This preview shows pages 1–3. Sign up to view the full content.
Chapter 3: Discrete Random Variables and Their Probability Distributions
35
Instructor’s Solutions Manual
3.33
a.
.
)
(
)
(
)
(
)
(
b
a
b
Y
aE
b
E
aY
E
b
aY
E
+
=
+
=
+
=
+
µ
b.
.
]
)
[(
]
)
[(
]
)
[(
)
(
2
2
2
2
2
2
σ
a
Y
E
a
a
aY
E
b
a
b
aY
E
b
aY
V
=
−
=
−
=
−
−
+
=
+
3.34
The mean cost is
E
(10
Y
) = 10
E
(
Y
) = 10[0(.1) + 1(.5) + 2(.4)] = $13.
Since
V
(
Y
) .41,
V
(10
Y
) = 100
V
(
Y
) = 100(.41) = 41.
3.35
With
FS
SS
B
∪
=
,
P
(
B
) =
P
(
SS
) +
P
(
FS
) =
( )
( )
4999
2000
5000
3000
4999
1999
5000
2000
+
= 0.4
P
(
B
first trial success) =
4999
1999
= 0.3999, which is not very different from the above.
3.36
a.
The random variable
Y
does not have a binomial distribution.
The days are not
independent.
b.
This is not a binomial experiment.
The number of trials is not fixed.
3.37
a.
Not a binomial random variable.
b.
Not a binomial random variable.
c.
Binomial with
n
= 100,
p
= proportion of high school students who scored above 1026.
d.
Not a binomial random variable (not discrete).
e.
Not binomial, since the sample was not selected among all female HS grads.
3.38
Note that
Y
is binomial with
n
= 4,
p
= 1/3 =
P
(judge chooses formula
B
).
a.
p
(
y
) =
()
y
y
y
−
4
3
2
3
1
4
,
y
= 0, 1, 2, 3, 4.
b.
P
(
Y
≥
3) =
p
(3) +
p
(4) = 8/81 + 1/81 = 9/81 = 1/9.
c.
E
(
Y
) = 4(1/3) = 4/3.
d.
V
(
Y
) = 4(1/3)(2/3) = 8/9
3.39
Let
Y
= # of components failing in less than 1000 hours.
Then,
Y
is binomial with
n
= 4
and
p
= .2.
a.
P
(
Y
= 2) =
2
2
)
8
(.
2
.
2
4
= 0.1536.
b.
The system will operate if 0, 1, or 2 components fail in less than 1000 hours.
So,
P(system operates) = .4096 + .4096 + .1536 = .9728.
3.40
Let
Y
= # that recover from stomach disease.
Then,
Y
is binomial with
n
= 20 and
p
= .8.
To find these probabilities, Table 1 in Appendix III will be used.
a.
P
(
Y
≥
10) = 1 –
P
(
Y
≤
9) = 1 – .001 = .999.
b.
P
(14
≤
Y
≤
18) =
P
(
Y
≤
18) –
P
(
Y
≤
13) – .931 – .087 = .844
c.
P
(
Y
≤
16) = .589.
3.41
Let
Y
= # of correct answers.
Then,
Y
is binomial with
n
= 15 and
p
= .2.
Using Table 1
in Appendix III,
P
(
Y
≥
10) = 1 –
P
(
Y
≤
9) = 1 – 1.000 = 0.000 (to three decimal places).
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document36
Chapter 3: Discrete Random Variables and Their Probability Distributions
Instructor’s Solutions Manual
3.42
a.
If one answer can be eliminated on every problem, then,
Y
is binomial with
n
= 15 and
p
= .25.
Then,
P
(
Y
≥
10) = 1 –
P
(
Y
≤
9) = 1 – 1.000 = 0.000 (to three decimal places).
b.
If two answers can be (correctly) eliminated on every problem, then,
Y
is binomial
with
n
= 15 and
p
= 1/3.
Then,
P
(
Y
≥
10) = 1 –
P
(
Y
≤
9) = 0.0085.
3.43
Let
Y
= # of qualifying subscribers.
Then,
Y
is binomial with
n
= 5 and
p
= .7.
a.
P
(
Y
= 5) = .7
5
= .1681
b.
P
(
Y
≥
4) =
P
(
Y
= 4) +
P
(
Y
= 5) = 5(.7
4
)(.3) + .7
5
= .3601 + .1681 = 0.5282.
3.44
Let
Y
= # of successful operations.
Then
Y
is binomial with
n
= 5.
a.
With
p
= .8,
P
(
Y
= 5) = .8
5
= 0.328.
b.
With
p
= .6,
P
(
Y
= 4) = 5(.6
4
)(.4) = 0.259.
c.
With
p
= .3,
P
(
Y
< 2) =
P
(
Y
= 1) +
P
(
Y
= 0) = 0.528.
3.45
Note that
Y
is binomial with
n
= 3 and
p
= .8.
The alarm will function if
Y
= 1, 2, or 3.
Thus,
P
(
Y
≥
1) = 1 –
P
(
Y
= 0) = 1 – .008 = 0.992.
3.46
When
p
= .5, the distribution is symmetric.
When
p
< .5, the distribution is skewed to the
left.
When
p
> .5, the distribution is skewed to the right.
3.47
The graph is above.
0
5
10
15
20
0.00
0.05
0.10
0.15
y
p(y)
3.48
a.
Let
Y
= # of sets that detect the missile.
Then,
Y
has a binomial distribution with
n
= 5
and
p
= .9.
Then,
P
(
Y
= 4) = 5(.9)
4
(.1) = 0.32805 and
P
(
Y
≥
1) = 1 –
P
(
Y
= 0) = 1 – 5(.9)
4
(.1) = 0.32805.
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '10
 yu

Click to edit the document details