It is estimated that 16.6% of all adults in the U.S. are uninsured. You take a random sample of 250 adults seen by a certain clinic and nd that 50 are uninsured. In such groups of 250 U.S adults, what is the mean number of those that would be uninsured? round to 1 decimal place U= What is the standard deviation? round to 2 decimal places 0=
It is estimated that 16.6% of all adults in the U.S. are uninsured. You take a random sample of 250 adults seen by a certain clinic and nd that 50 are uninsured. In such groups of 250 U.S adults, what is the mean number of those that would be uninsured? round to 1 decimal place U= What is the standard deviation? round to 2 decimal places 0=
A First Course in Probability (10th Edition)
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ISBN:9780134753119
Author:Sheldon Ross
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Chapter1: Combinatorial Analysis
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question 7
![### Question 7
It is estimated that 16.6% of all adults in the U.S. are uninsured. You take a random sample of 250 adults seen by a certain clinic and find that 50 are uninsured.
In such groups of 250 U.S. adults, what is the mean number of those that would be uninsured? Round to 1 decimal place.
- \(\mu =\) [Input Box]
What is the standard deviation? Round to 2 decimal places.
- \(\sigma =\) [Input Box]
### Explanation:
This problem involves calculating the mean and standard deviation for a binomial distribution. The mean (\(\mu\)) is calculated as follows:
\[
\mu = n \times p
\]
where \(n\) is the sample size (250) and \(p\) is the probability of being uninsured (16.6% or 0.166).
The standard deviation (\(\sigma\)) is calculated using the formula:
\[
\sigma = \sqrt{n \times p \times (1-p)}
\]
These calculations help in understanding the expected number of uninsured adults and the variability in different samples.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F588df426-7a7b-4d8c-8d66-8d52e2cca6e0%2F23c0a115-e445-47b7-a188-c8b83a0912b6%2F5eww1nt_processed.png&w=3840&q=75)
Transcribed Image Text:### Question 7
It is estimated that 16.6% of all adults in the U.S. are uninsured. You take a random sample of 250 adults seen by a certain clinic and find that 50 are uninsured.
In such groups of 250 U.S. adults, what is the mean number of those that would be uninsured? Round to 1 decimal place.
- \(\mu =\) [Input Box]
What is the standard deviation? Round to 2 decimal places.
- \(\sigma =\) [Input Box]
### Explanation:
This problem involves calculating the mean and standard deviation for a binomial distribution. The mean (\(\mu\)) is calculated as follows:
\[
\mu = n \times p
\]
where \(n\) is the sample size (250) and \(p\) is the probability of being uninsured (16.6% or 0.166).
The standard deviation (\(\sigma\)) is calculated using the formula:
\[
\sigma = \sqrt{n \times p \times (1-p)}
\]
These calculations help in understanding the expected number of uninsured adults and the variability in different samples.
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