t kfnh li6n
lenses is qualifling a new grinding
trdng muSn tt6nh gi5 ch6t luqng
machine and will qualiry the
m6y mii mdi vd s€ chdp nhfn ch6t
machine if the percentage of
Iusng cria m6y ni5u fi le l6i kh6ng
defects does
a) Ph6t birSu gin thuy6t
contains six defective lenses.
quy trinh sin
.ph$n tlugc chgn
su6t cfi vd mdu ngAu nhi6n nz=25
kiit het vuqt qria 25 ikrn vi
interocular
polished lenses that contain surface
lenses
life to 5050 hours and decreases
trung binh
A
of 250
the standard deviation to 30 hours.
Suppose that a random sample of
nr =
components is selected
c) Tim s&c m4nh c0a ki6m dlnh
ni5u gi6 tr! thgc cria lugng mua
4.27**.
randorn sample
sri miu ngdu nh.i6n.
kh6ng?
c) Compute the power of the test
A
appropriate set of hypotheses to
determine if the machine can be
kiiSm dlnh
H9 : p = 3711,
p=3/H1 :p*3 (u=0.05).
b) What is the power of this test
cho sp b6o hinh. M6u ngiu nhi6n
15 trgc tluo. c kiiSm tra vi th6y
*.=2,78. Bi6t ring o = 0,9 vd d0
b9 phAn chgn
is
tt
quy trinh
cii
tirin.
selected from the
"improved" process. What is the
gio. Gi6 sri ring, quy'trinh sin xu6t
cfi vd quy trinh cii ti6n c6 th6 xem
probability that the difference in
the two sample means X, - X, i9
nhu ld nhfing t6ng thti ttQc lAp.
at least 25 hours? Assume that the
old and improved proc'esses can be
HD:
regarded as
-N(50 ; +02 tt6+lo2 tzs1;
independent
Z=X-Y
P=P(zl>zs).
populations.
'
D&
0,9938.
273
4.29,"'. A consumer
company
is
electronics
comparing
for use in
tri gia
dr,rng so s6nh
of
nhau sir dpng trong m6y thu hinh.
its
Lo4i b6ng A c6 tlQ s6ng trung binh
100 vi d0 lOch chuAn l6; trong khi
d6 logi b6ng B c6 c6 tlQ sdng trung
binh chua bi6i, nhung c6 th6 coi c6
tlQ lQch chuAn gi6ng nhu cria lo4i
A. Miu ng6u nhi6n n : 20 b6ng
mdi loai duqc rtt ra vi tinh
A has
100 and
of
standard deviation of
ttiQn
dQ s6ng cria hai lo4i tldn hinh kh6c
television sets. Tube type
mean brightness
t. Hing
the
brightness of two different types
picture tubes
4-)pt
4.31*. The burning rates of two 4.31*. Nguoi ta nghidn criu vfn t6c
different solid-fuel propellants
ch6y cta 2 bq .fAy dirng nhi6n liQu
used in aircrew escape systems are
rin trong hQ th5ng tho6t hi6m cho
being studied. It is known that both
propellants have approximately the
same standar.d deviation of burning
16, while
tube type B has unknown mean
brightness, but the standard
deviation is assumed to be
identical to that for type A. A
random sample of n = 20 tubes of
each type is selected, and Xs; X^
is compr:ted. If pe exceeds p4,
nhi s6n su6t sc chgn loei B tlem sti
dpng. Sg kh6c bi€t quan s6t tluo. c
le fa-fe =3,5. B?n sC dirng
the manufacturer would like to
quyiSt dinh ndo vd t4i sao?
adopt type B for use. The observed
a) Test the hypothesis that both
propellants have the same mean
HD:
burning rate. Use o = 0.05.
rate; that
is fs -X^ =3.5. What
decision wouid you make, and
z
=
for use by an
electronics
component manufacturer. The
breaking strength of this plastic is
important. It is known that
ol =o2 =l'0 psi. From a random
sample of size n1 = l0 and fr2=12 ,
we obtain
ir
The company
I
=163 and
x,
=155.
will not adopt plastic
unless its mean breaking strength
exceeds that ofplastic 2 by at least
.
.
l0 . psi. Based on the
sample
information, should it use plastic
I ? Use cr = 0.05 in reaching a
decision.
cm/s. Hai mdu ngiu nhi6n kich
per second. Two random samples
of n1 = 20 and nZ =20 specimens
n2=20 ttuqcki6m
tra, v$n t6c ch6y trung binh Id
*l = l8 vi f2 = 24 cm/s.
b) What is the P-value of the test in
parta)?
b6c b6 khi
c)
Construct a 95Yo confidence
interval on the difference in means
\.
4.30**. Two types of plastic are suitable
centimeters
centimeters per second and
72 =24 centimeters per second.
(x -Y) t thaz t zo + 162 t 20 ;
-N(O;l), gi6 thuytit bi
vi chi khi Z> ro.os.
why?
=3
are tested; the sample mean
burning rates are El = I 8
Xs;Xa. Ntiu ps vugt qu6 pa,
difference
is o1 = 62
D& Kh6ng.
4.30*t. MQt c6ng ty thi6t b! di€n
Itl
nh$n
- rt2. What is the
practical
meaning of this interval?
th6y c6 th6 dirng ttugc 2 d4ng chSt
d6o. Sric b6n chiu va ttfp li quan
1,0 Psi. T& mdu ngdu
nhi6n kich thudc nl =10 vi
nZ=12 ngudi ta nhfln dugc
Xl =163 vit x2 =155. C6ng ty sE
kh6ng chgn chSt d6o I trir khi sric
quyiit ttlnh.
DS:
2=-4,67 <-1,645;
kh6ng.
t6c chdy trung binh.
b) P - gi6 tr! cria ki6m dinh d
phin a) li bao nhi€u?
c) Xdy dgng kho6ng tin cQay 95Yo
cria sU kh6c biet trung binh
nghTa thr.rc t6 cria
It1 - pZ. f
khoing niy
li
gi?
D& a) lZl=6,lZq=zo.Mt2
b)P-gi6tri =9,696'
G -Y tzol2lo!
r
t n1 +
ol / n2)
1,959).
432*r'. Two chemical companies can +Yz**. Hai hiing hod chdt c6 th6 cung
supply a raw material. The
c6p nguy€n li€u th6. N6ng ttQ cfia
concentration of a particular
lo4i ch6t ndo tl6 trong nguy6n liQu
element
in
material is
important. The mean the same,
*
this
ln di6u quan treng. NOng ttQ trung
concentration for both suppliers.is
but we suspect that the variability
hai nhi cung cdp ld
nhu nhau,' nhrmg. chtng ta ngpi
ring ss b6t 6n tlinh vd n6ng dQ c6
in a
random sample of size
nl =10 and n2=12, we obtain
thi! lA kh6c nhau vdi hai nhd cung
between the two companies. The
trong m6u ngSu nhi6n
standard deviation of concentration
274
a) Kii-lm tlinh gi6 thuy6t (cr = 0,05)
ring ci 2 hQ thiSng tISy c6 ctng vfn
= (6
bAn chlu va tffip cria n6 trQi hon so
v6i cria chdt d6o 2 it nh6t l0 psi.
Dga tr6n th6ng tin vO m6u, phii
chlng ch6t I s€ dugc dem vio sri
dgng? Dirng cr = 0,05 d6 dua ra
thu6c n1 =20,
c)
trqng. Ngudi ta bilit ring
Gl = 02 =
phi c6ng. ni6t ring c6 hai bQ rt6y
c6 ttQ l€ch chuin cria vfn ttic ch6y
gin nhu nhau, d6 ld o1 = a1=i
binh cria
ci
c6p. DQ lQch chu6n
cia
n6ng
ttQ
l0
16 sin
xu6t Uoi hiing ttu? nhdt ta s1 = 4,7 gll;
275
in a random sample of
= l0 batches produced by
company I is s1 = 4.7 grams per
nl
s2
=5,8 dl. C6 ttri ch&ng c6 d6
liter, while for company 2, a
random sample of n2 =16
th6 ld khec nhau?
batches yields s2 =
D& Kh6ng:
5.8 grams per
liter. Is there sufficient evidence to
conclude that the two population
variances differ? Use
manufacturing interocular
lenses
used in the human eye following
cataract surgery. Three hundred
lenses were tumble-polished using
the first polishing solution, and of
this number 253 had no polishing-
induced defects. Another
300
lenses were tumble-polished using
the second polishing solution, and
196 lenses were satisfactory upon
completion. Is there any reason to
believe that the two. polishing
solutions differ? Use cr
=
0.01.
Discnss how this question could be
ffiered with a
interval on pl
kiSt luQn
ring hai phuong sai tiing
(f6.e75(1
(cr:
0,05).
5;9) < fs,e5(l 5;9) < F
= 1,523 < f6.e, (l 5;9) < f6,62,
(1
5;9))
cr:0.05.
4.33*. Two different types of
polishing . solution are being
evaluated for possible use in a
tumble-polish operation for
4.33*. Hai d4ng kh6c nhau cria
ch6t b6i tron dugc xem x6t d,i
ding cho m6 thay thui tinh th6.
300 thu! tinh th6 dirng ch6t b6i
tron tht nhdt vi trong s,i5 d6 253
kh6ng c6 tryc trfc gi. 300 thuj'tinh
th6 tin ring hai chAt b6i tron ld
kh6c nhau hay kh6ng vdi mtc f
nghTa cr : 0,01 . Xdt xem vfn d6
.
, ..,1 .,.
t
niy
c6 th6 gi6i quy6t du-o. c th6ng
qua khoing tin cfly cta p1 - p2 .
lzl=
lr,
Itt
-Il(t
I n1+t I n2)
.
average weight loss
of at
least
3
ligu suy gi6m trgng luqng ghi lai 6
<16y. Sir dgng thrh tuc
gi6
ki6m dlnh
thuy(5t d6 tra ldi ceu
a) Do the data support the claim of
the producer ofthe dietary product
a) Phni chdng sii ligu ln phi hgp
v6i khing tllnh cria nhi sdn xu6t
with the probability of a type I
,
T.
' pnam
'I
gram Deo vor xac suat sal
san
error set to 0.05?
lAm lo4i I bing 0,05.
b) In an effort to improve sales, the
b) Trong mQt n5 lUc tt6 ndng cao
mric ti€u thg, nhh sin xu6t xem xdt
loi dim bio cfia hg tir "it nh6t :
pound" thinh "it nh6t S pound".
Ki,5m tlinh ldi tl6m bio m6i.
4.34**. Ngu&i ta quing c6o cho mQt sin
phim gi6m b6o dang l6ng ring st
dqng sin phim trong vdng 1 thring
eey ra gi6m trgng lugng it nh6t 3
phAm ndy trong vdng
I
th6ng vd s6
bing du6i
h6i sau:
1
165
161
5'
155
150
2
201
195
6
143
141
3
195
192
7
150
146
4
198
193
I
187
183
HD: a)
Tu =2,565 >
-t6.65(8- l)
;
thing.
D
d t?-( fo=ffil8
=-0,6982
> -to.os (8 - l) ; tiling.
/ \)
4.35**. Let X denote the number of 4.35**. Gqi X lA s6 v6t nut quan s6t
flaws observed on a large coil of
duqc t.en mQt cuQn l6n th6p ma.
galvanized steel. Seventy-fi ve coils
,
-p2.
product for one month results in an
resulting weight loss data are
reported below. Use hypothesistesting procedures to answer the
following questions
-El
=5,362>2,58=zop
C6; 0 * (0,19t 0,09)
pound. T{m ttdi tugng sri dpng s6n
claim.
vd th6y c6 196 tt4t y6u cAu. LiQu c6
DS:
pounds. Eight subjects use the
product for one month, and the
producer is considering changing
its claim from "at least 3 pounds"
to "at least 5 pounds". Test the new
th6 kh6c dirng ch6t b6i tron thri hai
confidence
4.34**. A liquid dietary product implies
its advertising that use of the
276
khi d6 vdi hdng thri hai, mdu
ngiu nhi6n n2 =lf 16 cho ra
r trong
are inspected and the following
data were observed for the values
of X:
Values (gia tri)
1
2
3
4
5
6
7
I
1
11
I
13
11
12
10
I
Obs. Frequency
(tAn s6)
75 cuQn tld dugc kh6o s6t vi thu
ttugc sti liQu sau v6 c6c gi6 tri
cria X:
277
a)
Does the assumption
Poisson
of
. a) Ph6i chlng gii thuy6t ring
the
distribution
ph6n br5 Poisson ld m6 hinh x6c
appropriate as a probability model
su6t cho s5 ligu
for this data? Use a
c6
:
0.05.
b) Calculate the P-value for this test.
niy
b) Tinh P - gi6 tri cho ki6m dlnh
nay.
HD: a) 7'= 4,907,9h8p lai;
(n,-nP,)'
=f
i]i
nPi
A
B
c
D
1
41
20
11
16
2
31
12
I
15
3
15
11
15
11
xem nhu ld
lj? Dirng cr:0,05.
b)xz
Machines (m6y)
Shift (ca)
Test the hypothesis (using o = 0.05)
that breakdowns are independent of
the shift. Find the P - value for this
=6,63
:
test.
f i5* tra eie thuyiSt (dnng
|
I
I
I
= 0,05) ring sli IAn h6ng h6c ld
UO: l{p v6i ca. Tim P - gi6 tr! cho
cr
ki6m tllnh niy.
X as the number of
4.36. Gqi X ld s5 chai tt6ng thi6u trong
a filling
operation in a carton of 24 bottles.
24 chai.75 thtng du_o. c
ki,3m tra ve si5 lipu sau dugc ghi
X2 = 9,532
Sixty cartons are inspected and the
lei:
9,532=af,.1ao(6)=+ P = 0,146.
4.36. Define
underfilled bottles from
following observations on
X
HD:
mQt thirng
< Xt,osQ.3) = 12,592
are
'recorded:
Values
Frequency
0
39
2
1
23
3
Gia tri
0
1
2
3
lan so
39
23
12
1
$6.s.nAr
a) Dga vio 75 quan s6t ndy, phii
12
1
ching ph6n b6 nhi th&c li
m6
hinh c6 th€ ch6p nhQn tiugc? Tgo
a) Based orr these 75 observations is
ra mQt tht tqc ki€m ttinh pht
a birromial
distribution an
appropriate model? Perform a
v6i
goodness-of-fit procedure with
b) Tinh P - gi6 tr!.
ho.
p
cl = 0,05.
o = 0.05.
b) Calculate the P-value for this test.
4.37**.
A
>*
breakdowns are collected:
278
cHuoNG
v
IIdi quy tuy6n tfnh tlon
5.1*. Roadway surface temperature (.r) 5.1*. Nhiet tIQ mat tlulng (x) tlugc coi
ln li6n quan d6n d0 bitin dang b6
is thought to be related to
mat O).sii tlCu t6m tit Ii
pavernent deflection (y). Summary
Simple Linear Regression
quantities were
Iy, = tz,ts,lv? =8,86,
Iy, = tz,ls,\v? =8,86,
I*,= A78, lx! =143215.8
I*,=1a78,
f
xf =143215,8
I*,Y,=1083'67 'n:20'
I*,Y,
'n=20'
a) Calculate the least squares
' estimates of the slope and
a) Tinh u6c lugng binh phuong
cgc ti6u cria hQ siS g6c vi hQ sti
intercept. Craph the regression
b) Dnng phuong trinh tlucrng h6i
=1083'67
company operates four 4.37**. MQt hdng sri dpng 4 :nlrhy 3 ca
m5i ngiy. Tir nh{t kf sin xu5t, dir
From production records, the
lipu sau vd s5 su c6 du-o. c thu
following data on the number of
thfp:
machines three shifts each day.
r4r
line.
b) Use the equation of the fitted
line to predict what pavement
chfln. Lflp ad th! Auone h6i quy.
,
quy udc lugng a6 au uao d0 bi6n
d4ng mflt tluong s6 quan s6t dugc
n6u nhiQt tl$ b6 mat le 85oF.
279
deflection would be observed when
the surface temperature is 85oF.
c) What is the mean pavement
deflection when the surface
temperature is 90oF?
b) Use the equation of the fiUed
line to predict what percent yield
would be observed wlren the
c) DQ bii5n @ng b6 m{t trung binh
ra sao khi nhiet dq bA m6t te 90oF?
d) D0 bitin dang bd m{t trung binh
biiin dOi cO bao nhi€u khi nhiQt d0
bC mat UitSn a6i tof'z
temperature is 1600.
c) What is the mean percent yield
d) What change in mean pavement
deflection would be expected for a
loF change in surface temperature?
Find
c) HiQu su6t trung binh bing bao
nhi6u khi nhi€t d0 ld l70o?
d) Find the estimate of o2
d) Tim u6c lugng cria s2.
.
D& a) Y =-4,473+0,496x+e;
linear regression 5.2. Tim MHHQ tuytin tinh don cho s6
model to the oxygen purity data in
Iieu d0 s4ch oxy d B6ng 5.1; ki6m
Table 5.1; test for significance of
tllrrh 1i nghla cfra vi_6c dirng
b)74,95%; c)79,9\Yo;
d)62=11,98.
5.4* . The
MHHQ.
,,S; Y = 7 4,284 + 1 4,947 x + e;
Consider the following
pairs (x;,y1
), i =
l0
data
1,..., 10, relating
y, the percent yield of a laboratory
experiment, to x, the temperature at
which the experirnent was run.
fiber is stored in a location without
(xi,yi), i = I,...,
content (y) of a sample of the raw
l0 li6n h€ bi6n y, hiQu su6t (%) cta
mQt thi nghiQm v6i x, nhiQt tlQ t4i
material were taken over 15 days
with the following data (in
tl6 thi nghi€m thgc hiQn.
percentages) resulting.
5.3. Xet I0 c{p
st5
tieu
X1
Yi
1
100
45
2
110
52
7
160
3
120
54
I
170
76
4
130
63
o
180
92
a)
ua
Calculate
62
the least squares
estimates of the slope and intercept.
Graph the regression line.
6
10
X1
Vi
150
68
190
a) Tim udc lugng binh phuong cuc
vi hC s6 nhqn.
Lap d6 thi cfra tlud'ng hOi quy.
tii3u cria h€ sti g6c
kho l5 ngdy th6 hipn d bdng sau.
X
46
53
29
61
36
39
47
49
v
12
15
7
17
10
11
11
12
x
52
38
55
32
57
54
44
v
14
o
16
I
18
14
12
75
88
noi kh6ng c6 ki6m so6t d9 Am. D0
6m tuong a5l 1xy d ncri luu kho vi
d0 6m (y) cia m6u v{t tiQu th6 luu
a humidity control. Measurements
of the relative humidity (x) in the
storage location and the noisture
(l 9); c6.
i
5
280
16,625
raw material used in the 5.4*. Vat liQu th6 dtrng tlti sin xuAt m6t
of a certain synthetic
lo4i sgi hfru co duo.rc c5t vio kho 6.
production
ltrl=ffi=n,78e
> 2,093 =
sdt dugc sE bing bao nhi6u khi
nhi6t ttQ la 1600.
when the ternperature is l70o?
a simple
regression using the model.
b) Dirng phuong trinh h6i quy thuc
nghi€m a6 ag Uao hiQu sudt quan
Find the mean square
estimate.
I
I
I
rl
ti,
u6c
luo.
ng binh phuong cgc
tieu.
,Sry
=134,517 +147,600x+e;
rrlt.-. ,l:::1 ,"
5.5**. An article in the rournar
sound and vibration described a
I
I
rroise exposure and I
thanh
vi
tr6n rap chi Am
Dao dQng m6 ta mQt
study investigating the relationship
nghi€n criu vA m6i quan
between
luc ti6ns 6n
vi
hQ
gifra 6p
su gia
eia tlng
tlns huy6t
huv6t
281
