Pqt Question Bank

r

16 Probabi l i ty and Random pr

Solution

Let X: the milage'in thousands of miles of the twe.Then X is a R.V with p.d.f

,x)0

x<0

(i) Probability that one of these tyres will last for afinost 10,000 miles

(ii) Probability that one of these tyres will rast any where from 16,000 to l= p[16 <x:24]

r24 1 --:-= lu ;i" 'od*_ e_o.8 _ e_1.2

=0.4493 - 0.3012

= 0.1481

(iii) P[One of these tyres will last atleast 30,000 miles]

= plx >:o]= f I

"-ia*r3o Z0= g-1'5

=0.2231

= plx < roJ= Jor(*)a* =

"['1"-**| * l lo

=l-e-r lL l ,

= 1- e{'5 = 0.3935

,r Example 13J A random variable X has the distribution function

[0,F(x) =Lk(x-1)4

i fxSli f l<x<3i fx>3

Find (i)P[2. X. 3J and (ii) Mean ofX.

Random Variables L.L7

Solutionrhe p.d.f of X is f (x)=*t(lr)

fo, i fx <1

.'. r(x) = | ;ntl<

-')' iil:i.,

[,- t(4* = r = +rcf (* - 1)' dx = 1

ouf (* '')''l '='L4l,

; r(x)=[i,'.-r'

116k=1=k=G

for l<xS3

otherwise

( i)p[2.x.3]= f r(x)ax

=l|r*-1)'d*=*[t,.-'l']'

15=-t6

(ii) Mean E(x) = f,,.xr(x)ax

=i f .(,.-r)'d"r Ix(*-r)o (*- t) ' ' l '+[ 4 20 .1,

=i[io')-rPa]Mean= l-!=z.a

= !f t, -9.l4L sl

Example | 4

The diameterofanelectric cable Xis acontinuous RVwitfrn'a'f {(x)

= tot(t - x)'0 < x < 1'

Find the (i) value of k. (ii) rhe c.d-f of x (iii)e[x =;l ;'*t;)(iv)

a number b such that

r (x<u)=r(x>t)

F-

Solution(i)To findk

Probabil i tY and Random

f_r(x)ax =t

nX(, . -")ax=t

kf l- ! ' l=1L2 3l

=!=1. ' .k=66

r(*) =[3"0-.)iffi;:"='

(ii) c.d.f of X : F(x) = f o* (t - *) ax

=u[,4-4][2 3 )

=3x2 -Zxt ,for0 < x <l

...r(*)=[3;'-r.' l::L'I.,

( 1t1 of l .* . ! l

1ii,yr[x =il :. x. ?l =-$,-]5 ) tL; ' " ' ; l

_ fii(.-.')*- fii(,.-"')d*

L. 19Random Variables

(iv) To find b such ttrat P (X '

b) = P (X > b)1

" 'P(x<b) =:

rb/ r \ . t+6[ (x -x ' )dx=-

(rr'- z*')or=l

+6bz -4b3 -1=0

6 = ! satisfies the above equation'

, l; . o=i

ExamPle l5

A cont inuous R'v X has u p'd ' f f (x)=3x2'0<x<1' Find a and b such that

(i)r(x < a) = r(x > a)and (ii)r(x > b)= 0'05

Solution( i )P(x<a)=r(x>a)

.'. f,r*ta*= fl*'a*1

. . .at =l_a3 + ut =1,

" 'a=0'7937( i i )n(x>b)=o'os

A+ flx'a*=0'05

"' b3 = 0'95

= b = 0'9830

*:rltt""

lu:r,*.*r*" random variabre with c.d.f F(x) = p[x < x]. Show that

E(x) = f [t-ttlrl]*'Solution

Since X is non-negative,

E(x)= f *(4*

.39 Probabi l i ty and Random proce

1.3 Moment Generating Function and properties

/ Moment Generating Function

\-/The moment generating function of a r.vXis defined Mr=o(t)=81",,1 ,)

origin ) Similarly the M.G.F about the mean p is defined by M *=r(t)= Ol",(x--

Notat ion

1. M.cF about origin uoQ)= ul/,)

2' M'G.F about mean lt u r(t)= tle'@-a)lIf Xis a continuous R.V. then,

M,(t)= fa.y6ya*If Xis a discrete R.V. then,

Mr(t)=le"r(x = x)

Note

U,(t)=e- 'Mo(t) .

Moments about the or ig in r . . . . , - - )M*(t)=nl"' ' l=slt*+

"" t 'x' I' t -uL' J-"1 ' l u - a

* . . . .+-- , +. .J

. . .u * (t) =r + L,n (x1+ln@,)*.. . . .*L r l*,1*.. . .

= M*(r)=t* lp i* !p i* . . t '

, l ! . , 21. . " .+-- ; .1t"+. . .

Hence the rn moment about the origin

It, = coefficient or Lnu, (t)

Note

r) p: =#lM-(,)1,=,

2) M*(t)(about x =a)= nld6-)lana

u r ( t) =r * f ,ai .*r;

+,. . . .* \ p,, *. . . where

Random Variables 1.39

- tPi = nl(x - o)' J= r'h moment about x = a

3) M r(t)(about mean t) = zl"'(*'llona

M x (t) = r + f,.n

**o, *.....*\ o, *... where

P, = El(x-t)'] =r'h central moment

Limitations of M.G.F

l. A random variable X mayhave no moments although its m.g.f exists.2. A random variable X canhave m.g.f and some or all moments; yet the m.g.f does not

generate the moments

3. A random variable X can have all or some moments, but m.g.f does not exist exceptperhaps at one point.

Properties of M.G.F

I . M*=oQ)="-*M*(t)

Solution

M *=, (t\ -- nl;v -"11 = "-"' El"o )= e-'' M x (t)

2. T\e M.GF of the sum of a number of independent random variables is equal to theproduct of their respective moment generating functions.

i.e., If X1,X2,...,Xnare z independent random variables, then

M *r+x2+.,......+xo (t) = M *, (t)M "r(t).....at

,^ (t)

3. If M x (t) = zle'* f then M o Q) = M * (Ct)

Solution

u u Q)= r[el*11 = nle{"v 1= M * (ct\

4.If y= *+ b,then Mr(t)="*M*(al) where U"(t)=M.G.Fof.X.

ur(t )= u[" ' " ]

I rx-o\ l=sl" ' l o ) l= '

LJ

I

s. rr Y = +,

then M, (t) = "-""

o M "[f)

r"u*t "f change of origin and

Solut ion

L.40 and Random Pro

Uniqueness Thoerem on M.GFThe M.G.F of a distribution, if it exists, uniquely determines the

M * (t\ = M, (t) = X and I are identically distributed.

--'^u1"til.1T'"(;)

Solved Examples

Example I

f rFind the M.G.F of a random variable having ttre p.d.f ;(") = ] 3'-t

< x <2

10, otherwise

Solut ion: _r r , [^ , , - l ,Mr(t)= f , ! " ' 'd '=: l? |' utJ J l t I

L Jx=_l

) l

-" ' ' -"- ' for t +o

3t

For / = o, t t , t * (o) =l | ,* =t

f " t t

_"-t

. . .ur( t )=l 3t , t*u

Ll , t =0

Example 2

Find the M.GF of a r.v Xwhose moments are given by pi =(r +t)tZ'

Random Variables 1 A1Io. ! I

SolutionMXQ)=f!,:

=i"L't'+t)t2'

=i(" +t't(zt)'r=0

:.T\e m.g.f ur(t)=(t-zt)-' - I\ /

(t-zt),

Example 3

, Find the m.g.f of a r.v Xwhose p.d.f is defrned by

v fx , for o<xslf (x)=lz-x , for l<xS2

L0 ,otherwiseHence find mean and variance of X.

Solution

M*(t)= Lf f @)dx

= fxsdx+ f tr_,1**| (a') ,"- l ' l - " "-12=L'l.?j 7),.1'-ai.;],e' e' I e, e2, et=___+___J-___t t ' t ' t t2tz

l -2e' + e2' ( t - " , \ '=---v^ =[. , Jh-"'l '

\.- .M*(t)=Y

To find mean and variance:

M*(t)=gl=(r.?t. l'u=l*,r.0)],=, =M',(o)

L.42 Probabi l i ty and Random P

M "(t)=z(r*+.+. l[+*l*.. ..], [ 2 6 ) \23 )

.'. Mk @)= rft)(i) = I + Mean = 1

ui(t)=rll.l. l[+.' I ^t' ' " * lt+' L2 3 )Lz l*" l+zl '+t* 6 lL3

Mrp)=r[i.i] =l='F')=Ivarianceor x = n(x') - {a(x)\' =I-, =I

Hence Mean=t ana variance=*

Example 4

A random variable Xhas probability function p (r) = |,

* =1, 2, 3,.... Find

andVariance.

Solutionu*(t)=ir".*

r=l

=ir4l'2t1lx=t\ o

)

. l - . r , :2 I=1lr*** [1] . r21 2 \2) )

' .u*(t)=++=*2

d=Mk(q=*l*),="

l1r-"'1"' -r'(*')l=L--pe-1,,


...p'r=l=a-1 -2'r' LQ-"')' ),=,

,t=l#"" (,)],=,I p - n )' (2", ) - +", (z - " )(-", )fLE],=,

-lQ-"')za .+""1L--P4-),=,

PL =z+ 4=6... Variance = p; - (A)' = 6 - 4 = 2

Hence Mean = 2 and Variance = 2

Example 5

Find the m.g.f for the distribution with

p(*)=

atx =I

atx=2

otherwise

?3

IJ

0

Also find Mean and Variance"

Solution

Xtakes the values I and2with probabilities f

and 1 respectively.

u"(t)=le"r(x =x)

" 'u*(t)=?"' *!" ' ''3 3

uk@=?"'*?" ' ''3 3

uIQ)=?" '*1" ' ''3 3

1. Probabi l i ty and Random P

:. pi = Mi p) =t *u n = Mi(o) = z

Hence Mean fi=l *OVariance = p'r-@)'

i.e., variance l r-f, =1

Example 6

Find the m.g.f of a R.V with p.d.f f (r)= ae-*,x> 0 and =0, otherwi

first four moments about the origrn.

Solutionu *(t)= f e'*ae-*dx

t@ -(a-r\x ''o t r" ' 'ax

=ol -"-("4.1*I o- t lo

; .M*(t \=* i r t <u

Arso u*(t)=[t-;) ' =r*I.r.*.....

The r"moment about the origin is given by

trp',=co-efficientof . in M*(t)

, r l" 'P"

=; i ' r =l '2 '" '

, r , 2 , 6 , , , , . , 24= Pr=-| l tz =

t ,h =7

^o Fo=}

Example 7

i Find the m.g.f of the distribution defined by dF =!"-l*16*,..o0 < 'tr < €'v variance.

' 2

Solutionur(t)= f*e" f (x\dx

=! fe"e+ax

L.45Random Variables

=ilLo 'ddx+ ff t."'a*)

= i[ Lrt'.')' d, + f e+'-'t' *)

={ld9f *{-"t'-u}-lz[ r+t l * L l - t J, ]

r [ r t I- - l -+- |

2l l+t I - t . )

. .M"(t)=|7,Vt t

Now M" (t) = (t-"1)-'

=I+ t2 + to + . . . . .

Pi=o;PL=21=2

.'.Variance h= 6-Q4' =). andMean=0

Example 8

Find the m.g.f of X whose p.d.f is givenby f (')=1-l'xl'-1<x<1

SolutionM"(t)= fta'71'1a*

= f,{ (l+x)dx+ Xr'(t -r\eT .- , , - lo | , , r r l l

=l4,1*.\-*l . l {1r- ').+ |Lt ' ' t ' f_, L l ' t ' )o

d\e' 'e 'e '1=T-7-7-;-7-7

"'u*(t)=t?Example I

The random variable xassumes the value x with the probability

p(x = x)= q*-r p,x =I,2,3,...Find the M.GF ofxand find its Mean and variance.


Example I

Xis arandomvariable followingbinomial distributionwithm eanz.4andvariance 1.44. FindP(x >s) and r(l . x < 4)Solution

Let X - B(";p). Then p[x =xf=nC,p"q, * , x=0,1,2,. . . ,n.'. Mean = np and Variance = npq .

+ np =2.4 andnp7 =1.44

. ' . q =nPq =l '4 =0.6np 2.4

p=l-q+ p=0.4 and ,=2.4 =60.4

Hence Pfx = xf = 6c,(0.q'(0.6)u-', x =0,1,2,..,6

p[x > s]= p[x = s]+ r[x = o]= 6Cs (0.4)5 ( 0.6) + 6C u(0.4)6= 0.0369 + 0.0041

.'. P[x >5J= 0.0410

P[t < X < 4]= p[X = 2]+ p[X = 3]+ e[x = +]= AC z(0.4)2 (0.6)a + 6q @.q3 e.6)3 + 6 C o(0.4)o (0.0),= 0.3 I 104 + 0.27648 + 0J3824

. ' .p[r<x <4f=0.72sgExample 2

A perfect cube is thrown a large number of times in sets of g. The occurence of 2 or 4 is;alled a success. In what proportion of sets wourd you expect 3 successes?Solut ion

pf2 or 4l=3=+'6 3

1)Hence O=;,q =f andn =8

Pfx =31 = 8c, f1)'f3)' = 0.2730\3 i \ .3 i

i.e.. In nearly 27.3 percent of cases, one would expect successes.

Probabi l i ty and Random

Example 3

A arri' B play aganne in which their chances of winning are in the ratio 3:?-

of winning at least 3 games out of 5 games played.

Solution

Probability forl's *it"i"S = J

Probability for B's *irrrrirrg = |

IEt XNumber of games in #cn I wins.

PIX =rl=nC,P'qn-'

=sc-[1)" (?\*'' \s / \s. /

P[atleast 3 games out of 5 forl's winning]

= plx>jl

= P(3) + P(4)+P(s)( z\3 ( t\2

sc [1).(.?)*r*[1)'=r.,[;J [;J

. _\5/ \si -\s/= {1t0,. 27 x 4 +5 x 81x 2 + 2a3l

5" '

= 0.6823

Example 4

Six dice are thrown 729 lines. How many times do you expect atleast 3 dicc

SolutionLetX the number of times the dice shown 5 or 6.

P[5 or 6,=l*l=It663

12; .p=1^o4=1

Here n=6

To evaluate the frequencY of X >3.

By Binomal theorem,/ r \r /,r \6-r

rlx =rl=6C,ttJ t;J where r =0,r,2,....,6

Random Variables 1.5 g

p[x >3J = p(3) + p@)+p(s) + p(6)

/ r \3/e\3 -^( t \o(Z\ , ( )s(2\ / r \6=ac, l : I l : | +6cot: l t : l +oc-- ' ( r / ( :J \3/ \3 ) ' l i ) l ;J.uql . j

= 0.3196.'. Expected number of times atleast 3 dice to show 5 or 6

= uxp[x >z]= 729 x 0.3196 = 233

Example S

In a precision bombing attack, there is a 50o/o chance that any one bomb will strike thetarget' Two direct hits are required to destroy the target completely. How many bombs must bedropped to give a99%o chance of completely destroying the target?

Solution

If z bombs are to dropped, then out of these atleast two should hit the target to destroy itcompletely. Hence we require

p(2) + pe) +...+ p(n) > 0.99 where

P@ =(:)p' Qn- ' ,x = o,t ,2, . . .nwith p = n - !\r,/ 2

Hence 1- ^P(0) - p(l) > 0.99

_ /r) , / r ) ,. . . r_t ; l_,1: l >o.rr\ - t \z/

n+l:+ < 0.012',

:+ 100(z +t) < 2,

The least value of z is 11.Hence, 1l bombs should be dropped,

Example 6

The probability of a man hitting a target i. |

. co If he fires 7 times, what is the probabiliryof his hitting the target at least twice? (ii) How many times must he fire so that the probabilityof hitting the target at least once is greater tnan

lt

1.60 and Random P

Solution

Probability of hitting the target p = 1n4

: .q=l- P=j

Let Xdenote the number of times of hitting the target. Here n =7

By Binomial distribution, rlx = rl= 7c,(i)' (;)'" ,r = 0,r,2,...,7

(D P[hitting the target atleast twice]

= plx >21

=r-P[x <2]

=t-[p(o)+r(r)]

- [ rr \ ' - / r \ / : \u ' l=l- l l : l +71: l l : l IL\4/ \4 i \4 i _ l

=l- to i3 '=0.5551(app\

(ir) Plhitting the target at least once]

=t- p lx=0l=l-qn

: . r -q ' r?-t-(1\ ' ,?3 \4/ 3

=r -?= [1)"3 \4, /

/E \ ' I+ l_ I <_\4, / 3

)n-1,

Example 7

Comment on the following: "The mean of abinomial distribution is 3 and

Solution

In binomial disffibution, mean ) variance. But variance < mean.Since variance = 4 and mean = 3, the given statement is wrong.


Example 8

If the probability -. Ior success * too

, how many bials are necessary in order that probabilty

ofat least one success is greater than 1.2

Solution

p, hobablity of success = -f-100

g, hobabilityof failure = ,19 =0.99100

kt the required number of trials : zBy B.D. the probability of atleast one success

=t_eo =t_(0.99) 'Bythe given condition,

t-(0.s9)" >12

=+(o.r)".],2

:+ zlog(0.99) < log0.5

=n(T.sssa).T.oseoi.e., z[-t +o.eeso] < _t + o.eroi.e., -o.oo44z < _o.3olo

+ 't

\ o.3olo _.o ttm=()6.4

. . .n=69

Example g

Ifxand rare independentbinomial variates ,[ti) ^anft,l), fina p[x+r=3].

sofution \ z'/ \

X +Y is also abinomial variate withparameters \ + n2 =12 and p =

... e[x +y = 3] = pc, fll' f1)'J r \2) \2)

=512r0

I

2

1.62 Probabi l i ty and Random p

Example t OTen coins are tossed 1024 times and the following frequencies are obsened.frequglcres with the expected frequencies:

Solution

Here z = l0

p = the probability of getting a head in one toss.I. ' . p =; and, q =l_ O =:

The expected frequencies are the respective terms ofthe binomial distribufim

The frequen cy of rheads is ,oro.ro".( !)'*'f t )"

Hence we have *" ""-o*;;-''"''lu

) l') ' r =0'r'2'"''10

38 106 188 2s7 226 r28 5s

o | 2 3 4 s 6-- j -

106 188 2s7 226 t28 59

Random Variables 1_.6 g

Proof

Let the m.g.f of { be Mu(t). Then

Mx, (t) = "*n[4 ("'

-r)],i =r,2,...,n

M*(t) = Mx,*xr*...*xn (t)

= M*, (t)M", (t)......Mx, (t)

= "a("'

-t) "u1"'

-t)....."t ("' -t

_ "(A+tz-.+U\ct-r)

="'("'-t ="*r[,t(",-r)] where n= 4+k+.....+4This is the m.g.f of Poisson distibution with parameter ).= xr+ 1, +.....+ \. Hence the

result.

Solved Examples

Example I

If X and Y are independent Poisson variates such that p[X=4=p[X=21 andP[v =z] = r[v =3] find v[x-zv]

Solution

Let X be Poisson distibuted with paxameter 2 and Y be Poisson distributed with parameterp.

Then n[x=l ]=plx =2f =+="-r !r ' l ! 2 l

:.2":2

E(X)=2andV(X)=2

P[Y = 2] = ply = 3l=r {:! = "-1-l'r r213!

: .p=3. ' .V(Y)=3

vlx-2",=111;,151

L.7 0 Probabi t i and Random

Example 2

If X is a poissor

E(x') . Arso find ;t"l! -ch that efx = 21=sp[x= 4]+ eor[v = o]'

Solution

p[x=r l = ' -^! '' r !

.'. P[x = 2J= ep[x = a]+ eop[x = o]; ! l-9e-AAa

*eo"-^tru= -2, 41 6r

= t=] . t ' +L

=,x"4 +3)"2 -4=0

: . ( t " , ++)(* _)=o

: .22 = -4 or ) ,2 =l

Hence )"=l (Since 12 carrrctbenegative)

Mean, ), =I .

Ef* ' ]= ) .2 +).=2

P[x>"=,-;t;;],rr,

=l-e-r [ t+4=1- 2

Exampre B e

A manufacturer of cotton pins knows that S%oof his product is defective"

:T::::111,T_og*i""..:h1T, more than 4 pins wlr be defective, rrrmate probability that a box wilr fail to meet the guaranteed quality?Solution

Here o=j' ,o; 'n = loo. . . A=np=5

r[x = ' ]=+,,

=0,1,2,.. . . where X: No. of defectrr.c

r[d

Random Variables L.7 |

Probability that a box will fail to meet the guaranteed qualrty

=p[x>4]

= 1-P[x < 4]

= t - [p(o) + P(1) + P(2) + r(:) + r(+)]l- .s tzs 62s1=1-e-511+5+3+| 2 6 24)

= 0.5620

Example'4

A car-hire firm has two cars, which it hires out day by day. The number of demands for a

car on each day is disfibuted as a Poisson distribution with mean 1.5. Calculate the proportion

of days on which neither car is used and the proportion of days on which some demand is

refused.

Solution

x\No. of demands for a car.

Then X follows Poisson dishibution with parameter mean X.=1.5

Proportion of days on which neither car is used = p[X = 0]= e-r'5 = 0.2231

Proportion of days on which some demand is refused

=plx>2]

=1-P[x<2]

= 1- [P(o) + P(r) + P(2)]

- 1 - e-r'5 [t + t.s + t. tzs]

= 0.1913

Example 5

Six coins are tossed 6400 times. Using the Poisson distribution, what is the approximate

probability of getting six heads x times?

SolutionProbability of getting one head with one coin : 1

2 ( t \u rThe probabilty of getting six heads with six coins = | : | - ^

\2) & rAverage number of 6 heads with six coins in 6400 tbrows = nP = 6400 x - = l9g

Mean of the Poisson distribution = ). =100

Probabi l i ty and Random P

By Poisson distribution, the approxirnate probability of getting six heads

r[x = *]= += gs11,x =0,r,2,....

Example 6

If X is a Poisson variate with mean - l, , show that n [x'?] = ,ae [x + r]

t rrat s( lx-r l) =2.\ l l , f

e

Solut ion ^_t 2,

ByPoissondisffibution PIX = r] =;,t =0,1,2,"'

Then E(X) = )'andB(X2) = ).2 + 1.

EIX+1]=E(x)+1

= 1+l

" ' LE(X+l)= 717+l)= 72 a 7

It follows that E(X') = 2E(X +1)

E(lx-tl)=i.. fr{t '-tt). l " - t l

="-t) l ' t - - t s ince 2=Ivl

x=0

_,[ . o 1Z*. . . . ]=e ' l I+ t t* a* zt Ii -

=e r l l * t - - l

- L fl--(n+1)!l

=! [ ,* i (n+1)-1- lel i* (n+1)! I

=![ ,* i t - i l - lel- 3. ' ! f t (n+1)! l

= 1 J l*e-(e-1; l=Z

e' - e

Example 7

IfX and Y are independent Poisson variates, show that the conditiond

given (X+Y) is binomial.

Random Variables L.7 3

Solution

Let X be Poisson distributed with parameter 2 and Y be Poisson distributed with parameterp . Then X+Y follows poisson distribution with parameter ).+ p .

r [x= r /x+y=n]= t [ l==:^f =n_-4P[x+Y=n]

=. l I= 'Jp[y=n-r lFfX + y =;]- since X and Y are independent

"-LAt "-pOn-t_ r! (n-r)!

"_<t*r,) 7,L + lty-i+ n!

= ot (2) ' f o.)"- .r ! (n-r) ! [ ,?+p) ln+p)

. ' . f [X=r/X+Y=n] =nC,p'qo-, where, = +and q - I -p =- ! -A+p X+p

'"';" the conditional distribution is a Binomial distribution with parameters n and p where

-v =-

- 1+p'

Example 8

A random variable has the m.s.f. "*o[4,

. r'r

p[rt-2o<x<p.r";;;r;:;"":t!1t;. ')J ' Find its mean' standard deviation and

Solution

Since the m.g.f. of a poisson random variable X is exn[;(", _ r)], we get ). = 4..'. Mean = Variance = 4

* F=4ando=2

P[p - Zo <X < p + 2ol= efg. X. S]

=pl l<x<7J

="-of+ ***+*!* !* !* ! ]L zt 3! 4! sr 'at ' t r l

= 0.931

ilI

1.7 4 Probabi l i ty and Random p

Also,

Example g

In a book of 520 pages' 3g0 typographical errors occur. Assuming poisrnumber of errors per page, frnd the probability that a random sample of 5 pae:etTor.

Solution

Average number of errors per page = 390 = 0.75s20

:. )" = 0.75

Let X: Number of errors per page

. '. P[x = rf= 92; = 0,1,2,...

P[No enor in a page] = p[X = 0]

=g-1 ="4'75

=0.472... probability that there is no error in 5 pages = (0.+lZ)t = 0.0235

Example I O

Suppose that the number of telephone calls an operator receives from 9:00a day follows poisson distribution with mean -- 3. Find the probability tha:receives no calls in that time interval next day. (ii) In the next three days the opea total of one call in that interval.

Solution

Let X: The number of phone calls received by the operator during th*.Then X is poisson dishibuted with mean Z = 3

. ' . P[x = r]= +,r = 0,t,2,...

Plr - 2o <x < ttl= p[0 < X. 4J= p[X = 1]+ plx = z]+ e[x = :J

="*[+*s * '2fL 3J

= 0.4131 .

Random Variables 1.7 5

(i) P[operatorreceives no calls] =P[X=0]="-' =0.0497

(ii) Probability of receiving a call = 1- e-3

The operatorreceives a maximum of one call in the time interval is described as below.

Day First Second Third

No. of

Calls

I

0

0

0

1

0

0

0

I

.'. The required probability = r("-' )' (t - t' )

Example | |

Fit a Poisson distribution to the following data which gives the number of yeast cells persqrxre for 400 squares.

No. of cellsper souare (x)

0 I 2 3 456 7 8 910

No. of Squares (f) 103 143 98 42 842 0 0 00

Find the expected frequencies.

Solutionte, {ro

Mean = 2L _ "o. =1.32Ir 400

The parameter of Poisson distribution )" =1.32 .

poisson distribution is N " '2' = 4go e-' " (1 '32)'

r ! r !

_ 106.96(r.32) 'r !

The expected frequencies are the values of toe .se , (1'3?)'

when r = 0,1,2,.....,10r!

The expected frequencies are 107, l4l, 93, 4I, 14, 4, l,0,0, 0, 0.

Random Variables

Proof

rhe p.d.f, of X is f(x) = {^\^

"o|#r"..' plx t t] = ['r"-'rxdlx = [-"'rxl- -

"-rtL r Jt L - Jt

.'. P[x > s + t/X t r] = P[x t:?:+:t ^-x?

t]- r

e[x>s]

1.105

- "-1:;" =e-tt =r[x>tJe-rs

I 'l

Another Form

If x has an exponential distribution, then for every constant a)0,P[Y < x/X >u]= plx < x] or al l x, where y = X_a.

Note

The converse of this result is also true. i .e. , i fP[X t s + t/X > rJ= p[X > sl, then X fottows an exponentialdistribution.

Solved Examples

Exannple IIf X has an exponential distribution with mean 2, find p[X . I/X <21.

Solution

If X has parameter )", thenits p.d.f. is f(x) ={1"-u x > 0

Mean =1 =2 = ).=l I o other"wise

12

Ir -+f(x)=11 " x>0

otherwise

X>s+t

and Random

P[x. ux<zl=plx<rnx<z]

e[x < z]a1 - Iplx<r l I ; " '&

_-"L

f [x<z] ez1I

t" 'd*/ * \ r| - ; lt -e ' l

_\ / ._ l -" ' '( -+ ' ) ' l -e- '

[-" 'J,. ' . P[x < l/x < 2l=9!2!= 0.5686

' 0.6321Example 2

A continuous random variable X has the p.d.f. f(x) given byl -x

f (x)=]Ae5 x>0

I o otherwise'Find the value of A and compute (i) p[X <6/X> 3l (ii) p[X > 6]

Solution@x

f ,r1";c*=t+Af,e 3d"=1

=+ -SA[O - t] = t

[ t - :...f(x) =

tt;

. ' .5A=1=+A=15

x>0

x<0

Random Variables 1.10 736

. e s _e 5=_

J-:e '

e-{.5 _ e_r.2

et '

_ 0.5488_0.3012o.sAt8 -

p[x <6/x>3f=o.qstz

P[x '3J-r-plx<3J

=r_ [r1x;ax&

= e{6 = 0.54ggExample B

. .t]o:*" that thc life ofan industrial lamp, in thousands ofhours, is exponentially dishibuted

wrtn rarlure rate )" =i (one failure every 3000 hours on the average). Find (D the probabilitythat the lamp will last lbnger than its mean life of 3000 hours (ii) the probability that the lamp willlast between 2000 and 3000 hours and (iii) the probability that the lamp will last for another1000 hours given that it is operating after 2500 hours.Solution

Let X: the life of the industial lamp in thousand hours.Then X is exponantially distributed with parameter ). =!.

f r+ 3. ' . Thep.d. f . ofXis f (x)={ l 'e ' x>0

I o x<o(, P[The lamp will last more than 3000 hours]

= PIX t 3]= t -r [x <:J

=0.3679

=r- fr(r)o*= 1- f1"-i6*{3

| 3 l

=t- l l - " ' l="- 't_J

.10 Probabi and Random

(il) P[The lamp will last berween 2000 and 3000 hours]

=p[2<x<3J

= F(3) -F(2)( =\r -z)=l l -e, l - l r_"T I ,u( / ( . ,

-2

=e3 -e- l

= 0.5134 -0.3679Required probability = 0. 1455

(iii) Probability that the ramp w'l last for another 1000 hours given2500 hours is p[X >3.s/x>2.sf=p[XrrJ oy *"_o!,rrr.

... p[X > 3.5/X > 2.sl=l _ plx < tJ

Exampfe 4

Solution

Let X: the time that the camerawill run with out having to be reset.Then X is a r'v' with exponentialry distributed wit,. e= 50 days. The:

The amount of time that a camera wilr run with out having to be rese

li:::^.:r"""ltial,distributiol wrth 0 = S|days. Find the probabilitl. *t,t(i) have to reset in ress than 20 days. (ii) not have to be reset in at reas:

[ - r l r=t-Lt-e ,J=e-I = O3r6s

f r +f(x)=

lA" ' x>o

L 0 x<0(i) P[camera will have to be reset in less than 20 days]

= pfcamera will run for less than 20= plx < 2oJ

ezo 7 -x= I ,i" "a*l=1=11-e'

l=0.5297LJ

Random Variables2-6

(ri) P[camera will not have to be reset in at least 60 d.y=]= p[camera will run for atleast 60 days]= P[X t 60]

=f1"#a*60 50

= "f

= 0.3012Example 5

The daily consumption of milk in a city in excess of 10000 gallons is approximately expo-nentially distributed with parameter g=1000. The cify has a daily stock of 20000 gallons.What is the probability that the stock is insuflicient for both days if two days are selected atrandom?

Solution

If the random variable X denotes the daily consumption o-fmilk (rn gallons) in a city, thenY = X - 10,000 has exponential distribution with parameter B = 1000.

The p.d.f. of y is g(y) = #"*,0

< y < a

since the daily stock of the city is 20,000 gallons, the probability that ttre stock is insufficenton a particulr My = p[X > 20,000]

= PIY > 1o,ooo]l.= lo.*s(r)dv'&t4

= | ' ermdv.,ro,mo 1000

[ -r -l-

l -" '* |L jo.ooo

= -[o - "-to]

= "-to

The probability^that the milk is insufficient for both the days if two days areslecErl atwrdqn = (e-to )'

= "-zo

Example 6

Suppose that during rainy season on a fropical island the length of the shower has an expo-nential distribution with paf,ameter 2' = 2 ,time being measured in minutes. What is the probablitythat a shower will last more than three minutes? If a shower has already lasted for 2

-irr.rr"r,what is the probability that it will last for at least three more minutes?

1.1 Probabi l i ty and Random

Solution

Let X: length of the shower.

Then X follows exponential distribution with parameter ,r = 2 .t l+

... P.d.f. of X is f1x; = J;e ' x > 0

L 0 otherwiseP[a shower will last more than 2 minutes]

= f,r1*;a*F2=l- I f(x)dx

= r-1 f .Ta*2h

/ =r)'= l - l -e ' I

t t\ ,/o

= e-f = 0.3679

P[the shower will last for at least 3 more minutes given that it=P[x >stx>27

=plx>3]l -X

a6 |

- | - :e2dx, t l 2

r f@I-xl=l*t It tLJ3

-3

=ez =0.223I

Example 7

If X is a continuous random variable with p.d.f. f(x) = €-*,0 ( x < €,that the roots of the equation t2 - x(4t - 5) + 2l= 0 may be real.

Solution

The equation ax2 +bx*c =0 has real roots if b2 _4ac >0..'. For real root of the equation t2 - 4xt + 5x + 2r =0 , we must have

t6x'z-4(5x+21)>0

i .e., 4x, -sx-zl> o

by memoryless


solving 4x2 -5x-21=0, we get * - 5t Jlot

85+19

-x= 8 -I

. . .x=3 Or-4

.'. The roots of t' - 4xt+5x + 2r=0 maybe real for x >3 (x cannot be negative)

Requiredprobability = f [X > :]

r@ _". f "_* l -

=tr e ' - { t ) (=L- J3= e-3

Example 8

A component has an exponential time to failure distribution with mean of 10, 000 hours.(a) The component has already been in operation for its mean life. What is the probability

that it will fail by 15,000 hours?(b)At 15,000 hours the component is still in operation. What is the probability that it will

operate for another 5000 hours.

Solution

[,et X denote the time to failure of the component.

Then X has exponential distibution with fir€an = 10,000hours.

. ' .1=10.00O+ )"). 10,000

The p.d.f. of X is f(x) = - L"ffi,x ) 0 and : 0 otherwise.10,000

(a) Probability that the component u/ill fail by 15,000 hours given it has already been inoperation for its mean tife = P[X < 15,000/X > 10,000]

_ plto,ooo<x<ts,ooo]P[x>ro,ooo]

rd5,000

1.. f(x)dx=-

f,.*r1"P"e-l _ e-1.5 0.3679-0.2231

e-r 03679

= 0.3936

F

1.L20 Probabi l i ty and Random

Distribution Function

F(x) = f,r1*P*=

f apx\ae-o*r dx

Put u = axq -dv-

apxq-tdx

F(X) = f,./ e-"du

= [-"- ' ]0" '

= l - s-ax|

=+ F(X) = P(X '

*) = l-e-o"e

Note

(l\ From distribution function, we have

P[x<ul=t- e-o^P and PIXt af="-{

(2) A r.v. X follows Weibull distribution witka,F>0 i f the r .v. Y =aXP fol lowsdistribution with density function f"(y)=e-'

(3) When f =1, Weibull distribution reduces r:distribution with parameter d .

Solved Examples

Example l

Suppose that the lifetime of a certain kind of an emergency back up

r.v. X having the Weibull disfibution with parameter a = 0.1 and' B = 9.5

time of these batteries. (b) The probability that such abattery will last

Solution

For the p.d.f. of X in the form f(x) - dpxf-1"-"^p,X ) 0

Mean ='7.l,!*t]\.8 )

(a) Meanlifetime = r =(o.r;;i.(t.*)

Random Variables1.121

=10.r)- ' f (3)= roox2

=299 hours(b) probab'ity that abatteryw'l rast rnore than 300 hours

= p[X > 3oo]

= f;or1*; ax=

f*apxo-te-o*f dx

Put y-(0.')xos

= f;oto'rxo'5)x{'5"-{o't)*os*

Then dy = (0. l)(O.5)x{sdx

.'.PIX>300J= f"-ay where y, =(0.1X300)0.5

= g-Yt - "-{o.txroo)os

= 0.177

Example 2

Suppose that the time to failure of a certra:as a r.v. having weiburr distribution *,h oJ::,T:T:3:T"f':Tnutes

is considered

;::ffi"liTfcted to rast? (b) what is *," p,ouuuifrylii,".,n;ffi;:::ff il1;

Solution

(a) The expected time to last for the component

= E(X) = "r r(; . r)

= 1o.z)-' f(4) = t25 x 6= 750 mimrtes(b) Probabilify that such a component will fail in less than 10 ho

we know plx = a!= 1- "-,a

r rqu r' ress rnan 10 hours = PIX < 600J

. '. r[x < 6001= 1-.*,'*,i

=l-s.r '6e =0.8149

Probabil i tY and Random P

ExamPle 3

Suppose that the service life (in hours) of a semi-conductor is a r'v' nar

distribution with a = O.Ort and p --0'5 ' What is the probabilrty that such a

will be in working condition after 4000 hours?

Solution

Let X: the service life of the conductor'

Then X has Weibull distribution with parameter aandp and whose denst:1

f(x) = aPxf-t s-o*P'x ) 0 '

Here a --0.025 and P =0'5

Pfconductor will be in workng after 4'000 hours]

Now

. , .e-rro =e4.6 *o=#

E(x)= -rrli.r) *a v(x)= "flr(X.')-{'

= f r1x;ox=P[x>4ooo]

We know that P[X 2a]=s-"'P

.'. Required probability - "-{o

ozsx+ooo).s

= s-r sr =0'2057

ExamPle 4

If the life X (in vears) of a certain tvpe of car n* " Y"l:lt

U:TO*""

0 =2 ,find the value of the parameter a ' giventhe probability that the life

5 vears is e-0" . For these uulot' of a and p '

findthe mean and variance

SolutionThe densitY function of X is grven bY

f(x)= Zdxe-o" ,x>0 since P=2

P[xt sf= f,zaxe-o*'dx-"(t ' I

=g

= g-25a

Giventhat PIXt5]="'"

FortheWeibulldistributionwithparametersaandp,

1. L23

For a=# *0, =2,thevarues of mean and variance are obtained as follows

/ r \ ; / r \Mean =[r.);.l.r)

\ 100/ \2 )t - f t ) / r \=t0, . - r l

; l=sJr using r l : l=G' : " ' \z)

variance = [#)' |",

- {,[;)]'l

| / r : ,2 1=roolr- lg I I

I (2Jl

= toolr-alL 4l

Example 5

Each of the 6 tubes of a radio set has the life length (in years) which may be considered asa r.v. that follows a Weibull distribution with parame ters a = 25 arri f = 2 . Ifthese tubesfunction independently of one another, what is the probability that no tube will have to bereplaced during the first 2 months of service?

Solution

Let X: life length of the tube

Then the p.d.f. of X is given by f(x) - apy!-t"-o,r,x ) 0i.e., f(x) = 5Oxe-rt"',x > 0

P[a tube is not be replaced during the first 2 months]- iI l l

=Pl Xt: l since 2 months=1u*L ol 6r-*

f,soxe-rr",dx

=(-"-"*'): ="#\ / f

... P[all 6 tubes are not Ue ,eptacet during fust 2 months]( -zs\6 -2s

I t ' I=tT=o'olss\ /

Random Variables

ull

lor

srils

r.L.24 Probabil i tv and Randorn P

Example 6

The life time x inhours of a component is modeledby aWeibull di

Starting with a large number of components, it is observed that 15 o/o ot

that have lasted 90 hours fail before 100 hours. Find the pararrrcter a

Solution

P. d. f of x , f (*)=aP xQ-t e-o" , x>0

=2a x e-o" ,x>o

Given that P(.r < 100 I * r%)= 0'15

. P (90 <.r < 100) = 0.15

P(x > e0)

t00

lza * e-o" dx+ eo' =0.15€

lza x e-o'- dx90

put u=d*2 du=2axdx

r . r l (X)

l-"-"" l^^ o-Nro _ o-tu.zo' \ rm =0.153' :o =Q.lJ

/ r \@

"-* 'ol-"-"- '

\ /SO : ' . : : , , , r : : i ; : . r ' : . : - : :

This gives a =0.00008554.

Probabil i tY and nendarn

The conditional mean of lgiven X = x is definedby

ElYtx=xl=LYf(r tx)dt

The conditional variance of lis given X = x is defined by

vLY I x = xl= EIY' t x = x7-\rlY t x ='l\'

Stochastical Independence of Random Variables

TworandomvariablesXandlaresaidtobestochasticallyindependem

rlX = x,aY = Y;)= flX = *'l 'elf = Y'f

i.e., Pii = (A XP , ) for all i andT

Incaseofcontinuousrandomvariable,therandomvariablesXandY

f (*,r1= g!\h(t)

where f (*,Y)= johlP'd'f of Xand I

s(x) = MaryinA densitY of X and

n(l)=Marginal densitY of I'

Solved ExamPles-

ample I

Given the following bivariate probability distibution obtain

(i) Marginal distributions ofXand I

(iD Conditional disfiibution ofXgiven Y=2 and

(iii) The conditional disnibution of IgivenX= 1'

-D

J r tx-> 10-1

0

1

2

t2115 t5 15

32115 15 15

2t215 15 15

Two Dimensional Random Variables

Solut ion

(i) Marginal Distribution of XandY:

r fx = x]=ln(x,t)andp[r = yf =ZpQ,y)

(ii) Conditional Distribution of Xgiven Y = 2:

rlx =xlY =2]='l*:=*rnt-='l where .r=-1,0,1' PLY =21

efx =-r /Y =r]=?' , ! .? =?' 5/15 5

elx =o/Y:r1=#=i *o rlx =r/y =21=?tr!=?; /15 5

I1

*

J T\, [ -+ -1 0 I 2pQ,v)

0

I

2

12115 15 1532115 15 152t215 15 ls

4

15

6

15

515

Zp@,v)v

6s415 15 15

I

v

r. I 0 a_l

r (x) :654

15 15 ls

v. 0 I 2

n(t) :465

15 15 ls

x 0 I

P(X /Y =2): 21?1' ! \

Y: 1 20

P(Y t x =r) :112

444

Probabi l i tY and Random Pr

(iii) Conditional Distribution of fgivenX= |

PIY = y r x =tS=W for v =s,1, 2

Example 2

l-nt X andlbe two random variables having the j oint probability function I i

where x andy can assume only the integer values 0,1 and 2. Find the marpsnEl

distributions.

Solution

IX\ I -> 0 I 2 I n(*,t)

0

1

2

02k4k

k3ksk

2k 4k 6k

6k

9k

tzk

ln(*'t\ 3k 9k lsk 27k

ZZpQ,v)=ryx

> 27k =l

: . t =L27

(i) Marginal Distribution of XarrdY

(ii) Conditional Disftibution of X gSven Y = y

pIy-xny=t lplx=xty=tl='ff i

x 210

p(,) ,691227 27 27


(ii) Conditional Distribution of lfor given X = x

Example 3Two discrete random variables Xand lhave the joint probability density frrnction:

, \ A,'e-x . pv (l- o\*-YPlx,Y)

11; , -, v=0,1,2,...x; x=0,1,2,....

where ),,p are constants with 2 > 0and0 < p <I .

Find (i) the marginal proability density functions ofXand L

(ii) conditional distribution of rfor a givenxand ofxfor a given r.

Solution

(i) Marginal probability functions of X and y.

2.8

*

lj

*1

J

I

1

P r $) =t P @'') - f ')" e "^

P' lr - P)"'.v=o , l y l ( t -y)t

-e-^+' ; " -xt _ p,(r_p) ' - ,x'! ,=4, yl(x- y\(

I x\ r+ 0 I 2

0

I

2

24o--9ls

1153915

z!63915

{ x\r-+ 0 I 2

0

I

2

I

9

2

12

2466359946L2 T2

Two Dimensional Random Variables 2.10

,Example 4

J If X and f are To random variables having joint p.d.f11.

f (* ,Y)=] ; tu- x- Y) ' o<x< 2'2<Y <4

f0, Otherwise

Find (i) r[x<tnr<:] (ii)plx + r < 3]andSolution

(ii i)PIx <r/Y <3f

(i) r[x <r ny < t]= f, t f e, il dyd,c

=*f fru -x-v)avdx

=*l[t' -.\r-+];,*=! l (u- ' - t )*g

" \ 2)

t l t * ' l ' t l t r l=-l-x -- | =- t ___ |

8L2 2 )" 8L2 2)

; .r [x <tnr.: ]=i

(ii)r[x +r < :] = f f ']tu-" - y)aya,

=* I[(' -a,-lf)',' *= * f[,.-')t, - 4-Q :)' .rf*

=*{ ' -7x+x2ry]-

=!ls*-t*' *4* (:-')' l '8L 2 3 61.

=1[3_Z*1*1_2. l8L 2332J

=f;ltn1=a

and Random p

(iii) Plx < | / Y4l = P[x=;lr-' I,'

3]ply 4l

p(y4)= f fifr_ x_y)ayax

=* rlru --\-(2-r\1*801. ' \z

- ) )*"

= * f [1- ')0, =]tz - zt= |. . .p[x <t /y <r]=#=;

Example 5

If the joint p.d.f of (X,y) is given by f (r, y) = 2,0 < x< y < 1, findO M arginaldensiqz f,urctbns of X and. y .

(ii) Conditional densities f (x I y)andf (l / x'.(iii) Conditional Variance of Xgpveny=2.

Solution

(i) Marginal densiry tunctions fr(r) and f"(y,):

f*(*)= Lf e,ia,= !)ro, =z(r- x),os.r s 1

f '(Y)= ff (*,Y\a,

= (r*=2y,0<y<r

(ii) ConditionalDensityFunctions:

f (x/ t )=!k l=Z=L,0.* . ,

-r (t r x) = Hl

= ii= fr ,o 3 x < v sl


(iii) Conditional mean of Xgiven y = y

zfxr y]= fx|(x/y)dx

Solution

(t P[x.4=Xfrn"*idyd*

=f"- ' l f , Iro ta "dl la'

= I'-.l-"-li*=

[e-,dx=t-L

P[Xrr]= f f e-@t)ayax

= I'r-'l-"rli*= f"-.lr_rb

[ -- "-t*1-=l -e '+- lL 2JO,l t

=l--=-

22

(ii)

" ( t ' \ rf x' f '= | r l - l t rx=- l - l =!{ \y) yLz l " 2

nlx'/y)= (x,f (x/y)*=1f+l' =4/L 3 J. 3

Conditional vari ance V fx / yl= Elr' t yl- {nfx t yll'

_y' _yt _y '3412

Conditional variance ofXgiven y =z=tl = 1

Exampre 6 l2l'=' 3

/ I t f (* ,y)-eQ*' l ,o<x,y<oand=0, elsewhereisthejointp.d.fofrandomvariablesX

, and I, find (i)r[x<r] (ir)rlx >rl Qtlrlx+r<\

2.12

2rL Probabi l and Random P

(iir) rlx +r < 1l = ll-' e+rafix

= J "-' l-n'I^' *= X"--[, - "-t

."')a,

= fe*ax-at ldx

=(-t") i-a'( ') i =r-?

Example 7

TWo dimensional R.V (X,Y) have the joint p.d.f

f (r,Y)=8r/,0 < x < Y <l and = 0,elsewhereI r r l

(i) Find Pl X .; ^

y.; | (ii) Find the marginal and conditional

I z +J(iii) Are Xand lzindependent?,

Solutionl - r t l et4ey - ,

(D Plx . ;nY.;)= f , | f (* , t )dxdyL Z +J

= l'^ (uyarattat t

= f'osrl il a,J0 '1.) |

L-Jo

=4I'o y'dy =(r^) ' l '

=1256

(ir) MarginalDensities:

g(") = lf (*,flat,= f3*vav=qxl l - r ' ] ,0<x<1

n(v)= ff (r,ila*= ('uva*

=su[4.l'' l -2 l ,

'=4Y3ro<.Y<1.


Conditional Densities :

f (x/y l=f ! : ' ! ) =\ . \ =4,0.x<y<r. , n(y) 4y' y2 ' - "- ' /

f ( t rx)=' ! : t ( )=-P 2Y ̂ " ' .' * )= s(4

=;44=fr i ,0<'x<Y<l '

(ur) g(r)' ' ' 4x 'h(y)= ,ffi* f (*,y)

:. X andY are dependent.

Example 8The joint p.d.f of the two dimensional RV (X, f) is given by

^, l t t .1<.r <yszf(r ,y)=l s 'L

L0 ,elsewhere

IT

(i) Find the marginal densities of XandY.(iD Find the conditional density tunctions f (x / y)nd f (y / ,)

Solution(i) Marginal density functions ofXand I:

s (.r) = f,l ro, = l,lt)',,=*G-r,) , r<x<2

9\

h(y)= fl,*=Yltli,=*b, - t ) , t<y<z

9\-

(ii) Conditional Density functions : ,

r (* r y) = #

= 4 U= h,,

< x s y < 2

=#,rax<y<28-w9-r(yr*)=H=

!{+-.')

2.tL Probabi l i ty and Random

Example I :

, The joint p.d.f of two random variables X and I is/ , . , \

r / , , 1f t"( r -y) , o<x<2, ly l<*J\x,y)=1^ '

, ,|.0, elsewhere

Evaluate the constant fr. Find the marginal and conditional density of i'

Solution

ly l<*- -x <y <x.

(D To find fr

t f"u(* -w)dydx=t?2^'?2

- t I x' (2x)dx - k I xlO)dx =r

[ -o l ': .ZklLl =r

L 41.

. ' .8k=1+fr=18

(ii) Marginal densities of X arrd Y :P ^tg(") = lJ \x,YYY

1: 8 J*(t' -'Y)aY

=" r dv-! f vdvgr ' - gt"-

=t- lul , -{(o)=48L'r-x 8" 4

.'. The marginal density of Xis1

g(t)= L '0t"2

r'(r)= Lf @,y)a*

:. h(y\ =l f G' - ry\d*(-z<.y < o)\ r / gty\ - ,

1andh(y) =. l l(* ' - *y)dr(o . * .2)

8Jrt ' /

Two Dimensional Random Variables 2.L6

: n(y)=*[(+)", _,(*)1, ,]

=lfg*Y'- zr* t1813 3 2J

=:-t..f f-r" -2<y<o

andh(v)=*[[+), ,(+)")

=llg -v' -zu*Lf813 3 2)

=:-I*f i*o < y <2

.'. The marginal densrty of Iis

l ! -+.* in-2<y<ar(y)=l ? o o:

L;-i*fri"0 < v <2

Conditional density function

f etx)='!;:P' glx)

I ( r - u);. 7 (t, / x) = U;# =

#,-* I ! 1 x,

Examnple 10

The jointp.d.f of R.V.sXand lis givenby

f (''Y)=lj"-'''"-';x > o'Y > o

L0 ;otherwise

nina ̂ e[X >t/Y = t]

A<x<2


:.fr(*)=]:"-r,,1+ f "+ *l,l2z l.l2zr

h jByArea property of standard normal distribution,

* f e-" ' 'dz- l42r k

"' f- (') = l"-" '',-@ (.tr < cc.

'J2zMarginal density of I:

Writting x' - 2pxy r y' = (* - pyf * (I - pr) y,

-u2 tt -rl$-orfl'

r'(i l=ffiL"'tGla*/ \2

r r l l ' -PY l

=_! "-r,, * f !"-ulE) a,,l2n JZr r*

Jt- p'

Using z =#, we get! r - p '

f, (y) = _f, "_,,,t* 1e",,, a"f=ir"_r,, * h fe-,,,d2=l

.'. Marginal density of Z is

f, (Y) =,ftr"-"'',-@ ( / ( o

when p = o, f (,, y) = **l-VPlt [ (x '+v '11

Alsof, (-) f ,(y)=i"*pl _r- =, ) l= f G,y\zlTlz

.'.If p = 0, X and y areindependent.

.,, Example 17/' The joint distribution function of two random variables X and, r is given by

F(*,y)=l-e-' - e-v + e-6*t), x>0, y> 0 and= 0, otherwise. Findthemarginal andconditionaldensities ofxand r. Are xand rindependent? Also compute p[X< t,y < t] .

Probabi l and Random

Solution

The joint P'd.f f (x,Y) =W

'. f (*,y)=*[t -e-' -"-t *"-( '*i l foxoy -

=*1"-' -"-(*r)]=

"-('*t)

rhe joint p.di f (x,yr=[;u.' :ffi:*:"t

Marginal densities of X and Y:

g(")= f t@,iar= f,e1*idY= "-'l-"-'f]

= "-' ,* > 0

h(Y)= lf (*,fla*=/ f,e-'dx="-rl_e-,fi=e r ry>0

Conditional densitY functions :

f ( t lY)=H=e-"x>0nQ)

f(yrx)=' ! :*) =e- ' ,v>a. s\x)

Since g(x)ft (t)=s-@'l = 7(x,y),Xancl'Iare independent'

P[x <1;r<1] = ll"-<*'tafi*=(f"'*X |o'or)=0-""X,-"')=(t-J)'=0.3996


Note

2.42

(I) For finite numerical values of X and Ithe co-efficient of correlation is

l \ -:Lx"Y-xY

the formula for

fo=

f=

(or)

(2) The co*elation co'efficient between' two variabres isunaffected by a change of origin and scale.

i.e.,If (I =ffora, =+,thenrn=r- and is given by

ut(3\-

Example | ,/

Ifxand rare independent random variables, show that they are uncorrelated. Is the con-verse true? Justify your answer.

Solution

IfXand Iare independent variables, then

cov (x,r) = E (xy) - E (x) E (y)= a(x)n(y)_ E(X)E(Y)-0

... correlation coeffrcient r =cov(x,Y) -

+ x andyare uncorrelutec. gffi converse is not true.*The variables may be uncorrelated but they may be dependent.,,

Solved Examples

l;>,o _* ltlf _v,

nZv' -(Ir)'

Example

,Kis a variablewith densiry f (x) =1 ,1,1. ILet Y = X2. Xandlare dependent u"riait"..

E(x)= I,i*=i*l' =o- L+J_r

.'. E(x)E(y) =lE(y) =sFurther E (Xy) = t(x,x,) = E (X') = o

.'. cov(x, r) = E (xy) - E (x) E(y) = o... X andy are uncorrelated but dependent.

Elampte 2,/

\'/ Find Karl pearson's correration co-efiicient betweenxand rfromx78 89 97 69 59 79 61 6

.43 Probabi l and Rand

le 61 6tY, 125 n7 156 112 107 n6 n3 lOiSolution:

IEtU=X-75andy=y_125.

Tlten rr=r*

We have 2u = -7;2, = 4;Zu, = rrr@Correlation co-effi cient

12236i ' f^ , =., -=o'917 ' Hence6 =0.917

1i


Example 3 ,/

A computer while calculatin g r, from2lpairs of observations obtained the following con-stants: n = 25,1 x = 125;Z x' = e's};Z ! = ioor2 y, = ouo *i I r"! = 508A recheck showed that he had copied down two pairs,(6,14), (g,6) while the correct varueswere (8,12), (6,8). obtain the correct value of correlation co-effrcient.

Solution

Correct value of I" = 125 _ 6: g + g + 6 =125Correct ralue of I, = 100 - 14 _ 6 + 12+ g = 100Correct value of Z*' = 650 _ 36 _ 64 + 64+ 36 = 650Correct value of Zy, = 46A _l-96_ 36 + 144 + 64

= 436Correct value of |ry =50g - g4 - 4g + 96 + 4g

= 520.'. Correct value of co_efficient of correlation

Example 4

If o',,oj and, ol_"

Solution

ol_,

o? + o'zr__:1-."y =-----

zoxay

= var(r - l) = vm(x)+ var(y) - 2cov(x,y)= o: + oj - 2ro,o,

, ' .2ro,or=o:+o1-ol- ,

or_+o?_o,:= -=

' / x-v

2o,6,

r 0.667

be the variances of x, y and x _ y respectively, show that

NLry -

n2*'_(2.)'J"Zv'-(Zi'l25x1o0

2s)(6so)-(rzs)'2s)($6)-0oo),

L45Probabi l i ty and Random

,/ExamOIe 5J The followin

weights lpouna.yg table gives the number of students having different

Calculate co_efficie

SOfUtiOn

._-- "- vvrrvr<ruun Derween heights and weights.

Let x and y be the variates f^r rh- ,r.^:^,-.ates for the weights and heights respectively.

rnt u=x-705 and v=Z:gl0 2.5From the table, we have

2"t" =37;2fu' =t%;lfuv =2s;>.fv =tq 1fi,= 404 and yCorreladon co_effi cient

f ry=luu=

. ' . t=

,,1 r oo (t n) _ e7 ), / o(4oAE:0.09

50-55

55-60

60-65

65-70


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o, for '

tana = f r r - \

I +4mz

!o ' - ro '= ' o l

=o,:* t

o,'

+tarrd-r-r2. o,o,

f nterpretationr 6r '+ort

(i) Wtren r -0, a= E

,inesorregre.,.?;:;':Til::T';;::H*:ffi1,"::::ffi T":"(ii) luren /=t1' t*.,,=o+a=O or z -Thetwo regression lines coincide and there isperfect correlation between the variables x and y.

Regression Gurves for the MeansEIY / X = xl is called the regression function of y on X.nlX t f = yJ is called the regressionfunction .f*

"*,Regression curve of lon Xis l, = E[y / X = x]Regression curve ofXon Iis .r = E[X tf = yl

Solved ExarnpGEExample I

obtain the equations of the rines of regression for the fotowing data

Probabit i ty and R

Solution

Zu =o;Zr, =o;Zu, =36;Zv, = ++ -d IReg r e s s io n c o effrcien E :

u..='ZY.-(zu)(1v\

_,"i?_'0",_:::,8x3Eo

=o'667

u- ='ZY.-(Zu)(1r')

8x :?'' -(z')'

=g*44=0.545

Yr rrX=L" =68and V-- -69nn

Equation of line of regression of y on XY -F =t*(x - x)

.'.Y _69=0.re2(x _a\

=I=0.667X+23.64

Equation of line of regression of X on y:

x_x=r9z_(r_t)oy

+x_68 =0.54s(y _69)

X =0.545y +30.395

Two Dimensional. Random Variabtes 2.69

Example 2

In a partially destroyed laboratory record of an analysis of correlation data, the followingresults only are legible.

Variance of x : g.,Regression equations: gx_10y+66= 0 ; 4;x_lgy=2Id.Find (D the mean values of x andy.

(ii) the corelation coefficient between x and y and(iii) the.standard deviation ofy.

Solution

(i) Since both regression lines pass through the poiht (i,r),we have8t - 10t = -66 and 4Or -lB, = 214. Solving we get x = 13 and ! =lT .

(ii) Let 8'r-10IF66 = 0 and,40x-l8y = 2l4bethe lines of regression of ronxan d x on yrespectively.

8 - . .66 , 18 214Then . /= j : t+:- and.r= ̂ " y+ -- 't0 l0 40' 40

.r, =*=1-o u,=#=*Hence , t =4 .9 - 9

520 25

. ' . t=t ]=10.6

Since both *r" ,Jgr"..ion coefficients are positive, we take r = 0.6(iiD We have

614f .4=_

615

4 3o... ._ =; .*s ince o- = 35 5 3 -x -

)or=4

.'. Standard deviation of v = 4./,/

Example 3/

The equations of two regression lines obtained in a qorrelation analysis are as follows:3x + l2y = 19;3.y + 9x =46. Obtain

(i) the value of correlation coefficient (ii) mean value ofx and v.

I2 Probabil i ty and

the lines of

find.the means of

Solution

(i) Regression line ofy on x: 3x +I2! =19

=+ y _ _1r*12' t212

Regression line of.r ony: 3y +9x = 46

=).r- - ! r*{3'9

...bo =-* u"a b* =-!12ry3

Hence rt =bo.bo

=1t2

rJt=r=t;ft i .e., r=+-:-

Since both regression are negative

f=-

coefficients1

-2J3(ii) Both regression lines pass through ttre point (I,y-)

37+l2y =19and 97+3!=46

Solving, we get f =5 and I= I

'3

Example 4 /I

Given that X =4Y +5 andy =lcX +4 are

respectively, show that O <4k< l. If k =L

of correlation between them. 16'

Solution

X =4y +5?b, =4

y =kX +4-bo=ls

: . r2 =4k. But 0<r2 <l+0<4/r<l

I f f r=a. , , =!+r=11t6 ' 4 -2

I;., = i,

smce regression coefficients are positive

L

Two Dimensional Random Variables 2.V 0

For fr = j, ,rr. two lines of regression aret6 'x-4y =5 and x- l6y=-64

Solving, V =5.75 and-7- =28

Example 5

If the lines of regression of y on x and )c on y are respectivery arx+br!*cr=e ^4arx + bry * c, = 0, prove that arb, S arb, .

Solut ion

arx + bry+q = 0 ) bo = - f , | o.1

=_!26 on y)

Example 6

For two random variables x and y with the same niean, the two regression equations are

! = ax +band x = dy +B. Show that !=+. Find also the cornmon mean and correla-B l -a

tion coefficient.

Solution

Let the common meanbe m.

Then the two regression equations are

y _m= a(x_m)and

x _m= a(y _*\

But y = ax + b and x = d! + B arethe regression lines.

. . . b=m(t-a)and f =m(t_a)

b' l -a=- =.

B 1'-a

arx+brytcr=0+bo

. ' . r '=boxb,

/ \ /( o, \ ( b, )=t_+ ! t _- |

t 4,1[ . o,)arb,

4o,. ' . r t < l+arb,<bra,

Probabi l i t and Random

ExamPle l

The lifetime of a certain kind of electric bulb may be considered a-' r

mean 1200 hours and S'D 250 hours' Find the probability that the a

exceeds 1250 hours using central limit theorem'

Solution

Let X i (i =1,2,"'60) denote the life time of the bulbs'

Then p' =E(X,)=l200hrs and o' ' =Y(X')=2502hrs

= P=1200 and o' =250"

I-et Ndenote the average life time of 60 bulbs'By C'L'T' X follosr

tet Z =!a+ be the S.N.Vct l ln

Plx'1250] = rlz >t'ssf=0.5-P[0 <z 4'551

= 0'5 - 0'4394

.'. Requred ProbabilitY = 0'0606

ExamPle 2

lf X t, X z, X t' " "I n are Poisson variates with parameter

rna r[izo< S" 3160]where so = x' + x' + " "' + x "

n

SolutionS"=X, +Xr+"" '+X"n=75

Each X,is Poisson distribtrted with parametet )'=2'

var(X,) = )"=2

By CtT, S, follows Normal distribution wrth rnean n p = 7 5 x 2 =

variance = no2 =t50

. ' .,s' - N(t50'150)

To find Pllzo<s" <160I

r€t z =|ffibe the S.N.V

I

4a

This gtr"es

Iwo Dimensional Random Variabtes2,gg

plrzo<s, <t6oJ=r[_2 .4ssz <o.s2l= p[_z.qs < z < o]+ p[o < z < o.tz]= rfo < z < 2.+s]+ r[o < z < 0.82]=0.4929 +0.2939

Exampre B = o'7868

A dishibution with unknown mean p hasshould be taken from the disfribution ,r, or..o**ce 1.5. Using CLI, find how large a sampleprobablitv s*pr",n"* *r ;;;;;ffiffi"fiil::":':fl.-ur be atreast o]gs,h"i ,h"

Solution

tet X denote the samnle haah ^r^ --,,Given E (x,)=, #t;?fr ::i]J,"', ;o

n takenfrom the distriburion.By CLT,Xfo[ows normal disfibution with p and variance o,

t.".,8 ar( o,{) n

We have to find r, rrr" .) o# r*pre, such that

when X-p=0.5,2-S(o.O a .^^- r -=

:. rllP _ a!. o.r] = efl z k0.4082,,1;fSo we have to the least value of z such that

nl z kot.+ol2J-nl= o.e5From the area table of the norma-l curve,

r[l z 1<t.sa]= o.e5

.'.0-4082J;=1.96+ Jr - r-96

.'. The required sample size n = 24 0-4082

Pla -0.s . On _o,,*o.s]

> o.es (or)pnx - pl* 0.5] > 0.e5L€t Z =!(

o /,/n

t ln"-=;LX - al

J

ProbabilitY and Random P

ExamPle 4

A sample of size 100 is taken from "t"tt:t:1,:-t:t^""T"3,1t"t3"Ti

..r1:T;"ffffi;r* can we assert that the mean of the sample will

F=60 bY more than 4'

Solution

I'Et Xi,i= 1,2,3'"'100 be the sample values drawn from the given

.Then f(X,\=60 and f(X)=400'Herez='100" :

;;;;:;sample mean 7 follows noTal distribution with mean p =

o' 4oo Avar iance;=f f i=*

Probabilif that the sample mean X wi-ll not differ from p = 60 by more

"[x - pl=A)

.'. "[x

- r,l=4= "ilt ool< +]

- =r[so< X <a+)

r.et z =-I1-, =N tu'o l {n

.'. P[s6 < x 364]= rl-z<z <21

=2P[o <z<21

=2x0.4'7'72

.'. Required ProbabilitY = 0'9544

ExamPle 5

Suppose that in a certain circun' 'o:ttlt'o,:,T:,:11"::l"Tl;nff

"r":"TJ:":,H';* il;';; '"'n'"tf"rv'

what is the probabilitv that tlre

the circuit will exceed 98 ohms' assuming independence'

Solution

l-et X,rcptesent the resistance in the i'i resistor'

Tlren X,', X 2, "'X ro NQidentically distributed R'Vs

Since the resistors are connected in series'the fotal

resistance

RI S,- Xr+ Xr+ ""+ X'o

Here Elx,l=5 andl"d(x') =0'20for all f'

,[f;{


By C.L.T S, - N(nyt,noz)

To find P(.1, > 98) :

E(S") -np=20x5=100

var (5,) = oot = 20 x 0.20 = 4

I-r:t 7 = lo-JD

P[s, ' tt]= r[i >-t)

=0.5+p(-1 .z.o)= 05 + p(0 .2 .1)

= 0.5 + 0.3413.'. Required probability : 0.8413

Example 6

20 dice are thrown. Find approximately the probability that the sum obtained is between 65and 75 using C.L.T.

Solut ion

Let X, be the R.V representing the number shown on the ih dje.6, r1\ l -rhen E(x,)=I ; l i l= l t r +2+3+4+5+61=1

ul*t)=li, l= jr,, +22 +32 +42 *ri*1

Then

=916

"*(4) = nlxll-ln(x,)f'_9r _49 _35

64t2I-et S, be the total score.

So = Xr + X, + ....+ Xzo

t (s,) =20*Z=79 =(np)

"*(s") =20*f =\={"C)

2.101 Probabi and Random P

By CLI, S, - N(n1t,no'\

i",+-{zof)

Let Z =s'-70 oe the S.N.vl rTs

!3When S, =65,2 = -0.65

S^ =75,2 = 0.65

:. Pl65<,s" < 75] = r1-0.t5 < z < 0.65f

=zPlo <z <o.6sl

= 2x0.2422 = 0.4844

.'. Required probability = 0.4844

Example 7

What is the probability that the average of 150 random points from 6e

within 0.02 of the mid point of the interval?

Solution

Let Xi(i =1,2,...150) be the random points in the interval.

Then the random variable X, follows uniform distribution in the interval

1 11. for0<x<1The P.d.f of x, is f (4=

t_O=l O. o***

Mean = o * b *rdvariance -@

- o)' for turiform distribution rn

2t2

:.Elx,l=I *a vlx,l= .L^2 ' ' t 12

Let X denote the average of the random points X,, X2,X3,....X$,

. v Xr+ X, +.. . . . .+ Xrro"^ =-----36-

rnen r(X) =I^o*'(t)=#*= 1800

andom Variables

PL -'aJ

Ij

;

ro nnd p[lx-0.51

I

. /12 x 150

. '. r[0.+s < * <o.sz]= p[-r,..

=zPlo'<z <0.8+.

=2Plo<z so.ssl=2(w023)= 0.6046

... The required probability : 0.6046

Example 8

If I/i ,i =1,2,..,20are independent noise voltages received in an adder and Zis the sum of thevoltages received, find the probability that the total incoming voltage Z exceeds 105, usingcentral limit theorem. Assume that each of the random variables ( is uniforrnly distributed over(0,10)

Solution

The random variable ( follows uniform distribution over (0,10). Then E(v,)= 5 and(10-n\ ' 1nn

vU/,1=. u/ = tuu = vf i r , \=T12 12 \ ' / 3

The total incoming voltage is [/ =V, +V, + .....+yn

By CtjI, Zfollows Normal distribution with mean= nF =20x5 = 100 and

variance = not =t?oJ

To find rltr >rcSl:

I-et Z =l/ -1oo be the S.N.V

1500r/;

2.104

2.103 Probabi l i t and Random Pr

Pfv1105]= P[z>a39]=0.5-P[o <z <0:,9]

= 0.5 - 0.15i7 = 0.3483

.'. P [Total incoming voltage Zexceeds 105]:0.3483

Example 9

A coin is tossed 10 times. What is the probability of getting 3 or 4 or 5 headsl

limittheorem.

Solution

ktXdenote the number of heads in 10 tosses.

Here z = I0, p= probability of getting head in a trial = ]

: .q =L- P =+^2

Meanof X=nD=10x1=5-2

Variance X = npq= 19*1*1 = 2.522

:.p=sando=ff=1.58

To find Pf3 < X =5] . Since Xis a discret€ RV we find P12.5 < X <551

t-" . -5 5.5-51: . Plz.s< x < s.5l = Pl

-

< z <# |' L 1.58 1.58 I

=P[-1.58 <z <0327 where z =

=P[-r.58 .z.ol+r[o< z <w2]= P[o < z <r.587+r[o < z <B2l

= 0.4429 + 0.1255

Hence required probability = 0.5684

Example I O

The buming time of a certain tlpe of lamp is an experimental random variable

hours. What is the probability that 144 of these lamps will provide a total of rnmr

hours of burning time?

Two Dimensional Random Variables 2.t04

Solution

Let X, be the buming time of the i'h larcrp.

." X1,X2,X3,....Xtu are identically distributed independent random variables.

For exponential distribution,

E ( Xl= 1 *d ,,-i*"" = I\ / ) "At

.'. Mean of X, = 30 and var(X ) = eOO

Let S, = Xr I X, +....+ Xr* bethe total bumrng time.

By CLT, S, - N(np,no'\

I-et 2 = s,:nlt = 4 -(t++)roJi,o 30.,1144

= sn - 4i20360

To find P(S, > 4500)

PIE t45oo]= plz>o.sl

=0.5-P[0 <z <o.sf

= 0.5 - 0.1915 = 0.3085

? _4500-4320360

= 0.5

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