M3 Exam
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School
Portage Learning *
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Course
110
Subject
Industrial Engineering
Date
Jan 9, 2024
Type
Pages
9
Uploaded by MinisterOctopusPerson886
M3:
Exam
-
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Due
No
due
date
Points
50
Questions
7
Time
Limit
90
Minutes
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Respondus
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Instructions
You
may
only
have
the
following
items
when
taking
an
exam:
computer,
1-2
pieces
of
blank
scratch
paper,
a
pen/pencil,
and
a
calculator.
You
may
ONLY
use
the
equation
sheets
that
are
provided
WITHIN
the
exam.
The
use
of
printed
versions
will
be
considered
a
violation
of
the
Academic
Integrity
Policy.
Attempt
History
Attempt
Time
Score
LATEST
Attempt
1
76
minutes
44
out
of
50
Score
for
this
quiz:
44
out
of
50
Submitted
Aug
8
at
5:58pm
This
attempt
took
76
minutes.
Question
1
513
pts
You
may
find
the
following
files
helpful
throughout
the
exam:
Statistics_Equation_Sheet
&
Find
the
answer
to
each
of
the
following
by
first
reducing
the
fractions
as
much
as
possible:
a)
P(490,2)=
b)
C(670,665)=
Your
Answer:
.
490!
490!
—
P
(490,2)
=
90
=
{38
=
490
(489)
=
239,
610
a)
b)
_
670!
__
670!
__
670(669)(668)(667)(666)
‘0(670’
665)
o
665!(670—665)!
~
670'5!
5(4)(3)(2)(1)
=
1.1(
4
G
4
4
D
4
a.
490!
490!
P(490,2)
=
0o
agal
490(489)
=
239,610.
b.
670!
670!
70(669)(668)(667)(666)
€(670,665)
=
665!(670-665)!
~
eesIs!
4
5(4)6(:)(2;3(1)
=
1.108
x
10
Question
2
5/5pts
You
may
find
the
following
files
helpful
throughout
the
exam:
Statistics_Equation_Sheet
&
(https://previous.nursingabc.com/upload/images/Help_file
picture/Statistics
Suppose
you
are
going
to
make
a
password
that
consists
of
6
characters
chosen
from
{3,4,5,7,d,f,g,k,q,r,s,w}.
How
many
different
passwords
can
you
make
if
you
cannot
use
any
character
more
than
once
in
each
password?
Your
Answer:
P(n,r)
=
(12"75)
=
12
_
12(11)
(10)
(9)
(8)(7)
=
665,
280
P(12,6)
=
(1213!6)!
=
16—2!'
=12(11)(10)(9)(8)(7)
=
665,280.
4
G
4
Question
3
6/9
pts
You
may
find
the
following
files
helpful
throughout
the
exam:
Statistics
_Equation_Sheet
=
(https://previous.nursingabc.com/upload/images/Help_file
picture/Statistics
Suppose
A
and
B
are
two
events
with
probabilities:
P(A)=.35,P(BC
)=.45,P(ANB)=.25.
Find
the
following:
a)
P(AuB).
b)
P(AS).
c)
P(B).
4
G
4
Your
Answer:
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a.For|[P(AUB)|use|
P(ANB)=P(A)+P(B)—
P(AUB)
and
rearrange
to|P
(AU
B)
=
P(A)
+
P(B)
—
P(AN
B)
P(A)=1-P(AY|=[P(4)=1—
.35
=
.65
P(AUB)
=
.65+
.45
—
.25
=
.85
b.
P(A)
found
above
as
.65
c.
ForlP(B)
=1
—
P(B°)|rearranged
to
|(B°)
=1
—
P
(B)
P(B°)=1-—
.45
=
55
a.
For
P(AUB).
Use
P(AuB)=P(A)+P(B)-P(ANB)
.
But
for
this
equation,
we
need
P(B)
which
we
can
find
by
using
P(B)=1-
P(B°
).
So,
P(B)=1-.45=.55.
P(AUB)=.35+.55-.25=.65
b.
For
P(A°).
Use
P(A)=1-P(A°
)
which
may
be
rearranged
to
(A°
)=1-P(A)
.
P(A®
)=1-.35=.65.
C.
For
P(B).
Use
(B)=1-P(B®).
P(B)=1-.45=.55.
Question
4
3/6
pts
You
may
find
the
following
files
helpful
throughout
the
exam:
Statistics_Equation_Sheet
&
Suppose
A
and
B
are
two
events
with
probabilities:
P(A°
)=.45P(B)=.45,P(ANB)=.20.
a)
What
is
(A|
B)
?
b)
What
is
(B|
A)
?
Your
Answer:
P(A°
)=.45
P(B)=.45,P(ANB)=.20.
P(ANB
a)
P(AB)=
;(;))
a.
For
P(AIB)
use
P
B
20
P(A|B)
=
g:g))
P(A|B)
=
=2
=
444.
b.
For
P(Bl4)
use
P(B|A)
=&
;‘22;’)
P(B|A)
=
z—g
—
3636.
Question
5
10/10
pts
You
may
find
the
following
files
helpful
throughout
the
exam:
Statistics_Equation_Sheet
&
(https://previous.nursingabc.com/upload/images/Help_file
picture/Statistics
Your preview ends here
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- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
In
a
tri-state
conference,
20%
attendees
are
from
California,
45%
from
Oregon,
and
35%
from
Washington.
As
it
turns
out
15
%
of
the
attendees
from
California,
5
%
of
the
attendees
from
Oregon,
and
19%
of
the
attendees
from
Washington
came
to
the
conference
by
train.
If
an
attendee
is
selected
at
random
and
found
to
have
arrived
by
train,
what
is
the
probability
that
the
person
is
from
Oregon?
Your
Answer:
|
P(O)-P(TRAIN|O))
‘P
(O|TRAIN
))H
20415745+
05735119
—
"109
4
G
4
P(Train
|
C)=.15..
P(Train|
0)=.05..
P(Train|
W)=.19..
P(C)=.20,P(0)=.45,P(W)=.35.
We
want
to
find
P(O|
Train),
so
use:
P(0)
»
P(Train|0)
FARa)=
(P(C)
«
P(Train|C))
+
(P(0)
»
P(Train|0))
+
P(W)
+
P(Train|W)
P(O|Train)
=
.45
+
.05
-
9
(OlTrain)
=
ST
45+
05+
35
19
5%
Question
6
10710
pts
You
may
find
the
following
files
helpful
throughout
the
exam:
Statistics_Equation_Sheet
&
(https://previous.nursingabc.com/upload/images/Help_file_picture/Statistics
The
probability
that
a
certain
type
of
battery
in
a
smoke
alarm
will
last
2
years
or
more
is
.85.
The
probability
that
a
battery
will
last
5
years
or
more
is
.15.
Suppose
that
the
battery
is
2
years
old
and
is
still
working,
what
is
the
probability
that
the
battery
will
last
at
least
5
years?
Your
Answer:
E=2
years
or
more
F=
5
years
or
more
IP(ENF)
=15
P
PENR)
_
a5
:
~
P(E)
.8
Define
E
to
be
the
event
that
the
battery
will
last
2
years
or
more.
Define
F
to
be
the
event
that
the
battery
will
last
5
years
or
more.
We
are
given
that:
P(E)=.85.
P(F)=.15.
Note
also
that
if
a
battery
lasts
more
than
5
years,
it
would
have
had
to
have
lasted
2
years,
so
P(ENF)=.15.
The
question
is
asking
“given
event
E
has
occurred,
what
is
the
probability
that
event
F
will
occur”?
This
may
be
found
by
using:
__
P(EnF)
15
P(F|E)
=
'
.1765.
Question
7
513
pts
You
may
find
the
following
files
helpful
throughout
the
exam:
Statistics_Equation_Sheet
&
(https://iprevious.nursingabc.com/upload/images/Help_file_
picture/Statistics
Suppose
that
7
out
of
17
people
are
to
be
chosen
to
go
on
a
mission
trip.
In
how
many
ways
can
these
7
be
chosen
if
the
order
in
which
they
are
chosen
is
not
important.
Your
Answer:
Since
we
do
not
want
to
count
all
of
the
possible
orderings,
we
use
combinations.
17!
17!
(17)(16)(15)(14)(13)(12)(11)
7MQA7-7)0
7110l
71
=
19448,
C(7.17)
=
Quiz
Score:
44
out
of
50
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