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Sagot :
Answer: The sampling distribution can be approximated by a Normal distribution with a mean of 0.75, a standard deviation of 0.0585.
P(p^ <= 0.582) = 0.0020
Explanation:
First, the mean of the sampling distribution is the same Probability for the population, you can find the standard deviation (after checking with the 10% condition) doing sqrt((P * (P-1))/n), then after finding the standard deviation you can find the Z-score doing sqrt(p^- p) / SD), giving us a Z-score of -2.867. Then using normalcdf, you do normalcdf(lower:-1e99, upper: -2.867, mean:0, SD 1) and then you get .0020.
Definitely not the best way it could have been explained but hopefully it made some sense
The probability that the proportion of patients who wait less than 30 minutes is 0.582 or less is 0.0020
What is probability?
Probability can be defined as the likelihood of an event to occur. In statistics, the mean of the sample distribution typically shows the probability of the population.
From the parameters given:
- The sample size (n) = 55 patients
- Let's assume that the mean (x) = 32 (i.e. 58.2%) of the patients
The sample proportion [tex]\mathbf{\hat p}[/tex] can be computed by using the expression:
[tex]\mathbf{\hat p = \dfrac{x}{n}}[/tex]
[tex]\mathbf{\hat p = \dfrac{32}{55}}[/tex]
[tex]\mathbf{\hat p =0.582}[/tex]
If the percentage of the probability of all patients in the emergency room = 0.75
The probability that the proportion of patients who wait less than 30 minutes is 0.582 or less can be computed as:
[tex]\mathbf{( \hat P \leq 0.582) = P \Big( \dfrac{\hat p - p}{\sqrt{\dfrac{p(1-p)}{n}}} \leq \dfrac{0.582 - 0.75}{\sqrt{\dfrac{0.75(1-0.75)}{55}}} \Big )}[/tex]
[tex]\mathbf{( \hat P \leq 0.582 )= P \Big( Z \leq \dfrac{-0.168}{\sqrt{0.003409}} \Big )}[/tex]
[tex]\mathbf{( \hat P \leq 0.582) = P \Big( Z \leq -2.88 \Big )}[/tex]
From the Z distribution table:
[tex]\mathbf{( \hat P \leq 0.582) = 0.00198}[/tex]
[tex]\mathbf{( \hat P \leq 0.582) \simeq 0.0020}[/tex]
Learn more about probability here:
https://brainly.com/question/24756209
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