stat 6201 midterm exam i solutions october 1, 2015 name · stat 6201 midterm exam i october 1, 2015...

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STAT 6201 Midterm Exam I October 1, 2015 Name ........................................................................... 1. (7 points) Assume that a random variable X has the following cdf Find: (a) Pr(0 <X 3) (b) Pr(0 X 3) (c) Pr(0 <X< 3) (d) Pr(1 X 2) (d) Pr(1 <X 2) (e) Pr(X 5) (f) Pr(X> 5) 1 SOLUTIONS = Fl 3) Flo ) = 0.8 - 0.2 = 0.6 = F (3) - Flo ) + P ( xo ) = 0.6+0.1 = 0.7 = FIH - Flo ) = 0.6-0.2=0.4 : = 0 = 0

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STAT 6201 Midterm Exam IOctober 1, 2015 Name ...........................................................................

1. (7 points) Assume that a random variable X has the following cdf

Find:

(a) Pr(0 < X 3)

(b) Pr(0 X 3)

(c) Pr(0 < X < 3)

(d) Pr(1 X 2)

(d) Pr(1 < X 2)

(e) Pr(X � 5)

(f) Pr(X > 5)

1

SOLUTIONS

= Fl 3) - Flo ) = 0.8 - 0.2 = 0.6

= F (3) - Flo ) + P ( xo ) = 0.6+0.1 = 0.7

= FIH - Flo) = 0.6-0.2=0.4

:= 0

= 0

2. Suppose that the joint pdf of two random variables X,Y is

f(x, y) =

(c(xy

2) if 0 x 1 2x y 2

0 otherwise

(a) (2 points) Derive the value of the constant c.

(b) (3 points) Derive g1(x | y), the conditional pdf of X given Y = y.

2

fjkgxyidydx =L

sjcxtfkd " = 's § x4dx= '÷ ( Ho'

- Yt )=8÷ . ,÷=t¥

at

kly ) = flaig( Y) e- marginal of Y

zxeyez ⇒

Yk0 ex e Is

fdH= ).cxgidx = cy

'. It µ = gI

qkly )=c÷f¥=sy÷ oe " ±

(c) (3 points) Find Pr(X < 1/2 |Y = 3/2).

(d) (4 points) Find Pr(X < 1/2).

3

Plxctl 'HW= Kg ,kI⇒dx= 't"

'g±yax=3÷ . ¥ 'K= 3÷ . f=±a

P(×< ⇒ = µ fibddx where find = marginal of X

fdx ) = {xcxyidy = ex . ftp. ='13 (8-8×3)=85 ( x . x

") oexei

Plath . Koski 't÷eH 't"

- ¥IY=o÷fI . stat 's .±, li . ⇒=Ei÷=i÷

3. Let Y be the rate (calls per hour) at which calls arrive at a switchboard. Let X be the

number of calls during a two-hour period. Suppose that the marginal pdf of Y is

f2(y) =

(e

�y

if y � 0

0 otherwise

and that the conditional pmf of X given that Y = y is

g1(x | y) = Pr(X = x |Y = y) =

(3y)

x

x!

e

�3yfor x = 0, 1, 2, . . .

(a) (3 points) Find the marginal pmf of X. You may use the formula

Z 1

0y

k

e

�y

dy = k!.

(b) (3 points) Find the conditional pdf of Y given that X = 0, that is, find g2(y |X = 0).

4

PRIX --x)=[oflxiy )dy=[at .1H . girly )dy= fete

.tk?I.e3ydy=to.fT3yYi4Ydy-=tff,tI.fl4yY.ehydy=fap,H.fx!=H.F4Px=o.i.z...

.÷x !

( Use change of

variables )

gdy1X=o)= fl0iY) =

fly ) . 9,1019 )=

et . e-39

-- ¥4.547 yzoPr(X=O ) 171*-0 )

Prato )=t ,. ( I ,T=±,

4. Let X be a random variable having the Uniform(0, 6) distribution.

(a) (2 points) Find f(x), the pdf of X, and sketch it.

(b) (3 points) Find F (x), the cdf of X and sketch it.

5

ftp.t !ME

:i÷÷antitheft;

it

If:{Y

¥G

(c) (3 points) Let Y = X

2. Find G(y), the cdf of Y , and sketch it.

(d) (3 points) Find f(y), the pdf of Y , and sketch it.

6

*

.nu#Hrlxih=fxltry,!%n it no

' ' an ftp.itaoaa.H.IT"

i÷+÷y )=dd6y# = To its . y

" "

if o<y< 36

=D

otherwise¥0