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STAT 350 – Fall 2008
Homework #4 – Solution
Corrected
through Lecture #8
Text Exercises
Chapter 5, Problems #18, 20, 46b&c, 50, 52, 54
5.18
Let A
1
denote the event that person 1 has blood type A (etc.)
a.
P(A
1
and A
2
) = P(A
1
)P(A
2
) = (0.42)(0.42) = 0.1764
b.
P(B
1
and B
2
) = 0.10
2
= 0.01
P(AB
1
and AB
2
) = 0.04
2
= 0.0016
P(O
1
and O
2
) = 0.44
2
= 0.1936
c.
P(two randomly chosen people have the same blood type)
= P(A
1
and A
2
) + P(B
1
and B
2
) + P(AB
1
and AB
2
) + P(O
1
and O
2
) = 0.3816
d.
This is the complement of the event in (c):
Answer = 1 – 0.3816 = 0.6184
5.20
Let A = the event that the top subsystem works, B = the event the bottom subsystem
works
P(A) = P(1 works OR 2 works) = P(1 works) + P(2 works) – P(1 works and 2 works) =
0.9 + 0.9 – 0.9
2
= 0.99
P(B) = P(3 works AND 4 works) = 0.9
2
= 0.81
P(system works) = P(A OR B) = P(A) + P(B) – P(A and B) = P(A) + P(B) – P(A)P(B) =
0.99 + 0.81 – (0.99)(0.81) = 0.9981
5.46
Let X = the inside diameter of a randomly selected piston ring in cm.
E(X) = 12, SD(X) = 0.04
b.
() ()
12
EX EX
==
()
0.04
0.005
64
SD X
SD X
n
=
c.
The mean of a random sample of size 64
Homework #4 – Solution
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View Full Document5.50
a.
() ()
50
EX EX
==
,
()
( )
1
0.1
100
SD X
SD X
n
=
49.75 50
50
50.25 50
49.75
50.25
0.1
0.1
0.1
2.5
2.5
(2.5)
( 2.5)
0.9938 0.0062
0.9876
X
PX
P
PZ
⎛⎞
−−
−
<<
=
<
<
⎜⎟
⎝⎠
=−<
<
=
Φ
−
Φ
−
=
−
=
b.
49.8
,
( )
1
0.1
100
SD X
SD X
n
=
49.75 49.8
49.8
50.25 49.8
49.75
50.25
0.1
0.1
0.1
0.5
4.5
(4.5)
( 0.5) 1 0.3085
0.6915
X
P
⎛
−
=
<
<
⎜
⎝
<
=
Φ
−
Φ
−
=
−
=
⎞
⎟
⎠
5.52
a.
50
,
( )
1.5
0.5
9
SD X
SD X
n
=
50
52 50
52
4
0
0.5
0.5
X
P
≥=
≥
=
≥
≈
On the standard normal table, the best you can say is the probability is approx. zero,
the actual probability is about 0.000032
b.
50
,
( )
1.5
0.23717
40
SD X
SD X
n
=
50
52 50
52
8.43
0
1.5
40
1.5
40
X
P
≥
=
≥
≈
Even statistical packages will return this probability as 0.
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 Fall '08
 Staff

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