M6 Problem Set
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Course
110
Subject
Mathematics
Date
Jan 9, 2024
Type
Pages
5
Uploaded by ChefOwl3912
M6:
Problem
Set
Due
No
due
date
Points
5
Questions
4
Time
Limit
None
Attempt
History
Attempt
Time
Score
LATEST
Attempt
1
120
minutes
5
out
of
5
Score
for
this
quiz:
5
out
of
5
Submitted
Aug
26
at
3:41pm
This
attempt
took
120
minutes.
Question
1
0/0pts
A
company
manufactures
a
certain
over-the-counter
drug.
The
company
samples
80
pills
and
finds
that
the
mean
amount
of
drug
in
the
pills
is
325.5
mg
with
a
standard
deviation
of
10.3
mg.
Find
the
90%
confidence
interval
for
the
mean
of
all
the
pills.
Your
Answer:
n=80
|z
=325.5]
s=103
The
formula
for
confidence
interval
is:
T=z2—=<pu<xT+2—=
n
n
NG
NG
For
90%
confidence
interval,
z
=
1.645.
10.3
10.3
325.5
—
1.645fi
<
p
<
325.5
+
1.645fi
323.61
<
p
<
327.39
We
are
not
told
the
number
of
pills
manufactured
but
we
may
assume
it
is
a
very
large
number.
Since
the
sample
size
is
greater
than
30,
we
should
use
Case
1:
Very
large
population
and
very
large
sample
size.
We
are
given
the
sample
mean
and
sample
standard
deviation.
So,
we
have
n=80
X=3255
s=103
We
will
use
these
values
in
the
equation:
_
s
_
s
e
Soad
el
SO
X
AR
G
Vn
vn
For
a
90%
confidence
level,
we
look
at
table
6.1
and
find
that
z=
1.645.
When
we
substitute
these
values
into
our
equation,
we
get:
10.3
10.3
3255—1645
—
<
u<
3255+
1.645
—
Jao
"
V80
When
we
do
the
arithmetic
on
the
right
and
left
hand
side,
we
get:
32361
<
u<
327.39.
Question
2
0/0
pts
A
total
of
200
rods
have
been
manufactured.
You
sample
50
of
the
rods
and
find
that
they
have
a
mean
length
of
15.5
inches
with
a
standard
deviation
of
1.73
inches.
Find
the
80
%
confidence
limit
for
the
mean
length
for
all
of
the
rods.
Your
Answer:
N—n
N-1
N—n
N-1
w—z—
<,u<:c—|—z
N=200
n=50
|x
=
15.5|
s=1.73
For
80%
confidence
z=1.28
200—50
200-1
200—50
1.73
00
<
<155
+1.28L8
iV
|155—128173\/
=(15.23
<
p
<
15.77|
The
population
is
finite.
So,
we
should
use
Case
3:
Finite
population.
Use:
In
the
statement
of
the
problem,
we
are
given:
N
=200
n=50
£=155
=173
For
a
80%
confidence
level,
table
6.1
gives
z=1.28
ek
_qopll3
[200-50
...
.
173
[200-50
W
e
J200-1
S
T
RERTES
1200
-1
1523
<
u<
15.77.
Question
3
0/0
pts
Suppose
that
you
are
a
nurse
and
you
are
assigned
to
do
checkups
of
people
one
day
per
week
in
a
certain
village.
You
have
a
total
of
300
patients
in
the
village.
You
have
the
option
of
doing
the
checkups
in
the
mornings
or
in
the
afternoons.
Therefore,
you
ask
35
patients
and
find
that
62%
prefer
afternoon
appointments
while
38%
prefer
morning
appointments.
Find
the
95%
confidence
limit
for
the
proportion
of
all
patients
that
prefer
afternoon
appointments.
Your
Answer:
p=.62
n=35
z=1.96
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