M4 Problem Set
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Course
110
Subject
Statistics
Date
Jan 9, 2024
Type
Pages
11
Uploaded by ERProfParrotPerson911
M5:
Problem
Set
Due
No
due
date
Points
5
Questions
7
Time
Limit
None
Attempt
History
Attempt
Time
Score
LATEST
Attempt
1
1,359
minutes
5outof
5
Score
for
this
quiz:
5
out
of
5
Submitted
Dec
10
at
5:10pm
This
attempt
took
1,359
minutes.
Question
1
0/0pts
Suppose
that
you
are
attempting
to
estimate
the
weight
of
600
parts.
In
order
to
use
the
infinite
standard
deviation
formula,
what
sample
size,
n,
should
you
use?
Your
Answer:
Sample
size
must
be
less
than
30
In
order
to
use
infinite
standard
deviation
formula,
we
must
have:
=
0.05
0
n
<
0.05(600)
n
<30
So,
the
sample
size
must
be less
than
30.
Question
2
0/0
pts
Suppose
that
you
take
a
sample
of
size
15
from
a
population
that
is
known
to
be
normally
distributed.
Can
the
sampling
distribution
of
X
be
approximated
by
a
normal
probability
distribution?
Your
Answer:
Yes,
the
sampling
distribution
of
X
can
be
approximated
by
a
normal
probability
distribution
because
the
population
is
normally
distributed.
Yes,
the
sampling
distribution
of
X
can
be
approximated
by
a
normal
probability
distribution
because
the
population
is
normally
distributed.
(Recall
that
if
the
population
is
normally
distributed,
the
sampling
distribution
of
X
can
be
approximated
by
a
normal
probability
distribution
even
for
very
small
sample
sizes.)
Question
3
0/0
pts
Suppose
that
you
take
a
sample
of
size
20
from
a
population
that
is
not
normally
distributed.
Can
the
sampling
distribution
of
X
be
approximated
by
a
normal
probability
distribution?
Your
Answer:
No,
the
sampling
distribution
of
X
cannot
be
approximated
by
a
normal
probability
distribution
in
this
case.
(Recall
that
if
the
population
is
not
normally
-
distributed,
the
sample
size
must
be
at
least
30.)
No,
the
sampling
distribution
of
X
cannot
be
approximated
by
a
normal
probability
distribution
in
this
case.
(Recall
that
if
the
population
is
not
normally
distributed,
the
sample
size
must
be
at
'
least
30.)
Question
4
0/0
pts
Suppose
a
pharmaceutical
company
wants
to
do
a
study
of
the
commissions
of
its
sales
force.
Let's
assume
that
there
are
4,300
sales
people
and
the
population
mean
for
the
sales
force
is
$52,400
in
commissions
and
has
a
population
standard
deviation
of
$3,500.
What
is
the
probability
that
a
simple
random
sample
of
50
members
of
the
sales
force
will
have
commissions
within
$400
of
the
population
mean?
Your
Answer:
o
3500
—
=
—
|
==
=
—|=494.97
9%
=~
&l
192
T
/m
.
__|
(52000—52400)
(52800—52400)
|
_
Z=(
X
-p)I0
3
=|
o7
ANA—5
o
=
-.808
and
.808
|P(—.808
<
Z
<
.808)
=
P(Z
<
.808)
—
P(Z
<
—.808)
~
.790.
=58.1%
4
C—
4
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We
calculate
the
standard
deviation
of
the
sample
distribution:
ou=
L
=
0
i
Ym0
We
now
have
the
necessary
information
needed
to
determine
the
probability
the
sample
mean
X
will
be
between
$52,000
and
|
$52,800
this
is
the
range
which
is
within
$400
of
the
population
mean
u
of
$52,400.)
Calculate
the
z-score:
X—p
Oz
zZ=
Using
this
formula,
we
will
calculate
the
two
z-scores
that
we
will
use
to
answer
our
question.
_X—p
52,000—-52,400
—400
_
T
ooz
494,97
T
49497
z
—0.8081
¥
—.808
and
_X¥—p
52,800
—52,400
400
o=
BDEST
A4Sk
Dousl.
808
4
We
will
round
these
values
to
-.81
and
.81,
respectively.
So,
we
want
to
find
P(-.81<
Z
<
.81)
on
the
standard
normal
probability
distribution
table.
Recall
that
P(-.81<
Z
<
.81)=P(Z<.81)-P(Z<-.81)=.79103
-.20897
=
.58206.
So
now
we
know
that
there
is
a
0.58206
probability
that
a
simple
random
sample
of
50
members
of
the
sales
force
will
provide
a
sample
mean
that
is
within
the
$400
of
the
sample
mean.
Conversely,
there
is
a
1
-
.58206=
.41794
probability
the
sample
mean
will
not
be
within
the
$400
range.
Question
5
0/0
pts
Suppose
that
you
have
a
large
number
of
potatoes.
The
mean
weight
of
the
potatoes
is
7.4
ounces
with
a
standard
deviation
of
2.2
ounces.
We
may
assume
a
normal
distribution.
a)
If
you
choose
25
potatoes
at
random,
what
is
the
probability
that
the
mean
of
this
sample
of
25
potatoes
is
between
7
and
8
ounces?
b)
If
you
choose
45
potatoes
at
random,
what
is
the
probability
that
the
mean
of
this
sample
of
45
potatoes
is
less
than
7
ounces?
c)
If
you
choose
40
potatoes
at
random,
what
is
the
probability
that
the
mean
of
this
sample
of
40
potatoes
is
more
than
7.5
ounces?
Your
Answer:
—
22
__
a)lo,
=
v
44
z—
7—7.4
z-score
for
7
ounces:
|z
=
(a_“)
_
|
T
)
_
91
w
3
z—
7.4
z-score
for
8
ounces:
|z
=
(wa_“)
=
(8.41
)
—
1.36
we
want
to
find
P(-.91<
Z
<
1.36)
on
the
standard
normal
probability
distribution
table:
P(-.91<
Z
<
1.36)=P(Z<1.36)-
(Z<-.91)=.91309-.18141=.73168.
There
is
a
.73168
probability
that
a
random
sample
of
25
potatoes
with
have
a
mean
weight
between
7
and
8
ounces.
b)
Standard
deviation
of
sample:
|0z
=
%
=
%.3280
_
(z=p)
(7-T4)
=%
amo
%2
Z-Score.
|2
P(Z
<
-1.22)
on
the
standard
normal
probability
distribution
table:
P(Z
<
-1.22)
=
11123.
11123
probability
that
a
simple
random
sample
of
45
potatoes
will
have
a
mean
weight
less
than
7
ounces.
c)
Standard
deviation
of
the
sample:
|0z
=
%
=
%.34785
_
(m—p)
_
(T-74)
_
—
To
34785
=
.29
Z-Score:
|z
P(Z
>
.29)
on
the
standard
normal
probability
distribution
table:
P(Z
>
.29)
=1-P(Z<.29)
=1-.61409=.38591.
.38591
probability
that
a
simple
random
sample
of
40
potatoes
will
have
a
mean
weight
greater
than
7.5
ounces.
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a)
We
calculate
the
standard
deviation
of
the
sample
distribution:
*Tyn
V25
Calculate
the
z-score
for
7
ounces:
SRR
TR
R
g
T
g
S
Calculate
the
z-score
for
8
ounces:
_X-u_
8-74
T=
.
=%
=
1.36
So,
we
want
to
find
P(-.91<
Z
<
1.36)
on
the
standard
normal
~
probability
distribution
table.
Recall
that
-
P(-.91<
Z
<
1.36)=P(Z<1.36)-P(Z<-.91)=.91309-.18141=.73168.
Therefore,
there
is
a
.73168
probability
that
a
random
sample
of
25
potatoes
with
have
a
mean
weight
between
7
and
8
ounces.
b)
We
calculate
the
standard
deviation
of
the
sample
distribution:
2.2
0
=—=—=.3280
Y
Vn
a5
Calculate
the
z-score:
_X-op_7-74_
o,
2=
T
3280
So,
we
want
to
find
P(Z
<
-1.22)
on
the
standard
normal
probability
distribution
table.
P(Z
<-1.22)
=
.11123.
Therefore,
there
is
a
.11123
probability
that
a
simple
random
sample
of
45
potatoes
will
have
a
mean
weight
less
than
7
ounces.
c)
We
calculate
the
standard
deviation
of
the
sample
distribution:
7~
_
0
__
0V
_
¢
s
.
LdalCulate
ine
Z-score.
So,
we
want
to
find
P(Z
>
.29)
on
the
standard
normal
probability
distribution
table.
Recall
that
P(Z
>
.29)
=
1
-
P(Z
<
.29)
=
1-.61409=.38591.
Therefore,
there
is
a
.38591
probability
that
a
simple
random
sample
of
40
potatoes
will
have
a
mean
weight
greater
than
7.5
ounces.
Question
6
0/0
pts
According
to
the
American
Diabetes
Association
(2013),
8.3
%
of
the
U.S.
population
have
diabetes.
a)
Suppose
that
you
take
a
random
of
80
Americans,
what
is
the
probability
that
7
%
or
less
of
these
80
people
have
diabetes?
b)
Suppose
that
you
take
a
random
of
120
Americans,
what
is
the
probability
that
8
%
or
more
of
these
120
people
have
diabetes?
Your
Answer:
1—
.083(1—.083
a)|op
=
/22
—
/SO0
_
43084
z-score
to
determine
the
probability
of
sample
size
yielding
a
proportional
mean
p
within
the
desired
range:
|z
=
%
=
'9356323
=
—.42‘
P(Z<-0.42).
From
the
standard
normal
table,
we
find:
P(Z<-0.42)=.33724.
33724
probability
that
the
percentage
of
the
sample
that
has
diabetes
is
less
than
7
%.
b)
|y
=
4/
ZE2
=
\/
0830
—0%9)
_
0252
z-score
to
determine
the
probability
of
our
sample
size:
2=
oy
T
T
022
P(Z>-0.12).
From
the
standard
normal
table,
we
find:
P(Z>-.12)=1-
P(Z<-.12)=1-.45224=.54776.
.54776.probability
that
the
percentage
of
the
sample
that
has
diabetes
is
greater
than
8
%.
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1-
0.083(1
—
0.083
afi=jp(
’”:j
(
)
_
0.03084
n
,
80
Now
we
use
our
old
friend
the
z-score
to
determine
the
probability
of
our
sample
size
yielding
a
proportional
mean
p
within
the
desired
range:
p—p
007-0083
T
OO
z=
We
want
P(Z<-0.42).
From
the
standard
normal
table,
we
find:
P(Z<-0.42)=.33724.
So
there
is
a
.33724
probability
that
the
percentage
of
the
sample
that
has
diabetes
is
less
than
7
%.
b)
=
0.0252
_
|p(1—p)
_
|0.083(1—0.083)
%
=
n
120
Now
we
use
our
old
friend
the
z-score
to
determine
the
probability
of
our
sample
size
yielding
a
proportional
mean
p
within
the
desired
range:
_p—p
_0.08—0.083
e
DORED.
T
ihe
z
We
want
P(Z>-0.12).
From
the
standard
normal
table,
we
find:
P(Z>-.12)=1-
P(Z<-.12)=1-.45224=.54776.
So
there
is
a
.54776.probability
that
the
percentage
of
the
sample
that
has
diabetes
is
greater
than
8
%.
Question
7
515
pts
As
a
reminder,
the
questions
in
this
review
quiz
are
a
requirement
of
the
course
and
the
best
way
to
prepare
for
the
module
exam.
Did
you
complete
all
questions
in
their
entirety
and
show
your
work?
Your
Answer:
yes
Quiz
Score:
5
out
of
5
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