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STAT 516
Midterm Exam 3
Friday, April 18, 2008
Name
Purdue student ID (10 digits)
1. The testing booklet contains 8 questions.
2. Permitted Texas Instruments calculators:
BA-35
BA II Plus
BA II Plus Professional
TI-30Xa
TI-30X IIS (solar)
TI-30X IIB (battery)
The memory of the calculator should be cleared at the start of the exam, if possible.
No other calculators are permitted.
3. Simplify the answers and write them in the boxes provided.
4. Some partial credit may be awarded at the discretion of the professor.
5. Show all of your work in the exam booklet.
Answers without justification may be viewed as exhibiting academic dishonesty.
6. Extra sheets of paper are available from the professor.
1. A point X, Y is chosen uniformly from the unit circle. In other words, X, Y are jointly
distributed continuous random variables with joint density
(
C 0 ≤ x2 + y 2 ≤ 1
f (x, y) =
0 otherwise
where C is constant. Find
√ the probability that the distance from (X, Y ) to the origin (0, 0)
is less than 1/2, i.e., P ( X 2 + Y 2 ≤ 1/2).
Answer. Method 1. The desired event happens if and only if (X, Y ) is found in a circle of
radius 1/2 centered at the origin. Since the random variable have a uniform joint distribution,
then the desired probability is the area of the circle centered at the origin with radius 1/2
(namely, (π)(1/2)2 = π/4), divided by the area of the circle centered at the origin with
radius 1 (namely, (π)(1)2 = π). So the desired probability is π/4
= 1/4.
π
Method 2. We can solve for C by computing
Z
1
−1
√
Z
1−x2
√
− 1−x2
C dy dx = 1
which yields C = 1/π (or just noting that C is the reciprocal of the area of the entire circle
of radius 1, centered at the origin). Then the desired probability is
R 1/2 R √ 14 −x2 1
√
dy dx
−1/2 − 1 −x2 π
1
1/4
4
R R √ 2
=
=
1
1−x
1
1
4
√
dy dx
−1 − 1−x2 π
1
2. Let X and Y denote the number of eggs laid by insects A and B, respectively, in a study.
It is known that X and Y are independent Poisson random variables, each with parameter λ.
If the total number of eggs laid is 150, what is the chance that insect A laid at least 90 eggs?
[Hint: Recall that, if X, Y are independent Poissons with parameters λ1 , λ2 respectively,
then the conditional distribution of X, given X + Y = n, is Binomial with parameters n and
1
.]
p = λ1λ+λ
2
Answer. As in the hint, the conditional distribution of X is Binomial with n = 150 and
λ
p = λ+λ
= 1/2. So the desired probability is P (X ≥ 90). Here are two methods for finding
P (X ≥ 90).
Method 1. Note that X is approximately normal with mean np = 75 and variance
np(1 − p) = 37.5, so the desired probability is
P (X ≥ 90) = P (X ≥ 89.5)
89.5 − 75
X − 75
≥ √
=P √
37.5
37.5
≈ P (N ≥ 2.37)
= 1 − Φ(2.37)
≈ 1 − .9911
= .0089
Method 2. Some students decided to give an expression for the exact answer. Since I did
not specify whether an approximation was desired, I gave full credit for this type of answer
too:
150−i
i 150 150
150 X
X
1 X 150
1
150
1
= 150
1−
P (X ≥ 90) =
P (X = i) =
2
2
2 i=90 i
i
i=90
i=90
Unfortunately, this is difficult to compute without Maple, but I still gave full credit for this
type of answer.
2
3. A fair coin is tossed 300 times. Let H100 denote the number of heads in the first 100 tosses,
and H300 the total number of heads in all 300 tosses. Find the correlation ρ of H100 , H300 ,
i.e., ρ(H100 , H300 ) = √ Cov(H100 ,H300 ) .
Var(H100 )Var(H300 )
Answer. Note that H100 is Binomial with parameters n = 100 and p = 1/2, so Var(H100 ) =
(100)(1/2)(1/2) = 25. Also, H300 is Binomial with parameters n = 300 and p = 1/2, so
Var(H300 ) = (300)(1/2)(1/2) = 75.
Now write Xi to indicate whether the ith toss
Xi = 1 if the ith toss
P shows heads, i.e.,P
300
shows heads, and Xi = 0 otherwise. So H100 = 100
X
and
H
=
i
300
i=1
j=1 Xj . So
Cov(H100 , H300 ) = Cov
100
X
Xi ,
300
X
i=1
!
Xj
j=1
=
100 X
300
X
Cov (Xi , Xj )
i=1 j=1
Note that Xi , Xj are independent if i 6= j. So Cov(Xi , Xj ) = 0 if i 6= j. So the sum in the
formula above simplifies to
!
100
100
100
X
X
X
Cov(H100 , H300 ) =
Cov (Xi , Xi ) =
Var(Xi ) = Var
Xi = Var(H100 ) = 25
i=1
So we conclude that the correlation is √
i=1
i=1
25
(25)(75)
3
√
=
3
3
≈ .5774
4. Consider independent standard normal random variables W, X, Y, Z.
Find P (W + X > Y + Z + 1).
Answer. We note that
P (W + X > Y + Z + 1) = P (W + X − Y − Z > 1)
and also W + X − Y − Z is a normal random variable with mean 0 + 0 − 0 − 0 = 0 and
variance 1 + 1 + 1 + 1 = 4. Thus, the desired probability is
W +X −Y −Z −0
1−0
√
= P (N > 1/2) = 1 − Φ(1/2) ≈ 1 − .6915 = .3085
P
> √
4
4
where N is standard normal.
4
5. Consider two jointly distributed continuous random variables X and Y with joint density
(
3x 0 ≤ y ≤ x ≤ 1
f (x, y) =
0
otherwise
Find P ( 15 ≤ Y ≤ 52 ).
Answer. Method 1. We compute using just one integral
Z 2/5 Z 1
1
2
3x dx dy
P
≤Y ≤
=
5
5
1/5
y
Z 2/5 2 1
3x =
dy
2 x=y
1/5
Z 2/5
3
=
(1 − y 2 ) dy
2
1/5
2/5
y 3 3
y−
=
2
3 y=1/5
3 !
3
2 1 2
=
−
−
2
5 3 5
1 1
−
5 3
3 !!
1
5
34
125
= .272
=
Method 2. We compute using two integrals
Z 2/5 Z x
Z 1 Z 2/5
1
2
3x dy dx
P
≤Y ≤
3x dy dx +
=
5
5
2/5 1/5
1/5
1/5
Z 2/5
Z 1
2/5
x
=
3xy|y=1/5 dx +
3xy|y=1/5 dx
1/5
2/5
2/5
Z
Z
3x(x − 1/5) dx +
=
1/5
=
3x
2/5
2 1
−
5 5
dx
1
2/5
3x3 3 x2 3 x2 −
+
3
5 2 x=1/5 5 2 x=2/5
2 2 !
3 3 !
3
3 2
2
1
2
1
−
+
1 − (2/5)2
−
−
5
5
10
5
5
10
=
1
34
125
= .272
=
5
6. A machine has a random supply Y at the beginning of a given day and dispenses a random
amount X during the day (with measurements in gallons). It is not resupplied during the
day; hence, X ≤ Y . It has been observed that X, Y have joint density
(
1/2 0 ≤ x ≤ y ≤ 2
f (x, y) =
0
otherwise
Find the conditional expectation of X, given that Y = y, for some fixed y with 0 ≤ y ≤ 2.
Answer. Method 1. Since X, Y have a joint uniform distribution, then once we are given
Y = y, we know that the conditional distribution of X is uniform in the range 0 ≤ x ≤ y,
so the conditional expected value of X in this case is y/2.
Method 2. We compute
Z ∞
Z y
f (x, y)
dx
E(X | Y = y) =
xfX|Y (x|y) dx =
x
fY (y)
−∞
0
We already know the joint density of X, Y , but we still need to compute the marginal density
of Y evaluated at y. This is
y
Z y
1 1
dx = x
fY (y) =
= y/2
2 x=0
0 2
Thus
Z
E(X | Y = y) =
0
y
1/2
x
dx =
y/2
Z
6
0
y
y
x
y2
x2 =
dx =
= y/2
y
2y x=0 2y
7. A complex machine is able to operate effectively as long as at least 3 of its 5 motors are
functioning. Suppose that each motor independently functions for a random amount of time
with density function
(
xe−x x > 0
f (x) =
0
otherwise,
For fixed t > 0, compute the probability that the machine functions until at least time t.
[Note: For simplicity, you do not need to simplify your answer on this problem. Just
circle your final answer.]
Answer.
For t > 0, the probability that a particular motor functions until at least time t
R∞
is t xe−x dx. We use u = x and dv = e−x dx, and thus du = dx and v = −e−x , to solve the
integral by parts:
Z ∞
Z ∞
−x
−x ∞
−t
xe dx = −xe x=t −
−e−x dx = te−t − e−x |∞
+ e−t = (t + 1)e−t
x=t = te
t
t
So the probability that exactly i of the 5 motors will function until at least time t is
5
((t + 1)e−t )i (1 − (t + 1)e−t )5−i
i
So the probability that the machine functions until at least time t is
5 X
5
((t + 1)e−t )i (1 − (t + 1)e−t )5−i
i
i=3
7
8. A standard deck has 52 cards, with values A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K (4 cards of each
value). Choose 5 cards randomly from the deck, without replacement. Given that exactly 2
Jack cards (i.e., value “J”) appear, find the expected number of Ace cards (i.e., value “A”)
that appear.
[Note: For simplicity, you do not need to simplify your answer on this problem. Just
circle your final answer.]
Answer. Method 1. Let X denote the number of Ace cards selected, and let Y denote the
number of Jack cards selected. Then the desired conditional expectation is
E(X | Y = 2) =
3
X
xpX|Y (x|2) dx =
x=0
3
X
p(x, 2)
x
pY (2)
x=0
44
(42)(483)
(x4)(42)(5−x−2
)
.
Also,
p
(2)
=
. So the desired
Y
(525)
(525)
44 3
4
X
1
=
x x 483−x
4
3
x=0
For 0 ≤ x ≤ 3, we note that p(x, 2) =
probability is
Method 2. Let X denote the number of Ace cards selected, and let Y denote the number
of Jack cards selected. Then X = X1 + X2 + X3 + X4 + X5 , where Xi = 1 if the ith
card selected from the deck is an Ace, and Xi = 0 otherwise. So the desired conditional
expectation is
5
X
E(X1 + · · · + X5 | Y = 2) =
E(Xi | Y = 2)
i=1
and Y =2) .
Each Xi is an indicator function, so E(Xi | Y = 2) = P (Xi = 1 | Y = 2) = P (Xi =1
P (Y =2)
(4)(48)
We note that P (Y = 2) = 2 52 3 . Also P (Xi = 1 and Y = 2) = P (Xi = 1)P (Y = 2 | Xi =
(5)
4
47
4 (2)( 2 )
1) = 52
. Thus
(514)
4
47
4 (2)( 2 )
51
52 ( )
1
4
E(Xi | Y = 2) = 4 48
=
(2)( 3 )
20
52
(5)
P
5
So the desired probability is E(X1 + · · · + X5 | Y = 2) = 5i=1 E(Xi | Y = 2) = 20
= 41 .
Method 3. Since we are given that exactly two of the five cards are Jacks, then the
remaining 3 cards are just chosen from 48 non-Jack cards. In this new setting, we let
X = X1 + X2 + X3 , where Xi = 1 if the ith card is an Ace, and Xi = 0 otherwise. So
E(Xi ) = P (Xi = 1) = 4/48 = 1/12, and thus E(X) = E(X1 + X2 + X3 ) = E(X1 ) + E(X2 ) +
1
1
1
E(X3 ) = 12
+ 12
+ 12
= 14 .
8