STAT 6201 Midterm Exam II

November 6, 2015 Name ...........................................................................

1. Let Xi denote the outcome of a clinical trial for the ith patient, where i = 1, 2, . . . , n. Weassume that Xi = 1 if the treatment was a success and Xi = 0 otherwise. Let Pi be thesuccess probability for patient i, which is unknown. The researchers believe that patients areindependent and that the variables Pi are iid with a common distribution with the followingpdf

f(p) =

(6p(1� p) if 0 < p < 1

0 otherwise

(a) (1 point) What is the conditional distribution of X1 given P1 = p ?

(b) (2 points) Find the joint distribution of X1 and P1.

(c) (4 points) Find the expected number of successful treatments in this trial.

1

Solutions

Xitloil } X,

~ Bernoulli PIX ,=l| Pep) =pXp Bernoulli (p) hH=p'

'

( I - p5"

n=o,

,

ha )=P(X=i ) =phlo )=P( Xto )= 1- p

.

g(x,,p)=f( p) .f(xi|p)= pflp) if x ,=i

marginalfinal | ( rp) .f( p ) if 04=0

X=X,

.tl/zt...+Xw = # of successes in this trial

E(x)=EC× , )t . . .+El×w) ✓

marginal pmf of X,

:

PAID- ftp.fcpldp = { 6p4tp)dp=§6p±6p3dp= 6.131.1 .

6.IT/.l=6.ts.6.ty=tPlx,=o)=tPlx,=D=t

thus, ECX , )=± ⇒ ECx)=÷

2. A woman leaves for work between 8 AM and 8:30 AM and takes between 40 and 50 minutes toget there. Let the random variable X denote her time of departure and the random variableY the travel time. Assume that these random variables are independent and uniformlydistributed.

(a) (3 points) Find the expected arrival time at work.

(b) (3 points) Find the probability that the woman arrives at work before 9 AM.Hint: this probability can be expressed as E(g(X,Y )) where the function g(x, y) isdefined as

g(x, y) =

(1 if arrival time is before 9:00

0 otherwise

2

denote 8:00 as time 0:00 ⇒ X~ Uniform ( 0,30 ) ( minutes)

Y ~ Uniform ( 40,50)Arrival : Xty El Xty ) = ECDTECY ) = 15+45=60 ( minutes past 8:00 )

Expected arrival time : 9:00

( Xty < 60 )

P( arrive before 9:00 )= E( g ( x ,Y)) where gl . ;) is given above

EC gay )) = { fsdigk ,y)fµDtdy ) dydx

= / Moga ,y)f,Hfyy)dydx=

= he fill . 's tooydit KEY .÷o . toaydx

= too . ( iokio ) + too . too fox)dx

= ft to.

. ( zook - Elio )=ts+ ! = ft -

- I

:-

go Klm X+Y< 60

.

xey.ioP(X+Y< 60) =L

30-

30

3. Suppose that X is the number of visitors to a webpage in any given day and that X followsa Poisson distribution with mean ⇤, however ⇤ is unknown. The webmaster believes that ⇤follows a continuous distribution with a pdf given by

f(�) =

(2e�2� if � > 0

0 otherwise

(a) (3 points) Find the mgf (t) of the random variable ⇤. Do not forget to specify forwhich values of t is the mgf finite.

(b) (3 points) Using the mgf in part (a) or otherwise, find the expected value of E(⇤)

3

Ylt )= E ( et ") = fdet ''

. z.ee 'd ,\=zf%' ( 2 't 'd ,\ < a if t<

2=2. tI"÷E=÷

El . ) = Ilo ) =u÷pl⇐.

= 's

or,

E ( 1) = €2 . aettdx = ( integration by parts) =L

(c) (3 points) Assume that only X = 1 person visited the webpage in a day. What is theconditional expected value of ⇤ given that X = 1.

4

For this we need the conditional pdf of A given XH

a joint

g. ( xlii ) .tk#x.marginal

flx , D= ftp.PLx.tl/D=2.e*.et#=zx.e3 '

Plx - i ) = file "dx = - }xe '' IT +3 Ee

' "dx =3 . tee''t

.

? ÷

f- Lx fkzgiei

"

g=- tee '

gdxlx .. D= k¥3

"

= qxesx

E( A- |X=l ) = los .gdx1x=Ddx=h?9x?e ' dx = ( integration by parts ) =3

4. Suppose that the measured voltage in a certain electric circuit has the normal distributionwith mean 120 and standard deviation 2. Assume that three independent measurements ofthe voltage are made, X1, X2 and X3.

(a) (3 points) What is the probability that all three measurements will lie between 116 and118?

(b) (3 points) What is the probability that X1 +X2 240.

5

P ( 116 e X,

e 118,

116 ± Xz e 118,

116 e X,

E 118) = P ( 116 ± X,

< 118 )3

P ( 116 ex,

< 118 ) =P ( -2 a ¥20 e . i )=P ( - ze Z e - D= OI ( t ) - § C- 2)

.z

§ ti ) = 1 - to (1) = 1- 0.8413 =.

1587

§ ( - 2) =1 - § (2) = 1 -

.9773 =

.0227

P ( 116 ± X,

± 118 ) =. 1587 -

. 0227 = 0.136

final answer : @. 136)3

P ( X ,tXz e 240) X,

txz ~ N ( 240,8)↳ variance

=P ( ×'tk÷40 = o) =P ( z ± o ) = I

6

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