Question
A survey asked 32126 people how much confidence they had in educational institutions. The results were as follows. Round your answers to four decimal places if necessary.
A great deal- 10,022
some- 17,813
hardly any - 4,291
(a) What is the probability that a sampled person has either a great deal or hardly any confidence in educational institutions?
The probability that a sampled person has either a great deal or hardly any confidence in educational institutions is [?]
A great deal- 10,022
some- 17,813
hardly any - 4,291
(a) What is the probability that a sampled person has either a great deal or hardly any confidence in educational institutions?
The probability that a sampled person has either a great deal or hardly any confidence in educational institutions is [?]
Answers
Answer
the correct answer is .4455
(b) Assume this is a simple random sample from a population. Use the Empirical Method to estimate the probability that a person has hardly any confidence in educational institutions.
Using the Empirical Method the probability is approximately [?]
(b) Assume this is a simple random sample from a population. Use the Empirical Method to estimate the probability that a person has hardly any confidence in educational institutions.
Using the Empirical Method the probability is approximately [?]
Answer
the correct answer is .1336
c) If we use a cutoff of 0.05, is it unusual for someone to have some confidence in educational institutions?
c) If we use a cutoff of 0.05, is it unusual for someone to have some confidence in educational institutions?
Answered by
GPT-4o mini
To find the probability that a sampled person has either a great deal or hardly any confidence in educational institutions, we first need to determine how many people fall into these two categories.
1. The number of people who have "a great deal" of confidence is 10,022.
2. The number of people who have "hardly any" confidence is 4,291.
Next, we add these two values together:
\[
10,022 + 4,291 = 14,313
\]
Now we calculate the probability. The probability (\(P\)) is given by the number of favorable outcomes divided by the total number of outcomes (the total number of people surveyed):
\[
P = \frac{\text{Number of people with a great deal of confidence or hardly any}}{\text{Total number of people surveyed}} = \frac{14,313}{32,126}
\]
Now we can perform the division:
\[
P \approx \frac{14,313}{32,126} \approx 0.445
\]
Thus, rounding to four decimal places:
\[
P \approx 0.4450
\]
Therefore, the probability that a sampled person has either a great deal or hardly any confidence in educational institutions is:
\[
\text{Probability} = 0.4450
\]
1. The number of people who have "a great deal" of confidence is 10,022.
2. The number of people who have "hardly any" confidence is 4,291.
Next, we add these two values together:
\[
10,022 + 4,291 = 14,313
\]
Now we calculate the probability. The probability (\(P\)) is given by the number of favorable outcomes divided by the total number of outcomes (the total number of people surveyed):
\[
P = \frac{\text{Number of people with a great deal of confidence or hardly any}}{\text{Total number of people surveyed}} = \frac{14,313}{32,126}
\]
Now we can perform the division:
\[
P \approx \frac{14,313}{32,126} \approx 0.445
\]
Thus, rounding to four decimal places:
\[
P \approx 0.4450
\]
Therefore, the probability that a sampled person has either a great deal or hardly any confidence in educational institutions is:
\[
\text{Probability} = 0.4450
\]
Answered by
GPT-4o mini
To estimate the probability that a person has hardly any confidence in educational institutions using the Empirical Method, we will use the number of people who reported "hardly any" confidence and divide it by the total number of surveyed people.
From the survey results:
- The number of people who have "hardly any" confidence is 4,291.
- The total number of surveyed people is 32,126.
Now, we can calculate the probability (\(P\)):
\[
P(\text{hardly any confidence}) = \frac{\text{Number of people with hardly any confidence}}{\text{Total number of people surveyed}} = \frac{4,291}{32,126}
\]
Now, let's perform the division:
\[
P \approx \frac{4,291}{32,126} \approx 0.133
\]
To obtain a more precise calculation:
\[
P \approx 0.1334
\]
Thus, rounding to four decimal places:
\[
P \approx 0.1334
\]
Therefore, using the Empirical Method, the probability that a person has hardly any confidence in educational institutions is approximately:
\[
\text{Probability} \approx 0.1334
\]
From the survey results:
- The number of people who have "hardly any" confidence is 4,291.
- The total number of surveyed people is 32,126.
Now, we can calculate the probability (\(P\)):
\[
P(\text{hardly any confidence}) = \frac{\text{Number of people with hardly any confidence}}{\text{Total number of people surveyed}} = \frac{4,291}{32,126}
\]
Now, let's perform the division:
\[
P \approx \frac{4,291}{32,126} \approx 0.133
\]
To obtain a more precise calculation:
\[
P \approx 0.1334
\]
Thus, rounding to four decimal places:
\[
P \approx 0.1334
\]
Therefore, using the Empirical Method, the probability that a person has hardly any confidence in educational institutions is approximately:
\[
\text{Probability} \approx 0.1334
\]
Answered by
GPT-4o mini
To determine if it is unusual for someone to have some confidence in educational institutions using a cutoff of 0.05, we first need to calculate the probability that a person has some confidence.
From the survey results:
- The number of people who have "some" confidence is 17,813.
- The total number of surveyed people is 32,126.
We calculate the probability (\(P\)) that a randomly selected person has some confidence:
\[
P(\text{some confidence}) = \frac{\text{Number of people with some confidence}}{\text{Total number of people surveyed}} = \frac{17,813}{32,126}
\]
Calculating this gives:
\[
P \approx \frac{17,813}{32,126} \approx 0.5544
\]
Now, since \(0.5544\) is significantly greater than the cutoff of \(0.05\), we conclude that having some confidence in educational institutions is not unusual.
**Conclusion:** Since \(P(\text{some confidence}) \approx 0.5544\) is much higher than \(0.05\), it is not unusual for someone to have some confidence in educational institutions.
From the survey results:
- The number of people who have "some" confidence is 17,813.
- The total number of surveyed people is 32,126.
We calculate the probability (\(P\)) that a randomly selected person has some confidence:
\[
P(\text{some confidence}) = \frac{\text{Number of people with some confidence}}{\text{Total number of people surveyed}} = \frac{17,813}{32,126}
\]
Calculating this gives:
\[
P \approx \frac{17,813}{32,126} \approx 0.5544
\]
Now, since \(0.5544\) is significantly greater than the cutoff of \(0.05\), we conclude that having some confidence in educational institutions is not unusual.
**Conclusion:** Since \(P(\text{some confidence}) \approx 0.5544\) is much higher than \(0.05\), it is not unusual for someone to have some confidence in educational institutions.
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