Transcript Probability Distributions

José Rizal University
Graduate School
Master in Public Administration Program
Exercises Report:
Probability Distributions
Presented to Mr. Rodolfo B. Gaerlan
Faculty Adviser
In Partial Fulfillment of
the Requirements of the Course
Subject: Statistics & Quantitative Analysis
By Group I:
John Ko, Amor Sande and Jesus Boadilla
2nd Trimester / School Year 2005-2006
Exercise 2-32
The mean output of cold-rolled steel pipe of the
Valley Tubing Corporation is 120,000 pounds
daily, with a standard deviation of 40,000 pounds.
Any output above 185,000 pounds daily would so
tax the materials-handling crew as to cause a
shutdown of the entire facility. What is the
probability of such an event if the output is
normally distributed?
Solution:
x-μ
z= σ =
185,000 – 120,000
40,000
= 1.625 standard
deviations
Check with the Standard Normal Probability
Distribution Table, we get the area from left end
to 1.625 standard deviations is 0.948
So the probability of shutdown event is:
1 – 0.948 = 0.052
Exercise
2-37
Lincoln County Construction Company
The
is
planning to place a bid on a construction contract.
After analyzing the work with a technique known as
program evaluation and review technique (PERT), it
has concluded that the construction project would
take 200 days to complete. This estimate carries with
it many uncertainties, however. When the
uncertainties are taken into consideration, the
company feels that the completion time is a random
variable, and its distribution approximates a normal
distribution having a mean of 200 days and a
standard deviation of 40 days. It is company policy
to quote in its bid a completion time which has a
95% probability of being met, to avoid late
completion penalties which are a provision of the
contract. What completion time should it quote in its
bid for this contract?
Solution:
Check with the Standard Normal Probability
Distribution Table, if a 95% probability of being met,
then the z value should be 1.65.
x-μ
z= σ → x=zσ+μ
→ x = 1.65 x 40 + 200 = 266 days
Exercise 2-42
The manager of the State Lottery Commission
plans to introduce a new game called Green-nGray. To play the game, a participant pays one
dollar and guesses the colors of five ping-pong
balls drawn sequentially and at random from a
very large container. Half the balls are green and
the other half are gray. In order to win, the
participant must correctly guess the colors of
each of the five balls in exactly the sequence they
are drawn. After a ball is drawn it is replaced, so
the probability remains at 50% that either green
or gray is drawn each time. The lottery
commission plans to pay about 60% of the
revenue from ticket sales. How much should a
winning ticket holder receive for a ticket costing
one dollar?
Solution:
Probability to win: 0.5 x 0.5 x 0.5 x 0.5 x 0.5 =
0.03125 So presumed ticket sales is 1 ÷ 0.0312 = 32,
then 60% of the revenue would be 32 x 0.6 = 19.2
dollars.
Therefore, a winning ticket holder for a ticket
costing one dollar should receive 19.2
dollars
Exercise 7-10
An airline claims that 95% of its flights arrive on
time. Use Table V to find the probabilities
that among 10 randomly selected flights
(a) At least 8 are on time
(b) From 7 to 9 are on time
(c) At most 9 are on time
Solution:
p = 0.95 n = 10
(a) x = 8 and more, then use table V, we find
0.075 + 0.315 + 0.599 = 0.989
(b) x = 7, 8, 9 then use table V, we find
0.010 + 0.075 + 0.315 = 0.400
(c) Since x = 1 to 9, then use table V, we find 1 – p(10)
1 – 0.599 = 0.401
Exercise 7-17
Find the probability that an Internal Revenue Service
auditor will get 3 tax returns with unallowable
deductions if she randomly selects 5 returns from
among 15 returns, 7 of which contain unallowable
deductions.
Solution:
a
b
x
n-x
p(x) = -----------------a+b
n
Using the above formula, we get:
7 8
p(3) =
3
2 = 35 x 28 =
15
3003
5
0.326