Name&Surname:
11 April 2016
STAT 240 SPRING 2016
APPLIED STATISTICS AND
PROBABILITY FOR ENGINEERS
Duration of this exam is 90 minutes. Turn off your cellular phones. All notes are closed.
Formulas are at the back side of this page.
1. (20 pts) A desk lamp produced by The Luminar Company was found to be defective (D). There
are three factories (A,B,C) where such desk lamps are manufactured. A Quality Control
Manager (QCM) is responsible for investigating the source of found defects. This is what
the QCM knows about the company's desk lamp production and the possible source of defects:
Factory
A
B
C
% of total production
0.35 = P(A)
0.35 = P(B)
0.30 = P(C)
Probability of defective lamps
0.015 = P(D|A)
0.010 = P(D|B)
0.020 = P(D|C)
The QCM would like to answer the following question: If a randomly selected lamp is
defective, what is the probability that the lamp was manufactured in factory C?
ANSWER:
( | ) is asked.
( | ) ( )
( )
( ) = ( | ) ( )+ ( | ) ( )+ ( | ) ( )
( ) = 0.015 ∗ 0.35 + 0.010 ∗ 0.35 + 0.020 ∗ 0.30
( ) = 0.01475
( | )=
Then,
0.02 ∗ 0.30
= 0.40678
0.01475
2. (20 pts) 2 fair dice are tossed. Let E be the event that “the first die is a 3”, F the
event that “the sum is 6” and G the event that “sum is 7”. Then show that E and F are
dependent, but E and G are independent.
( | )=
ANSWER: Independence is tested through the following equality. Two events A and B are
(
)= ( )∗ ( )
independent iff
= (3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
= (1,5), (2,4), (3,3), (4,2), (5,1)
= (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)
= ∩ = (3,3)
= ∩ = (3,4)
From above,
1
5
1
1
1
( )= , ( )=
, ( )= , ( ∩ )=
, ( ∩ )=
6
36
6
36
36
Therefore E and F are NOT independent but E and G are independent.
3. (15 pts) The following circuit operates only if there is a path of functional devices
from left to right. The probability that each device functions is shown on the graph.
Assume that devices fail independently. What is the probability that the circuit operates?
ANSWER:
P(Operates)=(1 − 0.1 ) ∗ (1 − 0.05 ) ∗ 0.99 = 0.986537
4. The phone lines to an airline reservation system are occupied 40% of the time. Assume
that the events that the lines are occupied on successive calls are independent. Assume
that 10 calls are placed to the airline.
a. (5 pts) What is the probability that for exactly three calls, the lines are
occupied?
b. (5 pts) What is the probability that for at least one call, the lines are not
occupied?
c. (5 pts) What is the expected number of calls in which the lines are all occupied?
ANSWER: This is a typical Binomial Distribution question where probability of occupied
0.40
10
a. ( = 3) =
0.4 0.6 = 120 ∗ 0.064 ∗ 0.027994 = 0.214991
3
10
b. 1 − ( = 0) = 1 −
0.6 0.4 = 1 − 0.000105 = 0.999895
0
c. ( ) =
= 10 ∗ 0.4 = 4
5. The average number of telephone calls per hour that arrive at a phone exchange is 10
calls.
a. (5 pts) What is the probability that there are exactly 5 calls in one hour?
b. (5 pts) What is the probability that there are 3 or fewer calls in one hour?
c. (5 pts) What is the probability that there are exactly 15 calls in two hours?
= 10 per hour.
ANSWER: This is a typical Poisson question where
(
)∗
a.
( = 5) =
b.
( = 0) + ( = 1) + ( = 2) + ( = 3) = 0.000045 + 0.00045 + 0.0022 + 0.0075 = 0.010336
c. Check that
= 0.037833
!
= 20 for 2 hours.
( = 15) =
(
)∗
!
= 0.051649
6. The waiting time for service at a hospital emergency department (in hours) follows a
distribution with probability density function f (x) = 0.5exp(−0.5x) for 0 < x. Determine
the following:
a. (5 pts) P(X < 0.5)
b. (5 pts) P(X > 2)
c. (5 pts) Value x (in hours) exceeded with probability 0.05.
ANSWER:
a.
( < 0.5) =
.
0.5exp(−0.5 )
b. ( > 2) = 1 − ( < 2) = 1 −
c. 1 − exp(−0.5 ) = 0.05
.
= exp(−0.5 ) |
0.5 exp(−0.5 )
= 1 − exp(−0.5 ∗ 0.5) = 0.221
= 1 − 0.6321 = 0.3678
0.95 = exp(−0.5 )
0.95 = −0.5
0.95
=
= 0.1025
−0.5
FORMULA SHEET

Bayes’ Formula
Pr( | ) =
Pr( | ) Pr( )
Σ Pr
Pr( )
Name&Surname:

11 April 2016
Binomial Distribution
Pr( = ) = ( ) =
(1 − )
=

=
( )=
=
(1 − )
Geometric Distribution
Pr( = ) = ( ) = (1 − )
=
−1
−1
1−
(1 − )
=
= , + 1, + 2, …
=
( )=
=
(1 − )
Hypergeometric Distribution
Pr( = ) = ( ) =

1
=
Negative Binomial Distribution
Pr( = ) = ( ) =

= 1, 2, 3, …
( )=
=

= 0,1, … ,
ℎ
= max(0, +
Poisson Distribution
Pr( = ) = ( ) =
!
= 0, 1, 2, …
− )
min( , )