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marta [7]
2 years ago
15

If the wait time is less than 30 minutes for 75% of all patients in the emergency room, what is the probability that the

Advanced Placement (AP)
2 answers:
Greeley [361]2 years ago
5 0

The probability that the proportion of patients who wait less than 30 minutes is 0.582 or less is 0.0020

<h3>What is probability? </h3>

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 \mathbf{\hat p} can be computed by using the expression:

\mathbf{\hat p = \dfrac{x}{n}}

\mathbf{\hat p = \dfrac{32}{55}}

\mathbf{\hat p =0.582}

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:

\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 )}

\mathbf{( \hat P \leq 0.582 )= P \Big( Z \leq \dfrac{-0.168}{\sqrt{0.003409}} \Big )}

\mathbf{( \hat P \leq 0.582) = P \Big( Z \leq -2.88 \Big )}

From the Z distribution table:

\mathbf{( \hat P \leq 0.582) = 0.00198}

\mathbf{( \hat P \leq 0.582) \simeq 0.0020}

Learn more about probability here:

brainly.com/question/24756209

solmaris [256]2 years ago
3 0

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

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The probability that the proportion of patients who wait less than 30 minutes is 0.582 or less is 0.0020

<h3>What is probability? </h3>

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 \mathbf{\hat p} can be computed by using the expression:

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\mathbf{\hat p = \dfrac{32}{55}}

\mathbf{\hat p =0.582}

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:

\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 )}

\mathbf{( \hat P \leq 0.582 )= P \Big( Z \leq \dfrac{-0.168}{\sqrt{0.003409}} \Big )}

\mathbf{( \hat P \leq 0.582) = P \Big( Z \leq -2.88 \Big )}

From the Z distribution table:

\mathbf{( \hat P \leq 0.582) = 0.00198}

\mathbf{( \hat P \leq 0.582) \simeq 0.0020}

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The sum of the area of the shaded region is given as follows;

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Using Microsoft Excel, or Wolfram Alpha, we have that the possible solutions to the above equation are;

x = 0, x ≈ 0.55 or x ≈ 1.85

The area under the line y = x/2, between the points x = 0 and x ≈ 0.55, A₁, is given as follows

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The area under the line y = sin²x, between the points x = 0 and x ≈ 0.55, A₂, is given using as follows;

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Therefore;

A_2 = \int\limits^{0.55}_0 {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_0 ^{0.55}

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A_4 = \int\limits^{1.85}_{0.55} {sin^2x} \, dx = \dfrac{1}{2} \left [x -sin(x) \cdot cos(x) \right]_{0.55} ^{1.85} \approx 1.00526

The shaded area, A_{2 shaded} = A₄ - A₃ = 1.00526 - 0.78 ≈ 0.22526

The sum of the area of the shaded regions, ∑A = A_{1 shaded} + A_{2 shaded}

∴ A = 0.023425 + 0.22526 = 0.248685

The sum of the area of the shaded regions, ∑A = 0.248685

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