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Solutions for Recitation Problems from Section 6.2
Section 6.2 for the normal distribution: Problems 18, 20, 23
18. Let
x
be the continuous r.v. of the price per share for companies on the
Given:
μ
= $30.00
σ
= $8.20
x
is normal
a.
Want
Prob
(
x
≥ $40)
At
x
= 40,
40 30
1.22
8.2
z

=
=
By equivalence,
Prob
(
x
≥ $40) =
Prob
(
z
≥ 1.22)
From
z
table, down the leftmost “z” column to “1.2,”
across the “1.2” row until under the column “.02,”
the table entry for “1.22” is 0.8888.
So,
Prob
(
z
≤
1.22) = 0.8888
But want
Prob
(
z
≥ 1.22)
The area under the entire normal is 1.
Also,
Prob
(
z
≥ 1.22) and
Prob
(
z
≤ 1.22) are “contiguous” areas, meaning that they adjoin each
other without overlap or gap.
Then
Prob
(
z
≤ 1.22) +
Prob
(
z
≥ 1.22) = 1
Then
Prob
(
z
≥ 1.22) = 1 –
Prob
(
z
≤ 1.22)
= 1 – 0.8888
= 0.1112
Returning to the original r.v.,
Section 6.2 recitation problem solutions
1
© Harvey Singer 2009
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View Full DocumentProb
(
x
≥ 40) = 0.1112
b. Want
Prob
(
x
≤ $20)
At
x
= 20,
20 30
1.22
8.2
z

=
= 
By equivalence,
Prob
(
x
≤ $20) =
Prob
(
z
≤ 1.22)
From negative
z
table, down the leftmost “z” column to “1.2,”
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 Fall '08
 SINGER

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