CEE 304 – Section 4 Solutions
1.
Transforming Variables Example
:
( )
x
X
f
x
e

=
for
x
> 0,
and 0 otherwise
Develop the density function for Y = X
½
Ans
. Y = g(X) = X
½
We have a onetoone relationship between X and Y, so we can use the following
formula to find the pdf of Y given the pdf of X:
( )
( )
Y
X
dx
f
y
f
x
dy
=
This is equivalent to:
1
1
( )
( )
(
( ))
Y
X
dg
y
f
y
f
g
y
dy


=
2
( )
(2 )
y
Y
f
y
e
y

∴
=
for y > 0,
and 0 otherwise
2. An engineering student got a summer job setting off fireworks on the 4
th
of July and other
summer events. For a particular display, the bomb is shot from the tube so that after
t
seconds
it is at height:
H = (100 m/sec) t – 0.5 (10 m/sec
2
) t
2
=
500 – 5(10 – t)
2
meters
Unfortunately the timers on the bombs are not accurate. The student estimates that the timer
will yield a time T to detonation after launch that is normally distributed with mean 10 seconds
and standard deviation 1.2 seconds.
T ~ N(
μ
= 10 sec,
σ
2
= (1.2 sec)
2
)
a.) What is the probability that the bomb explodes more than 495 meters above the ground?
]
495
)
10
(
5
500
[
]
495
[
2
≥


=
≥
t
P
H
P
= 0.5934
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View Full Documentb.) Above what height can you be 99% confident the bomb will explode?
99
.
0
]
[
=
≥
h
H
P
99
.
0
]
)
10
(
5
500
[
2
=
≥


⇒
h
t
P
Ans
. h = 452.26 m
c.) What is the pdf for the height at which the bomb explodes?
[We can easily derive the cdf for the height of explosion F
H
(h), but deriving the pdf f
H
(h) is
made difficult by the need to evaluate d
Φ
(h)/dh. When originally asked on an exam, this
question asked to outline the steps required to to obtain the pdf – it did not ask to do the
actual derivation. Here we will derive the cdf and then note how to obtain the pdf.]
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 Fall '08
 Stedinger
 Normal Distribution, Variance, Probability theory, Exponential distribution, additional waiting time

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