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The length of stay at a specific emergency department in Phoenix, Arizona, in 2009 had a mean of 4.6 hours with a standard deviation of \(2.9 .\) Assume that the length of stay is normally distributed. a. What is the probability of a length of stay greater than 10 hours? b. What length of stay is exceeded by \(25 \%\) of the visits? c. From the normally distributed model, what is the probability of a length of stay less than 0 hours? Comment on the normally distributed assumption in this example.

Short Answer

Expert verified
a) 0.0314; b) 6.56 hours; c) 0.0562, normal model might not fit well.

Step by step solution

01

Calculate the z-score for a stay greater than 10 hours

Given: mean \(\mu = 4.6\) hours and standard deviation \(\sigma = 2.9\) hours. First, we will compute the z-score for 10 hours using the formula:\[z = \frac{x - \mu}{\sigma} = \frac{10 - 4.6}{2.9}\approx 1.86.\]This z-score tells us how many standard deviations 10 hours is away from the mean.
02

Find the probability using the z-score

Using the z-score of \(1.86\), check a standard normal distribution table or use software to find the probability of \(Z < 1.86\), which is approximately \(0.9686\). Since we are interested in the probability of staying more than 10 hours, we calculate:\[P(X > 10) = 1 - P(Z < 1.86) = 1 - 0.9686 = 0.0314.\]
03

Determine the length of stay exceeded by 25% of visits

To find this threshold, we'll locate the z-score associated with the 75th percentile (since 25% exceed this value). From the standard normal distribution table, the z-score for the 75th percentile is approximately \(0.675\). Use the z-score formula in reverse to find the corresponding physical value:\[x = \mu + z \cdot \sigma = 4.6 + 0.675 \cdot 2.9 \approx 6.5575.\]Thus, 25% of visits exceed approximately 6.56 hours.
04

Evaluate the probability of a negative length of stay

Compute the z-score for 0 hours:\[z = \frac{0 - 4.6}{2.9} \approx -1.59.\]The probability \(P(X < 0)\) corresponds to the cumulative probability for \(Z < -1.59\), which is about \(0.0562\). This means there is a non-zero probability for a physically impossible event (negative stay), indicating that the normal distribution may not be a perfect model in this context.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculation
Probability helps us understand the likelihood of certain events occurring. In the context of a normal distribution, probabilities can be calculated by determining areas under the curve. The total area is equal to 1, with different sections representing different probabilities.
To find the probability of a length of stay greater than 10 hours in our example, we first calculated the z-score, which essentially tells us how far 10 hours is from the mean in terms of the standard deviations.
Once we had the z-score, we used it to find the probability of a stay less than 10 hours. This was subtracted from 1 to get the probability of a stay greater than 10 hours, as probabilities in a normal distribution are symmetric. This approach illustrates how z-scores and probability calculations are intertwined.
Z-score
A z-score represents how many standard deviations an element is from the mean of the dataset. It is a way to standardize scores across different distributions, making them comparable.
The formula for calculating the z-score is:
  • Subtract the mean from the data point.
  • Divide by the standard deviation.
In our scenario, a z-score for a 10-hour stay was calculated to be approximately 1.86, indicating that 10 hours is 1.86 standard deviations above the mean length of stay of 4.6 hours.
Calculating z-scores helps us understand how unusual or typical a certain observation might be within a given dataset. It transforms data into a standard normal distribution with mean zero and standard deviation one.
Percentiles
Percentiles are a measure used in statistics to indicate the relative standing of a value within a dataset. Simply put, a percentile is the position of a data point in the dataset when it is sorted in ascending order.
For example, to find the length of stay exceeded by 25% of the visits, we looked for the 75th percentile (since 25% of visits exceed this duration). A 75th percentile indicates that 75% of the data is below that percentile value.
In a normally distributed data set, percentiles can be easily determined using z-scores. Once we know the relevant z-score, we can convert it back into a data value using the mean and standard deviation.
Standard Deviation
Standard deviation is crucial in understanding the spread of values in a dataset. It tells us about the average distance of each data point from the mean, which gives insight into how spread out the observations are.
A lower standard deviation means that the data points tend to be close to the mean, whereas a higher standard deviation indicates data points spread out over a large range of values.
In the context of our example, the standard deviation of 2.9 hours shows how much the length of stay at the emergency department varies from the mean stay of 4.6 hours.
Calculating standard deviation is an important step in statistical analysis, as it plays a key role in z-score calculations and understanding the implications of various statistical measures like percentiles and probabilities.

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