2-dc_amp_nd_practice_test_key
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Phoenix College *
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Course
MISC
Subject
Statistics
Date
Nov 24, 2024
Type
Pages
5
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Density
Curves
&
The
Normal
Distribution
Test
MULTIPLE
CHOICE
).
/0253.
M&L
1.
The
heights
of
American
men
aged
18
to
24
are
approximately
normally
distributed
with
mean
68
inches
and
standard
deviation
2.5
inches.
Only
about
5%
of
young
men
have
heights
outside
the
range
_,_/
?é
=
){Z—;@’
Xb‘
~
63
A4
(a)
65.5
inches
to
70.5
inches
@
63
inches
to
73
inches’
(c)
60.5
inches
to
75.5
inches
(d)
58
inches
to
78
inches
(e)
None
of
the
above.
2.
The
amount
of
time
required
for
each
of
100
mice
to
navigate
through
a
maze
was
recorded.
The
histogram
below
shows
the
distribution
of
times,
in
seconds,
for
the
100
mice.
;
L
;
60
70
80
90
10
110
120
Time
{seconds)
Which
of
the
following
values
is
closest
to
the
standard
deviation
of
the
100
times?
(a)
2.5
seconds
10
seconds
(c)
20
seconds
(d)
50
seconds
(e)
90
seconds
3.
Zucchini
weights
are
approximately
normally
distributed
with
mean
0.8
pound
and
standard
deviation
0.25
pound.
Which
of
the
following
shaded
regions
best
represents
the
probability
that
arandomly
selected
zucchini
will
weigh
between
0.55
pound
and
1.3
pounds?
\
(c)
w2
wra
(@)
B
()
w053
e
(e)
pridse
pilde
Density
Curves
&
The
Normal
Distribution
Test
4.
The
weight
of
adult
male
grizzly
bears
living
in
the
wild
in
the
continental
United
States
is
approximately
normally
distributed
with
a
mean
of
500
pounds
and
a
standard
deviation
of
50
pounds.
The
weight
of
adult
female
grizzly
bears
is
approximately
normally
distributed
with
a
mean
of
300
pounds
and
a
standard
deviation
of
40
pounds.
Approximately,
what
would
be
the
g
weight
of
a
female
grizzly
bear
with
the
same
standardized
score
(z-score)
as
a
male
grizzly
bear
with
a
weight
of
530
pounds?
70
—
570
2007
=
06
(a)
276
pounds
@324
pounds
(c)
330
pounds
50
00
(d)
340
pounds
(e)
530
pounds
0
(p
-
)(1/
7
5.
A
distribution
of
test
scores
is
not
symmetric.
Which
of
the
following
is
the
best
estimate
of
the
zscore
of
the
third
quartile?
E
(a)
0.67
(b)
0.75
(c)1.00
(d)
1.41
@This
z-score
cannot
be
estimated
from
the
given
information.
6.
For
the
density
curve
illustrated,
which
of
the
following
is
true?
0
1
2
|.
The
mean
is
smaller
than
the
median.
O
Il.
The
proportion
of
outcomes
between
0.1
and
0.7
is
equal
to
0.6.
The
proportion
of
outcomes
exceeding
0.5
is
equal
to
0.75.
(a)
I
only
(b)
Il
only
@III
only
(d)
land
Il
(e)
lland
Il
7.
Scores
on
the
American
College
Test
(ACT)
are
normally
distributed
with
a
mean
of
18
and
a
standard
deviation
of
6.
The
interquartile
range
of
the
scores
is
approximately:
Py
A
—0.675
=
L~
(@s.1
(b)
12
©6
d10.3
@7
:
&
X,
2
/8,953
a2.05
8.
The
test
grades
at
a
large
school
have
an
approximately
normal
distribution
with
a
mean
of
50.
)/%”’
8
077
%
58.09
What
is
the
standard
deviation
of
the
data
so
that
80%
of
the
students
are
within
12
points
(above
or
below)
the
mean?
/g
(a)
5.875
9.375
(c)
10.375
(d)
14.5
(e)
cannot
be
determined
from
the
given
information
3
X,
Yy
—~/
982
=
S
g
369
#
Q/
E
Density
Curves
&
The
Normal
Distribution
Test
9.
If
a
density
curve
was
a
semicircle,
the
radius
would
be
approximately:
2
2
-
(a)
0.2
(b)
0.4
(c)
0.6
0.8
!
(e)1
e
r=
Jg
10.
A
market
research
company
employs
a
large
number
of
typists
to
enter
data
into
a
computer.
2~
)
&
The
time
taken
for
new
typists
to
learn
the
computer
system
is
known
to
have
a
normal
distribution
with
a
mean
of
90
minutes
and
a
standard
deviation
of
18
minutes.
The
proportion
of
new
typists
that
take
more
than
two
hours
to
learn
the
computer
system
is
o
=70
(@0.048
/éé7
(b)
0.394
(c)
0.452
(d)
0.548
(e)
0.952
—
/%
FREE
RESPONSE
11.
Use
the
density
curve
below
to
answer
the
following
questions.
-1
3
(a)
What
is
the
heignt
ot
this
density
curve?
Show
mathematically
how
you
know.
v
o
035
—
z(3-cn)
=/
/
(b)
What
is
the
interquartile
range?
2—0=
A
(c)
What
percentage
of
the
observations
lies
above
1.57
0.
375
(d)
What
percentage
of
the
observations
lies
between
0.5
and
1.2?
0175
/0['7,
>/
éfi)xaflfi
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Density
Curves
&
The
Normal
Distribution
Test
12.
Mr.
League
was
worried
that
the
test
you
are
currently
taking
(right
now)
would
either
be
too
easy
or
too
difficult,
so
he
wanted
to
do
a
“practice
run”
by
giving
it
to
his
collection
of
pet
spider
monkeys
(he
has 24
of
them
—
don't
ask).
The
scores
achieved
by
the
monkeys
took
on
a
normal
distribution
with
a
mean
score
of
72
and
a
standard
deviation
of
7.
Find
the
following,
including
sketches
and
calculations:
,,2/;
(@)
P(x>80)
L]’(b)
P(x<65)
,,Q/(c)
P(70<x<80)
PR
=
S
|
|
so-7k
Pz
=1143)
Zon7
Plec-l)zolsy
7
//,fl
26+
=
£/
17)
0.
456
13.
Mr.
League’s
smartest
pet
spider
monkey,
Pretty
Pete,
scored
in
the
90"
percentile
when
compared
to
his
colleagues.
What
score
did
Pretty
Pete
achieve?
(Show
sketches
and
calculations)
/
a?ffl.
=
}fi;_/zz/
X
»
7
-
o2
50
77/
(8
w
acap
14.
Mr.
League’s
favorite,
but
dumbest,
pet
spider
monkey,
Colonel
Fitzwilliam,
forgot
to
turn
his
test
in.
When
Mr.
League
found
it
on
the
floor,
he
had
to
re-calculate
the
summary
statistics
for
his
spider
monkeys
(it
was
still
approximately
a
normal
distribution
for
both
parts
aand
5
below).
=
/143
(a)
If
Colonel
Fitzwilliam’s
test
score
didn't
change
the
distribution’s
mean
and
decreased
the
standard
deviation,
what
was
Colonel
Fitzwilliam's
test
score?
2/
CF's
Test
Score:
7j/
(b)
If
the
Colonel’s
score
was
a
62
and
it
increased
the
standard
deviation
to
7.2,
and
he
scored
in
the
137
percentile,
what
was
the
new
mean
for
all
the
tests?
G
A
)y
e
———
7
A=
‘Q/
M
=
70
/(0
Density
Curves
&
The
Normal
Distribution
Test
15.
Mr.
League
is
considering
implementing
one
of
two
academic
requirements
(concerning
test
averages)
for
registering
for
the
AP
exam
for
his
course.
Over
his
entire
career,
student
test
averages
are
approximately
normally
distributed
with
a
specified
mean
and
standard
deviation.
s
Academic
Requirement
Plan
|
calls
for
not
allowing
AP
Statistics
students
to
register
for
the
exam
if
their
test
average
falls
more
than
2
standard
deviations
below
the
specified
mean.
e
Academic
Requirement
Plan
il
calls
for
not
allowing
AP
Statistics
students
to
register
for
the
exam
if
their
test
average
falls
more
than
1.5
interquartile
ranges
below
the
lower
quartile
of
the
specified
population.
(a)
What
proportion
of
students
will
not
be
able
to
take
the
exam
under
Academic
Requirement
Plan
[?
,
fles-r)
200228
220
%
of
it
vill
nit
Lo
A+
pb
S
thar
(b)
What
proportion
of
students
will
not
be
able
to
take
the
exam
under
Academic
Requirement
Plan
11?
f6fe
e
Fermactl
=
—>
04T
—(~0.676)
.35
o
—
()
35)
=
~
L7
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