Chapter 6
Continuous Probability
Distributions
Solutions:
1.
a.
3
2
1
.50
1.0
1.5
2.0
f
(
x
)
x
b.
P
(
x
=
1.25)
=
0.
The probability of any single point is zero since
the area under the curve above any single point is zero.
c.
P
(1.0
≤
x
≤
1.25)
=
2(.25)
=
.50
d.
P
(1.20
<
x
<
1.5)
=
2(.30)
=
.60
2.
a.
.15
.10
.05
10
20
30
40
f
(
x
)
x
0
b.
P
(
x
<
15)
=
.10(5)
=
.50
c.
P
(12
≤
x
≤
18)
=
.10(6)
=
.60
d.
10
20
( )
15
2
E x
+
=
=
e.
2
(20
10)
Var( )
8.33
12
x

=
=
5.
a.
Length of Interval = 310.6  284.7 = 25.9
1
for 284.7
310.6
( )
25.9
0
elsewhere
x
f x
≤
≤
=
b.
Note: 1/25.9 = .0386
P(x
< 290) = .0386(290  284.7) = .2046
c.
P
(
x
≥
300)
=
.0386(310.6  300) = .4092
d.
P
(290
≤
x
≤
305) = .0386(305  290) = .5790
e.
P
(
x
≥
290)
=
.0386(310.6  290) = .7952
Rounding up, we conclude that 80 of the top 100 golfers drive the
ball this far.
6.
a.
P
(12
≤
x
≤
12.05)
=
.05(8)
=
.40
b.
P
(
x
≥
12.02)
=
.08(8)
=
.64
c.
(
11.98)
(
12.02)
.005(8)
.04
.64
.08(8)
P x
P x
<
+
=
=
1 442 4 43
1 442 4 43
Therefore, the probability is .04 + .64
=
.68
17.
Let
x
= debt amount
μ
= 15,015,
σ
= 3540
a.
18,000
15,015
.84
3540
z

=
=
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Chapter 6
P
(
x
> 18,000) = 1
P
(
z
≤
.84) = 1  .7995 = .2005
b.
10,000
15,015
1.42
3540
z

=
= 
P
(
x
< 10,000) =
P
(
z
< 1.42) = .0778
c.
At 18,000,
z
= .84
from part (a)
At 12,000,
12,000
15,015
.85
3540
z

=
= 
P
(12,000 <
x
< 18,000) =
P
(.85 <
z
< .84) = .7995  .1977 = .6018
d.
14,000
15,015
.29
3540
z

=
= 
P
(
x
≤
14,000) =
P
(
z
≤
.29) = .3859
19.
We have
μ
= 3.5 and
σ
= .8.
a.
5.0
3.5
1.88
.8
z

=
≈
P
(
x
> 5.0)
=
P
(
z
> 1.88) = 1 
P
(
z
≤
1.88) = 1  .9699 = .0301
The rainfall exceeds 5 inches in 3.01% of the Aprils.
b.
3
3.5
.63
.8
z

=
≈ 
P
(
x
< 3.0) =
P
(
z
< .63) = .2643
The rainfall is less than 3 inches in 26.43% of the Aprils.
c.
z
=
1.28 cuts off approximately .10 in the upper tail of a normal
distribution.
x
=
3.5 + 1.28(.8) = 4.524
If it rains 4.524 inches or more, April will be classified as extremely
wet.
20.
μ
= 77 and
σ
=
20
a.
At
x
= 50,
50
77
1.35
20
z

=
= 
P
(
z
< 1.35) = .0885
So,
P
(
x
< 50) = .0885
b.
At
x
= 100,
100
77
1.15
20
z

=
=
P
(
z
> 1.15) = 1  .8749 = .1251
So,
P
(
x
> 100) = .1251
12.51% of workers logged on over 100 hours.
c.
A
z
value of .84 cuts off an area of approximately .20 in the upper
tail.
x
=
μ
+
z
σ
= 77 + 20(.84) = 93.8
A worker must spend 93.8 or more hours logged on to be classified a
heavy user.
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 Fall '10
 oba
 Statistics, Normal Distribution, Probability, µ, continuous probability distributions

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