The University of Hong Kong
Department of Statistics and Actuarial Science
STAT1801 Probability and Statistics: Foundations of Actuarial Science (1011)
Example Class 6
1.
Customers arrive at a travel agency at a mean rate of 9 per hour in accordance with a
Poisson process. Suppose the travel agency opens at 9:00 am.
a.
Find the probability that no customers arrived before 9:30 am.
Let
𝑋𝑋
be the number of customers arrived from 9:00 to 9:30 pm.
Then
𝑋𝑋
~
𝑃𝑃𝑃𝑃𝑃𝑃
(4.5)
[Since the period is half an hour.]
𝑃𝑃
(
𝑋𝑋
= 0) =
𝑒𝑒
−
4.5
= 0.0111
b.
Find the probability that more than five customers arrived before 9:30 am.
𝑃𝑃
(
𝑋𝑋
> 5) = 1
− 𝑃𝑃
(
𝑋𝑋 ≤
5)
= 1
− �
𝑒𝑒
−
4.5
4.5
𝑥𝑥
𝑥𝑥
!
5
𝑥𝑥
=0
= 1
− 𝑒𝑒
−
4.5
�
1 + 4.5 +
4.5
2
2!
+
⋯
+
4.5
5
5!
�
= 0.2971
c.
Find the probability that there were more than five customers arrived before 9:30
am, given that at least one customers arrived within that period.
𝑃𝑃
(
𝑋𝑋
> 5
𝑋𝑋 ≥
1) =
𝑃𝑃
(
𝑋𝑋
> 5
∩ 𝑋𝑋 ≥
1)
𝑃𝑃
(
𝑋𝑋 ≥
1)
=
𝑃𝑃
(
𝑋𝑋
> 5)
1
− 𝑃𝑃
(
𝑋𝑋
< 1)
=
0.2971
1
−
0.0111
= 0.3004
2.
Let
𝑍𝑍
~
𝑁𝑁
(0,1)
and
𝑋𝑋
~
𝑁𝑁
(80,100)
. Find the following probabilities,
𝑥𝑥
or
𝑧𝑧
.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '10
 MrChung
 Actuarial Science, Statistics, Normal Distribution, Probability, Standard Deviation, Poisson process, travel agency

Click to edit the document details