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Homework 9 (due November 24, 2009)
Problem 1. The join probability density function of X and Y is given by:
xy f (x, y) = c x2 +
0 < x < 1, 0 < y < 2,
2
where c is a constant.
(a) Find c so that this is a proper joint density.
Solution: To do this we integrate over x and y and set it equal to 1 Then solve for
c:
Z 1Z 2
Z 1
xy
2 1
7
2
2
x + dydx = c
2x + xdx = c
1=c
+
= c.
2
3 2
6
0
0
0
Thus c = 67 .
(b) Find the marginal density function for X
Solution: To find the marginal density we integrate over all y.
Z
6
6 2 2 xy
x + dy =
fX (x) =
2x2 + x .
7 0
2
7
(c) Find P {X > Y }
Solution: To do this we compute:
Z Z
Z
6 1 x 2 xy
6 15 3
6 5
15
P {X > Y } =
x + dydx =
x dx =
= .
7 0 0
2
7 0 4
7 16
56
(d) Find P {Y < 12 |X < 12 }
Solution
R 1 R 1 2 xy
2
2
x + 2 dydx
P {Y < 12 , X < 12 }
1
1
0
0
P {Y < |X < } =
=
1
R
1
2
2
P {X < 2 }
2
2x2 + xdx
0
R 1 x2
x
2
+ 16
dx
0 2
=
= 11/180
5/24
1
(e) Find E[X]
Solution:
6
E[X] =
7
1
Z
0
5
2x3 + x2 dx = .
7
(g) Find E[Y ].
6
E[Y ] =
7
Z
2
Z
y
0
0
1
xy
6
x + dxdy =
2
7
2
Z
0
1
y y2
68
8
+ dy =
=
3
4
76
7
Problem 2. A man and a woman agree to meet at a certain location about 12:30 P.M. If
the man arrives at a time uniformly distributed between 12:15 and 12:45 and if the woman
independently arrives at a time uniformly distributed between 12 and 1, find the probability
that the first to arrive waits no longer then 5 minutes. What is the probability that the man
arrives first?
Solution: If we let X be the amount of time after 12:00 that the man arrives and
Y be the amount of time after 12:00 that the woman arrives then X and Y are both
uniform random variables on the intervals (15, 45) and (0, 60) respectively. Since they
are independent their joint density function for X and Y is given by:
(
1
15 < x < 45, 0 < y < 60
f (x, y) = 1800
0
otherwise
To compute the probability that the person who arrives first waits no longer then 5
minutes we need to compute that:
P {|Y − X| ≤ 5} = P {−5 ≤ Y − X ≤ 5} = P {−5 + X ≤ Y ≤ 5 + X}
Z 45 Z x+5
1
1
dydx =
=
6
15
x−5 1800
To compute the probability that the man arrives first we must compute P {X > Y }
Z 45 Z x
1
1
dydx =
P {X > Y } = P {X − Y > 0} =
2
15
0 1800
2
Problem 3: Jill’s bowling scores are approximately normally distributed with mean 170
and standard deviation 20, while Jack’s scores are approximately normally distributed with
mean 160 and standard deviation 15. If Jack and Jill each bowl one game, then assuming
that their scores are independent random variables, approximate the probability that:
(a) Jack’s score is higher;
(b) the total of their scores is above 350.
Solution: If we let X be jacks score and Y be Jill’s score then X and Y are independent normal random variables with means µ1 = 160 and µ2 = 170 and variances σ12 = 400 and σ22 = 225 respectively. Then the answer to (a) is given by:
P {X − Y > 0}. However, since Y is normal, so is −Y with the same variance, and
−µ2 as its expected value. Also the sum of two normal random variables is normal
with mean the sum of the means and variance the sum of the variances. Consequently
1 +µ2
√ −µ
Z = X−Y
IS a standard normal random variable, thus:
2
2
σ1 +σ2
−µ1 + µ2
} = 1−Φ
P {X−Y > 0} = P {Z > p 2
σ1 + σ22
10
√
625
2
= 1−Φ
≈ 1−.6554 = .3446.
5
For (b) we solve it in a similar way to (a) except here we are interested in P {X + Y >
350}. Since both X and Y are independent normal random variables, thenn X + Y
√ −330
is normal with mean µ1 + µ2 = 330 and variance σ12 + σ22 = 625. Thus S = X+Y
625
is a standard normal random variable and we have:
20
20
P {X + Y > 350} = P {Z > √
}=1−Φ
1 − .7881 = .2119.
25
625
3
Problem 4. According to the U.S. National Center for Health Statistics, 35.2 percent of
males and 26 percent of females never eat breakfast. Suppose that random samples of 200
men and 200 women are chosen. Approximate the probability that:
(a) at least 110 of these 400 people never eat breakfast;
Let X =the number of men among the sample of 200 that never eat breakfast. and
let Y =the number of women among the sample of 200 that never eat breakfast. Let
S = X + Y . Assume that the likelihood of any one person never eating breakfast
is independent of any other person never eating breakfast, the E[X] = 70.4 and
E[Y ] = 52. Consequently E[S] = 122.4. Using Markov’s Inequality we can obtain
that:
P (S ≥ 110) ≤ E[S]/110 = 112.4/110
of course this is pretty useless information here since this number is bigger than 1
and any probability is less than 1. However we can do better by assuming that X
and Y are binomial with n = 200 p = .352 and n = 200 and p = .26 respectively.
Since 200(.352)(.648) > 10 and 200(.26)(.74) > 10 so both of these binomial random
variables can be approximated by normal random variable with mean np and variance
2
= 45.6192
np(1 − p). Thus X is approximately normal with mean µX = 70.4 and σX
2
and Y is approximately normal with mean µY = 52 and variance σY = 38.48. Since
the sum of normal random variables is normal then X + Y is normal with mean
2
µS = µX + µY = 122.4 and Variance σS2 = σX
+ σY2 = 84.0992. Thus we have:
P (X + Y ≥ 110) =
≈
=
=
=
109.5 − 122.4
X + Y − 122.4
√
≥ √
P
84.0992
84.0992
P (Z ≥ −1.352)
1 − Φ(−1.407)
1 − (1 − Φ(1.407))
Φ(1.407) ≈ .9207
Note, that since the sum of X and Y is a discrete random variable I have included a
continuity correction to compensate for this.
(b) the number of the women who never eat breakfast is at least as large as the number
of the men who never eat breakfast.
Hint: see example 3f in chapter 6.
4
Using X and Y from above then we are interested in the probability P (Y ≥ X) =
P (Y − X ≥ 0). The difference D = Y − X is also a sum of approximately normal random variables and since E[Y ] = 52 , V ar(Y ) = 38.48, E[−X] = −70.2
and V ar(−X) = 45.6192 (recall V ar(aX) = a2 V ar(X)). Thus we have that D is
approximately normal with µD = 52 − 70.4 = −18.4 and σD = 38.48 + 35.6192.
So,
−.5 − (−18.4)
D − (−18.4)
√
≥ √
P (D ≥ 0) = P
84.0992
84.0992
≈ P (Z ≥ 1.951)
= 1 − Φ(1.95)
≈ 1 − .9744 = .0256
Notice that since D is a discrete random variable being approximated by a continuous
random variable I have included a continuity correction.
5
Problem 5. Suppose that 3 balls are chosen without replacement from an urn consisiting
of 6 white balls and 9 red balls. Let Xi be equal to 1 if the ith ball sleected is white, and
let it equal 0 otherwise. Give the joint probability mass function (write out all the possible
values) for :
(a) X1 , X2 ;
(b) X1 , X2 , X3 .
Solution: For (a) we compute the values at the four possible pairs (i, j)
6 5
15 14
6 9
pX1 ,X2 (1, 0) =
15 14
9 6
pX1 ,X2 (0, 1) =
15 14
9 8
pX1 ,X2 (0, 0) =
15 14
pX1 ,X2 (1, 1) =
pX1 ,X2 ,X3 (1, 1, 1) =
pX1 ,X2 ,X3 (1, 1, 0) =
pX1 ,X2 ,X3 (1, 0, 1) =
pX1 ,X2 ,X3 (0, 1, 1) =
pX1 ,X2 ,X3 (1, 0, 0) =
pX1 ,X2 ,X3 (0, 0, 1) =
pX1 ,X2 ,X3 (0, 1, 0) =
pX1 ,X2 ,X3 (0, 0, 0) =
6 5 4
15 14 13
6 5 9
15 14 13
6 9 5
15 14 13
9 6 5
15 14 13
6 9 8
15 14 13
9 8 6
15 14 13
9 6 8
15 14 13
9 8 7
15 14 13
(1)
6
Problem 6: The joint density function of X and Y is
(
c(9x + 3y) 0 < x < 3, 0 < y < 1
f (x, y) =
0
otherwise
The constant c =
1
,
45
but you may leave your answers in terms of c.
(a) Are X and Y independent?
No it can not be factorized into the product of two. Also you could compute the
marginal densities and see that their product is not the joint density.
(b) Find the density function of X,
Z 1
3
9x + 3ydy = c(9x + ), 0 < x < 3.
fX (x) = c
2
0
(c) Find P {X + Y < 1}.
Z
1
Z
P {X + Y < 1} = c
1−y
9x + 3ydxdy = 2c.
0
7
0