Problem 90 - Applied Statistics and Probability for Engineers

Question

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.

Step-by-Step Solution

Answer

(a) 3.14%, (b) 6.56 hours, (c) probability is 0 (model limitation).

Step 1: Identifying the problem

We have a normally distributed variable—length of stay—with a mean (\( \mu \)) of 4.6 hours and a standard deviation (\( \sigma \)) of 2.9 hours. We need to find probabilities for certain lengths of stay and a specific percentile.

Step 2: Calculating the Z-score for part (a)

Part (a) asks for the probability of a length of stay greater than 10 hours. To find this, we first calculate the Z-score using the formula: \[ Z = \frac{X - \mu}{\sigma} \]Substitute \( X = 10 \), \( \mu = 4.6 \), and \( \sigma = 2.9 \):\[ Z = \frac{10 - 4.6}{2.9} \approx 1.86 \]

Step 3: Finding the probability for part (a)

Using a Z-table or standard normal distribution calculator, find the probability \( P(Z > 1.86) \). The table gives \( P(Z < 1.86) \approx 0.9686 \). Thus, \( P(Z > 1.86) = 1 - 0.9686 = 0.0314 \). The probability of a stay greater than 10 hours is approximately 3.14%.

Step 4: Calculating the Z-score for part (b)

Part (b) asks which length of stay is exceeded by 25% of visits. We need the 75th percentile of the distribution. From Z-tables, the Z-score corresponding to 75% is approximately 0.675.

Step 5: Finding the length of stay for part (b)

Using the formula \( X = \mu + Z \cdot \sigma \), substitute \( \mu = 4.6 \), \( Z = 0.675 \), and \( \sigma = 2.9 \):\[ X = 4.6 + 0.675 \cdot 2.9 \approx 6.5575 \] The length of stay exceeded by 25% of visits is approximately 6.56 hours.

Step 6: Probability for part (c)

Since the length of stay cannot be negative, the probability of a stay less than 0 hours in reality is 0. However, using the normal distribution model, calculate \( Z \) for 0 hours: \[ Z = \frac{0 - 4.6}{2.9} \approx -1.586 \]The probability \( P(Z < -1.586) \approx 0.0565 \), indicating the model predicts a small, non-zero probability.

Step 7: Comment on the normal distribution assumption

The finding in part (c) reveals a limitation of the normal distribution model: it predicts non-zero probabilities for impossible events (like negative stays). This highlights the model's inaccuracy at extreme values, though it may still be a reasonable approximation for most real-world data.

Key Concepts

Probability Calculation

In statistical modeling, probability calculation forms the backbone for understanding the likelihood of an event happening. In the context of a normally distributed variable, such as the length of stay in an emergency department, we calculate probabilities to predict outcomes.
The normal distribution is symmetric around its mean, which allows us to use techniques like standard normal distribution tables or calculators to find the probability of different events.

• For instance, to calculate the probability of a patient staying longer than 10 hours, we first convert this length of stay into a Z-score (more on that soon) and use it to find the cumulative probability.
• The probability of the occurrence of an event is always a number between 0 and 1, which could be translated to a percentage. In this case, a probability of 0.0314 implies a 3.14% chance.
• Probabilities help in making informed decisions in various fields including healthcare, by estimating risks and future needs effectively.

Z-score

The Z-score is a standard score indicating how many standard deviations an element is from the mean. It is a fundamental concept when working with normal distributions and helps in transferring data from a normal distribution to the standard normal distribution.
The Z-score is calculated using the formula: \[ Z = \frac{X - \mu}{\sigma} \] where:

• \(X\) is the value of the element you are examining,
• \(\mu\) is the mean of the distribution,
• \(\sigma\) is the standard deviation.

By transforming raw score data into a Z-score, we enable comparison between different datasets and make probability calculations more straightforward.
For example, determining a Z-score of 1.86 tells us that a stay of 10 hours is 1.86 standard deviations above the mean stay of 4.6 hours. Larger Z-score values indicate observations far from the mean, aiding in identifying outliers in the data.

Percentiles

Percentiles are a measure used in statistics to determine the relative standing of a value in a dataset. They are commonly used to understand how a particular value compares to the rest of the data. For normally distributed data, percentiles tell us what percentage of data points fall below a certain threshold.

• The 50th percentile, for instance, is the median of the dataset. Any percentile less than 50% indicates a score below the median, while those above 50% are above it.
• In the exercise, finding which length of stay is exceeded by 25% involves calculating the 75th percentile of the distribution. This is because 25% of visits exceed this threshold, leaving 75% below it.
• The conversion from a percentile to an actual observed value involves using the Z-score and mean to deduce the true data value.

Understanding percentiles can improve decision-making because they provide a context about where an observation stands compared to the general population distribution.

Statistical Modeling

Statistical modeling involves using mathematical frameworks to represent complex processes in a simplified manner. The normal distribution is frequently applied in these models due to its tractable properties and common occurrence in natural phenomena.
In the context of the exercise, the normally distributed model was used to predict probabilities and derive insights based on typical hospital stay durations.

• While it provides a baseline approximation, models have limitations. A key takeaway is that the normal distribution may predict impossible real-world scenarios, such as a negative length of stay.
• This calls for a critical assessment of model assumptions before applying them to extreme or boundary data points.
• Even so, statistical models are valuable for analyzing data, learning patterns, and making projections when correctly applied.

Effective use of statistical modeling involves balancing the robustness of mathematical logic with the nuances of real-world applicability.