Download Solution to Homework 5

Solution to Homework 5
1. [§5-2] A system consisting of one original unit plus a spare can function
for a random amount of time X. If the density of X is given (in units of
months) by
(
Cxe−x/2 x > 0
f (x) =
0
x≤0
what is the probability that the system functions for at least 5 months?
First we need to find the value of C. Since
R∞
−∞
Z ∞
xe−x/2 dx
Cxe−x/2 dx = C
0
∞ Z ∞0
−x/2 −x/2
= C (−2xe
−2e
dx)
−
0
0
∞
= C 0 − 4e−x/2 1=
Z
∞
f (x) dx = 1, we have
integration by parts
0
= 4C.
Therefore, C = 1/4. Then
Z ∞
1 −x/2
P(X ≥ 5) =
xe
dx
4
5
∞ Z ∞
1
−x/2 −x/2
(−2xe
=
−2e
dx)
−
4
5
5
∞ 1
10e−5/2 − 4e−x/2 =
4
5
14 −5/2
=
e
= 0.2873.
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integration by parts
2. [§5-4] The probability density function of X, the lifetime of a certain type
of electronic device (measured in hours), is given by

 10 x > 10
f (x) = x2

0
x ≤ 10
(a) Find P(X > 20).
(b) What is the cumulative distribution function of X?
(c) What is the probability that, of 6 such types of devices, at least 3 will
function for at least 15 hours? What assumptions are you making?
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(a)
P(X > 20) =
Z
∞
20
(b) Since
Z
P(X ≤ x) =
x
10
∞
1
10 10
= = 0.5.
dx
=
−
x2
x 20 2
x
10
10
10 dt = − = 1 − ,
2
t
t 10
x
provided x > 10, the cumulative distribution function of X is given
by

1 − 10 x > 10
x
F (x) =

0
x ≤ 10
(c) For each device,
P(X ≥ 15) =
Z
∞
15
∞
2
10 10
dx = − = = 0.667.
2
x
x 15 3
Let Y be the number of devices which function for at least 15 hours
among those 6. If we assume that the devices works independently,
then Y ∼ Binomial(6, 2/3). Therefore
P(Y ≥ 3) = P(Y = 3) + P(Y = 4) + P(Y = 5) + P(Y = 6)
6−3 4 6−4
3 2
2
6
2
6
2
1−
1−
+
=
3
3
3
3
4
3
6−5 6 6−6
5 2
2
6
2
6
2
1−
1−
+
+
3
3
3
3
6
5
= 0.8999.
3. [§5-5] A filling station is supplied with gasoline once a week. If its weekly
volume of sales in thousands of gallons is a random variable with probability density function
(
5(1 − x)4 0 < x < 1
f (x) =
0
otherwise
what must the capacity of the tank be so that the probability of the
supplys being exhausted in a given week is .01?
Let c be the capacity of the tank and X be the weekly volume of sales.
Then the event the supplys being exhausted in a given week is same as
2
{X ≥ c}. Thus we need to find c such that P(X ≥ c) = 0.01. Since
f (x) = 0 whenever x ≥ 1, we may assume c < 1. Then
P(X ≥ c) =
Z
1
1
5(1 − x) dx = −(1 − x) = (1 − c)5 .
4
c
5
√
Therefore, c = 1 − 5 0.01 = 0.6019.
c
4. [§5-6] Compute E[X] if X has a density function given by

 5 x>5
f (x) = x2
.

0
x≤5
Z
∞
xf (x) dx =
−∞
∞
= 5 ln |x| = ∞.
E[X] =
Z
∞
5
x·
5
dx =
x2
Z
∞
5
5
dx
x
5
This is an example of infinite expected value. In this case, we say the
random variable does not have finite expectation.
5. [§5-12] A bus travels between the two cities A and B, which are 100 miles
apart. If the bus has a breakdown, the distance from the breakdown to
city A has a uniform distribution over (0, 100). There is a bus service
station in city A, in B, and in the center of the route between A and B.
It is suggested that it would be more efficient to have the three stations
located 25, 50, and 75 miles, respectively, from A. Do you agree? Why?
Let X denote the distance to city A when the bus breaks down and Y
denote the distance to the nearest service station. Then X ∼ U nif (0, 100).
If the three stations located 0, 50, and 100 miles, respectively, from A,
then


0 < X < 25
X
Y = g1 (X) = |X − 50| 25 ≤ X ≤ 75 .


100 − X 75 < X < 100
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Thus
E[Y ] = E[g1 (X)] =
100
Z
25
0
1
x
dx +
100
Z
75
|x − 50|
25
1
dx+
100
1
dx
(100 − x)
+
100
75
Z 25
Z 75
Z 50
1
(x − 50) dx+
=
(50 − x) dx +
x dx +
100 0
50
25
Z 100
(100 − x) dx
+
75
"
25 50 75
1
1 2
1
1 2 =
x + 50x − x2 +
x − 50x 100
2
2
2
0
25
50
100 #
1
+ 100x − x2 2
75
Z
= 12.50.
On the other hand, if the three stations located 25, 50, and 75 miles,
respectively, from A, then


|X − 25| 0 < X < 37.5
Y = g2 (X) = |X − 50| 37.5 ≤ X ≤ 62.6 .


|X − 75| 62.5 < X < 100
Thus
E[Y ] = E[g1 (X)] =
100
Z
37.5
0
|25 − x|
1
dx +
100
Z
62.5
37.5
|x − 50|
1
dx+
100
1
|x − 75|
+
dx
100
62.5
Z 25
Z 50
Z 37.5
1
(50 − x) dx
(x − 25) dx +
=
(25 − x) dx +
100 0
37.5
25
Z 100
Z 75
Z 62.5
(x − 75) dx
(75 − x) dx +
(x − 50) dx +
+
75
62.5
50
"
25 50
37.5 1 2
1 2 1 2 1
25x − x +
=
x − 25x + 50x − x 100
2
2
2
0
37.5
25
75
62.5 100 #
1 2
1
1 2
+
x − 50x + 75x − x2 +
x − 75x 2
2
2
62.5
50
75
Z
= 9.375.
Therefore, the second plan for the service locations is better.
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6. [§5-18] Suppose that X is a normal random variable with mean 5. If
P(X > 9) = .2, approximately what is Var[X]?
Let σ denote the standard deviation of X. Then
4
9−5
X −5
=P Z>
,
>
0.2 = P(X > 9) = P
σ
σ
σ
where Z is a standard normal random variable. From the standard normal
table, we have P(Z > 0.84) = 0.2 (The table actually gives you P(Z ≤
0.84) = 0.8). Therefore, σ is approximately 4/0.84 = 4.76 and the variance
is approximately 4.762 = 22.66.
7. [§5-22] The width of a slot of a duralumin forging is (in inches) normally
distributed with µ = .9000 and σ = .0030. The specification limits were
given as .9000 ± .0050.
(a) What percentage of forgings will be defective?
(b) What is the maximum allowable value of σ that will permit no more
than 1 in 100 defectives (meaning no more than 1%) when the widths
are normally distributed with µ = .9000 and σ?
(a) Let X be the width of a randomly selected forging. Then this forging
is defective if and only if X > 0.905 or X < 0.895. Thus
P({a randomly selected forging is defective})
= P(X > 0.905 or X < 0.895)
= P(X > 0.905) + P(X < 0.895)
X − 0.9
0.905 − 0.9
0.895 − 0.9
X − 0.9
+P
>
>
=P
0.003
0.003
0.003
0.003
= P(Z > 1.67) + P(Z < −1.67)
= 2P(Z < −1.67).
= 0.095.
Here Z is a standard normal random variable. Thus 9.5% of forgings
will be defective.
(b) Since
0.01 ≥ P({a randomly selected forging is defective})
= P(X > 0.905 or X < 0.895)
= P(X > 0.905) + P(X < 0.895)
X − 0.9
X − 0.9
0.905 − 0.9
0.895 − 0.9
=P
+P
>
>
σ
σ
σ
σ
= P(Z > 0.005σ −1 ) + P(Z < −0.005σ −1 )
= 2P(Z < −0.005σ −1 ),
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and P(Z < −2.58) = 0.0049 and P(Z < −2.57) = 0.0051, we have
σ ≤ 0.005/2.58 = 0.0019. Therefore, the maximum allowable value of
σ that will permit no more than 1 in 100 defectives (meaning no more
than 1%) when the widths are normally distributed with µ = .9000
and σ is 0.0019.
8. [§5-39] If X is an exponential random variable with parameter λ = 1,
compute the probability density function of the random variable Y defined
by Y = log X.
The distribution function of Y is
FY (y) = P(Y ≤ y) = P(log X ≤ y) = P(X ≤ ey )
ey
Z ey
−x
−x e dx = −e = 1 − exp(−ey ),
=
0
0
for all −∞ < y < ∞. Therefore, the pdf of Y is
fY (y) =
d
FY (y) = exp(−ey )ey
dy
− ∞ < y < ∞.
9. [§5-41] Find the distribution of R = A sin θ, where A is a fixed constant
and θ is uniformly distributed on (−π/2, π/2). Such a random variable R
arises in the theory of ballistics. If a projectile is fired from the origin at
an angle α from the earth with a speed v, then the point R at which it
returns to the earth can be expressed as R = (v 2 /g) sin 2α, where g is the
gravitational constant, equal to 980 centimeters per second squared.
The distribution function of R is
FR (r) = P(R ≤ r) = P(A sin θ ≤ r) = P(θ ≤ arcsin(A−1 r))
arcsin(A−1 r)
Z arcsin(A−1 r)
θ 1
dθ = =
π
π −π/2
−π/2
=
arcsin(A−1 r) 1
+ ,
π
2
for −A ≤ r ≤ A and FR (r) = 0 for r < −A and r > A. Therefore, the
pdf of R is

 √ 1
, −A ≤ r ≤ A
d
.
fY (y) =
FR (r) = Aπ 1 − A−2 r2

dr
0
otherwise
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