This preview shows pages 1–3. Sign up to view the full content.
Problem Solutions for Required Homework #8
for ASW ESBE (5e) Chapter 6, Section 6.2
Required Homework #8:
Problems 45 and 47, in the endof chapter
Supplementary Exercises on pages 251 to 252.
Problem 6.45.
Let
x
be the continuous r.v. of the number of fatal crashes per year.
Given:
μ
= 1550 accidents/year
σ
= 300 accidents/year
x
is normal.
a. Want prob(
x
< 1000)
For
x
= 1000,
1000 1550
1.83
300
x
z
μ
σ


=
=
= 
From negative
z
table, down the
z
column to 1.8, across the row to under the
column for .03; the table entry for 1.83 is 0.0336.
prob(
x
< 1000) = prob(
z
< 1.83)
(by equivalence)
= 0.0336
b. Want prob(1000 <
x
< 2000)
For
x
=
2000
2000 1550
1.50
300
z

=
=
At
x
= 1000,
z
= 1.83, as already calculated in part a above.
Then
prob(1000 <
x
< 2000) = prob(1.83 <
z
<1.50) (by equivalence)
=
prob
(
z
<1.50) –
prob(
z
< 1.83)
ASW 5e Req HW #8 for section 6.2
1
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document = 0.9332 –
0.0336
= 0.8996
c. Find the
x
value cutting off 5% or 0.05 of the total area in the upper tail.
In
other words, the upper tail area is 0.05 for what
x
value?
Denote this special
x
value
x
*.
The procedure is to calculate the area to the right of the mean, find the
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 02/19/2011 for the course OM 210 taught by Professor Singer during the Spring '08 term at George Mason.
 Spring '08
 SINGER

Click to edit the document details