1. X is Poisson, with = 12.3
deviation: 5.2714 = 2. 296 .
3
7
Pr(X > 4) =
= 5. 2714. Its mean is 5.2714, standard
X i
i=5
i!
e = 60.58%
2. X is hypergeometric with N = 52, K =q and n = 7. Its expected value is
13
K
7
n N = 4 = 1.75, standard deviation: 7
MATH2F82
SECOND MIDTERM MARCH 19, 2003
Full credit given for three correct and complete answers.
Open-book exam.
Duration: 50 minutes
_
1. Consider a distribution having the following distribution function:
F (x|) = 1 exp x
0<x
where > 0. Find the Rao-Cra
MATH2F82
FIRST MIDTERM FEBRUARY 12, 2003
Full credit given for three correct and complete answers.
Open-book exam.
Duration: 50 minutes
_
In the next three questions, let X be a random variable with the following pdf:
1
1<x<1
f (x) = (1 x)
2
1. Find the
2
1. Since
Z
x3
3x3
exp( )dx = 0.89298 1/3
0
(X)3
= 1.40435 (X)3
=
0.892983
P
3
Optional: To compute the expected value of ( n Xi ) , we rst realize
i=1
that this expression equals
E(X) =
n
P
i=1
Xi3 + 3
P
Xi2 Xj +
i6=j
Secondly, we also need
Z
2
E(X ) =
1. Optional. F (x) =
R
xex dx = 1 (1 + x)ex .
X
Pr( 1+X < y) = Pr(X <
f (x) =
y
(1y)3
2. F (x) = c
y
)
1y
=1
1
1y
y
exp( 1y )
y
exp( 1y ), where 0 < y < 1.
x2
2
where 0 < x < 1 c = 2.
2
y
y
X
F (y) = Pr( 1+X < y) = Pr(X < 1y ) = 1y f (y) =
3. Optional.
2
R b bx n1 dx
ba
x ba
= na+b = a + n+1
ba
n+1
a
2 bx n1 dx
Rb
n(ba)2
= (n+2)(n+1)2 .
Var[X(1) ] = n a x na+b
n+1
ba
ba
1. E[X(1) ] = n
(a) No, since
na+b
n+1
6= a
na+b
n n+1
(b) Yes, since lim
=a
n(ba)2
(n+2)(n+1)2
n
(c) Yes, since also lim
=0
h
i
3
x3
2