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Precalculus
Use normal distributions to approximate binomial distributions
76.9
%
76.9 \%
76.9%
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Resources
\newline
Check
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Approximately
20
%
20 \%
20%
of newborns are born more than
1
1
1
week before their due date. A random sample of
20
20
20
newborns is selected.
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The standard deviation of the sampling distribution for the proportion of your sample that is born more than
7
7
7
days before their due date is
\newline
0
0
0
.
20
20
20
.
\newline
3
3
3
.
2
2
2
.
\newline
0
0
0
.
089
089
089
.
\newline
0
0
0
.
008
008
008
.
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Below is the problem statement for an exercise asking you to go through a hypothesis test. Which calculator function are you going to use to find the test statistic or p-value?
\newline
You wish to test the following claim
(
H
a
)
\left(\mathrm{H}_{\mathrm{a}}\right)
(
H
a
)
at a significance level of
α
=
0.05
\alpha=0.05
α
=
0.05
.
\newline
H
o
:
p
=
0.42
H
a
:
p
>
0.42
\begin{array}{l} \mathrm{H}_{\mathrm{o}}: p=0.42 \\ \mathrm{H}_{\mathrm{a}}: p>0.42 \end{array}
H
o
:
p
=
0.42
H
a
:
p
>
0.42
\newline
You obtain a sample of size
n
=
143
n=143
n
=
143
in which there are
73
73
73
successful observations. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution.
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T-test
\newline
1
1
1
-PropZint
\newline
Zinterval
\newline
1
1
1
-PropZtest
\newline
Z-Test
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High school students across the nation compete in a financial capability challenge each year by taking a National Financial Capability Challenge Exam. Students who score in the top
12
%
12\%
12%
are recognized publicly for their achievement by the Department of the Treasury. Assuming a normal distribution, how many standard deviations above the mean does a student have to score to be publicly recognized?
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If
f
(
x
)
f(x)
f
(
x
)
is an exponential function where
f
(
−
1.5
)
=
26
f(-1.5)=26
f
(
−
1.5
)
=
26
and
f
(
5.5
)
=
7
f(5.5)=7
f
(
5.5
)
=
7
, then find the value of
f
(
10
)
f(10)
f
(
10
)
, to the nearest hundredth.
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A survey found that women spend, on average,
$
146.21
\$ 146.21
$146.21
on beauty products during the summer months, with a standard deviation of
$
29.44
\$ 29.44
$29.44
. What percentage of women spend less than
$
160
\$ 160
$160
? Round to the nearest hundredth. Assume the distribution is normal.
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The graph of a normally distributed random variable is given. Use the graph to answer the questions that follow.
\newline
What is the mean?
\newline
27
27
27
\newline
0
∘
0^{\circ}
0
∘
\newline
What is the standard deviation?
\newline
3
3
3
\newline
σ
∘
\sigma^{\circ}
σ
∘
\newline
What percentage of values should fall between
27
27
27
and
30
30
30
?
\newline
%
\%
%
\newline
Which value has a
z
z
z
-score of
−
2
-2
−
2
?
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1
1
1
. Some common geometric terms are point, line, line segment, and ray. A point is an exact location in space. A line is a straight path of points that go on and on in epposite directions. A line segment is a part of a line with two endpoints. A ray is a pait of a line that has one endpoint and continues on forever in one direction. Label each figure with the correct term.
\newline
2
2
2
. Angles can be classified by their measures. A right angle forms a square corner. An acute angle is open less than a right angle. An obtuse angle is open more than a right angle butis open less than a straight angle. A straight angle forms a straight line. Identify each type of angle.
\newline
3
3
3
. Use a geometric term to describe and name each figure. Be as specific as possible.
\newline
The figure at the right is a
\newline
Name the figure using the points that are labeled.
\newline
4
4
4
. The figure at the right is an angle.
\newline
Name the angle with the points from each ray and the shared endpoint of the rays. The shared endpoint is the center letter.
\newline
On the Back!
\newline
5
5
5
. thesancen tric term to describe each figure. Be as specific as possible.
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[-/
1
1
1
Points]
\newline
DETAILS
\newline
BBBASICSTAT
8
8
8
7
7
7
.
6
6
6
.
006
006
006
.MI.
\newline
MY NOTES
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ASK YOUR TEACHER
\newline
Consider a binomial experiment with
15
15
15
trials and probability
0.55
0.55
0.55
of success on a single trial.
\newline
(a) Use the binomial distribution to find the probability of exactly
10
10
10
successes. (Round your answer to three decimal places.)
\newline
(b) Use the normal distribution to approximate the probability of exactly
10
10
10
successes. (Round your answer to four decimal places.)
\newline
(c) Compare the results of parts (a) and (b).
\newline
These results are almost exactly the same.
\newline
These results are fairly different.
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Mark noticed that the probability that a certain player hits a home run in a single game is
0.175
0.175
0.175
. Mark is interested in the variability of the number of home runs if this player plays
200
200
200
games.
\newline
If Mark uses the normal approximation of the binomial distribution to model the number of home runs, what is the standard deviation for a total of
200
200
200
games? Answer choices are rounded to the hundredths place.
\newline
0.14
0.14
0.14
\newline
28.88
28.88
28.88
\newline
5.92
5.92
5.92
\newline
5.37
5.37
5.37
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A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of
65
65
65
students in the high school and found a mean savings of
2700
2700
2700
dollars with a standard deviation of
700
700
700
dollars. At the
95
%
95 \%
95%
confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest whole number. (Do not write
±
\pm
±
).
\newline
Answer:
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A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of
142
142
142
residents and found the mean weight to be
164
164
164
pounds with a standard deviation of
26
26
26
pounds. At the
95
%
95 \%
95%
confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth. (Do not write
±
\pm
±
).
\newline
Answer:
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A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of
32
32
32
students in the high school and found a mean savings of
4300
4300
4300
dollars with a standard deviation of
800
800
800
dollars. At the
95
%
95 \%
95%
confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest whole number. (Do not write
±
\pm
±
).
\newline
Answer:
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A study by the department of education of a certain state was trying to determine the mean SAT scores of the graduating high school seniors. The study examined the scores of a random sample of
178
178
178
graduating seniors and found the mean score to be
487
487
487
with a standard deviation of
119
119
119
. At the
95
%
95 \%
95%
confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth. (Do not write
±
\pm
±
).
\newline
Answer:
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A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of
60
60
60
students in the high school and found a mean savings of
2500
2500
2500
dollars with a standard deviation of
1400
1400
1400
dollars. At the
95
%
95 \%
95%
confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest whole number. (Do not write
±
\pm
±
).
\newline
Answer:
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A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of
74
74
74
students in the high school and found a mean savings of
4500
4500
4500
dollars with a standard deviation of
1300
1300
1300
dollars. At the
95
%
95 \%
95%
confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest whole number. (Do not write
±
\pm
±
).
\newline
Answer:
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A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of
76
76
76
residents and found the mean weight to be
198
198
198
pounds with a standard deviation of
22
22
22
pounds. At the
95
%
95 \%
95%
confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth. (Do not write
±
\pm
±
).
\newline
Answer:
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A study by the department of education of a certain state was trying to determine the mean SAT scores of the graduating high school seniors. The study examined the scores of a random sample of
98
98
98
graduating seniors and found the mean score to be
492
492
492
with a standard deviation of
117
117
117
. At the
95
%
95 \%
95%
confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth. (Do not write
±
\pm
±
).
\newline
Answer:
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A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of
59
59
59
students in the high school and found a mean savings of
4000
4000
4000
dollars with a standard deviation of
1000
1000
1000
dollars. At the
95
%
95 \%
95%
confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest whole number. (Do not write
±
\pm
±
).
\newline
Answer:
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A study by the department of education of a certain state was trying to determine the mean SAT scores of the graduating high school seniors. The study examined the scores of a random sample of
199
199
199
graduating seniors and found the mean score to be
537
537
537
with a standard deviation of
82
82
82
. At the
95
%
95 \%
95%
confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest tenth. (Do not write
±
\pm
±
).
\newline
Answer:
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A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of
71
71
71
students in the high school and found a mean savings of
2600
2600
2600
dollars with a standard deviation of
1500
1500
1500
dollars. At the
95
%
95 \%
95%
confidence level, use the normal distribution/empirical rule to estimate the margin of error for the mean, rounding to the nearest whole number. (Do not write
±
\pm
±
).
\newline
Answer:
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A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of
32
32
32
students in the high school and found a mean savings of
2900
2900
2900
dollars with a standard deviation of
1300
1300
1300
dollars. Use the normal distribution/empirical rule to estimate a
95
%
95 \%
95%
confidence interval for the mean, rounding all values to the nearest whole number.
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A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of
113
113
113
residents and found the mean weight to be
159
159
159
pounds with a standard deviation of
34
34
34
pounds. Use the normal distribution/empirical rule to estimate a
95
%
95 \%
95%
confidence interval for the mean, rounding all values to the nearest tenth.
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A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of
133
133
133
residents and found the mean weight to be
165
165
165
pounds with a standard deviation of
25
25
25
pounds. Use the normal distribution/empirical rule to estimate a
95
%
95 \%
95%
confidence interval for the mean, rounding all values to the nearest tenth.
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A study was commissioned to find the mean weight of the residents in certain town. The study examined a random sample of
36
36
36
residents and found the mean weight to be
182
182
182
pounds with a standard deviation of
29
29
29
pounds. Use the normal distribution/empirical rule to estimate a
95
%
95 \%
95%
confidence interval for the mean, rounding all values to the nearest tenth.
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A house purchased
5
5
5
years ago for
$
100
,
000
\$ 100,000
$100
,
000
was just sold for
$
135
,
000
\$ 135,000
$135
,
000
. Assuming exponential growth, approximate the annual growth rate, to the nearest percent.
\newline
Work/Explanation
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$
400
\$ 400
$400
is invested in an account earning
9
9
9
.
8
8
8
\% interest (APR), compounded daily. Write a function showing the value of the account after
t
t
t
years, where the annual growth rate can be found from a constant in the function. Round all coefficients in the function to four decimal places. Also, determine the percentage of growth per year (APY), to the nearest hundredth of a percent.
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