Math 361/Stat 351 X1
Homework 10 Solutions
Spring 2003
Graded problems:
1(d), 3(c), 4(b), 7
Problem 1.
[4.2.4] Suppose component lifetimes are exponentially distributed with mean 10 hours. Find (a) the
probability that a component survives 20 hours; (b) the
median
component lifetime (see p. 165 for
a definition of the median of a distribution); (c) the SD of component lifetime; (d) the probability
that the average lifetime of 100 independent components exceeds 11 hours.
Solution.
(a) We need to compute
P
(
X
≥
20), where
X
is the life time of the component.
We are given
that
X
has exponential distribution with mean
E
(
X
) = 10 (measured in hours).
Since for an
exponential(
λ
) distribution
E
(
X
) = 1
/λ
, we must have
λ
= 1
/E
(
X
) = 1
/
10. Thus
P
(
X >
20) =
1

P
(
X <
20) = 1

F
(20) = 1

(1

e

(1
/
10)
·
20
) =
e

2
, using the formula
F
(
x
) = 1

e

λx
(
x >
0) for the c.d.f. of an exponential distribution.
(b) The median lifetime is the value of
t
for which
P
(
X
≥
t
) = 1
/
2 =
P
(
X
≤
t
). Now, by the same
argument as in (a), we have for
t
≥
0
P
(
X
≥
t
) =
e

5
/
10
, and setting this value equal to 1
/
2 we
get
e

t/
10
= 1
/
2 or
t
= 10 ln 2 = 6
.
93.
(c) Since the lifetime
X
is exponentially distributed with parameter
λ
= 1
/
10, its standard deviation
is, by the formula for the variance of an exponential distribution,
SD
(
X
) =
p
Var(
X
) =
p
1
/λ
2
= 1
/λ
= 10
.
(d) Let
X
1
, X
2
, . . . , X
100
be the lifetimes of the 100 independent components. The random variables
X
i
are i.i.d., each having exponential distribution with
λ
= 1
/
10 and
μ
=
E
(
X
1
) = 10,
σ
=
SD
(
X
1
) = 10. We need to compute the probability
P
(
A
100
>
11), where
A
100
=
1
100
100
X
i
=1
X
i
=
1
100
S
100
is the average lifetime (in the “naive” sense of average) of the 100 components. By normal approx
imation this is
P
(
A
100
>
11) = 1

P
(
A
100
≤
11) = 1

P
(
S
100
≤
1100)
≈
1

Φ
1100

100
·
10
√
100
·
10
= 1

Φ(1) = 0
.
1587
.
Problem 2.
[4.R:4] Let
X
be a random variable with density
f
(
x
) = 0
.
5
e

x

(
∞
< x <
∞
).
Find (a)
P
(
X <
1), (b)
E
(
X
) and
SD
(
X
), (c) the c.d.f. of
X
2
.
Solution.
P
(
X <
1) =
Z
1
∞
f
(
x
)
dx
=
1
2
Z
1
∞
e

x

dx
=
1
2
Z
0
∞
e
x
dx
+
1
2
Z
1
0
e

x
dx
(a)
=
1
2
e
x
0
∞
+
1
2(

1)
e

x
1
0
=
1
2
+
1
2
(

e

1
+ 1
)
= 1

1
2
e
.
1
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E
(
X
) =
Z
∞
∞
xf
(
x
)
dx
=
Z
∞
∞
x
1
2
e

x

dx
= 0
(since integrand is an odd function)
(b)
E
(
X
2
) =
Z
∞
∞
x
2
f
(
x
)
dx
=
Z
∞
∞
x
2
1
2
e

x

dx
=
Z
∞
0
x
2
e

x
dx
(by symmetry)
= 2
(from integral table or by integration by parts)
Thus,
SD
(
X
) =
p
E
(
X
2
)

E
(
X
)
2
=
√
2

0
2
=
√
2.
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 Spring '03
 aldous
 Probability, Probability theory, Exponential distribution, WI, integrand, unit square

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