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An article in the Journal of Cardiovascular Magnetic Resonance ["Right Ventricular Ejection Fraction Is Better Reflected by Transverse Rather Than Longitudinal Wall Motion in Pulmonary Hypertension" (2010, Vol. 12(35) ] discussed a study of the regional right-ventricle transverse wall motion in patients with pulmonary hypertension (PH). The right-ventricle ejection fraction (EF) was approximately normally distributed with a mean and a standard deviation of 36 and \(12,\) respectively, for PH subjects, and with mean and standard deviation of 56 and 8 , respectively, for control subjects. a. What is the EF for PH subjects exceeded with \(5 \%\) probability? b. What is the probability that the EF of a control subject is less than the value in part (a)? c. Comment on how well the control and PH subjects can be distinguished by EF measurements.

Short Answer

Expert verified
a. 55.74 b. 48.71% c. There is significant overlap; distinction based on EF may not be effective.

Step by step solution

01

Understanding the Problem

Given that the right-ventricle ejection fraction (EF) is normally distributed for PH subjects with a mean \( \mu_{PH} = 36 \) and standard deviation \( \sigma_{PH} = 12 \). For control subjects, the mean is \( \mu_{Control} = 56 \) with standard deviation \( \sigma_{Control} = 8 \). We need to calculate the EF value exceeded by only 5% of PH subjects and then find out how this value compares with control subjects.
02

Finding the EF Value Exceeded by only 5% of PH Subjects

To find the EF value exceeded by only 5% of PH subjects, we need to find the 95th percentile of the distribution of PH subjects’ EF. We will use the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} \]Set \( Z = 1.645 \) for the 95th percentile (since 5% in the tail):\[ X = Z \times \sigma_{PH} + \mu_{PH} \]\[ X = 1.645 \times 12 + 36 = 19.74 + 36 = 55.74 \]Thus, the EF value for PH subjects exceeded with 5% probability is approximately 55.74.
03

Calculating the Probability for Control Subjects

Now, we need to find the probability that a control subject's EF is less than 55.74. We calculate the Z-score for this value in terms of the control subjects:\[ Z = \frac{X - \mu_{Control}}{\sigma_{Control}} = \frac{55.74 - 56}{8} \approx -0.0325 \]Using the standard normal distribution table, we find the probability corresponding to \( Z = -0.0325 \), which is approximately 0.4871. Thus, the probability that a control subject has an EF less than 55.74 is about 48.71%.
04

Evaluation of Distinction Between PH and Control Subjects

The EF value that exceeds 95% of PH subjects (55.74) falls almost at the median for control subjects (50th percentile). Since there is significant overlap (control subjects have a high probability of having an EF less than 55.74), distinguishing PH and control subjects based solely on EF measurements may not be highly effective.

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Key Concepts

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

Normal Distribution
In the world of statistics, the normal distribution is a fundamental concept that is often used to model real-world data. This distribution is famously known as the bell curve due to its symmetrical, bell-shaped appearance when graphed.
For any normally distributed dataset, most tendencies of the data will cluster around the mean, with fewer instances appearing further away from the mean as one moves towards the extreme values—the tails of the curve. In health research, understanding normal distribution is crucial as it allows for predicting the probability of certain outcomes.

For the example given, the right-ventricle ejection fraction (EF) for pulmonary hypertension (PH) and control subjects are both modeled using a normal distribution with respective means and standard deviations. This approach helps in determining outcomes such as the percent of EF values above a certain threshold.
Z-score Calculation
The Z-score is a statistical measure that describes a data point's relation to the mean of a group of data points. It is measured in terms of standard deviations from the mean. Calculating a Z-score involves subtracting the mean score from the data point and then dividing the result by the standard deviation:
\[ Z = \frac{X - \mu}{\sigma} \]
For our problem, it's used to find the specific EF value for PH subjects exceeded by only 5% probability. This entails determining the 95th percentile, as only 5% fall beyond this threshold to the right. By setting our Z-score to 1.645 for this percentile, we compute the EF value using the mean and standard deviation values for PH subjects, leading to an EF of approximately 55.74.

This calculation highlights how Z-scores can be utilized to ascertain probabilities and percentiles within normally distributed datasets, offering a powerful tool for researchers and data analysts.
Probability Analysis
Probability analysis helps determine the likelihood of various outcomes occurring under certain conditions. In relation to our exercise, once we calculated the EF exceeded by only 5% of PH subjects, we then analyzed its significance in terms of control subjects.
To tackle this, we recalculated the Z-score using control subjects' statistical parameters to determine the probability of their EF being below the critical value of 55.74. This required using the formula:
\[ Z = \frac{X - \mu_{Control}}{\sigma_{Control}} \]
Here, the calculation resulted in a Z-score of approximately -0.0325, falling close to zero. This revealed a probability of about 48.71% that a control subject's EF is less than 55.74. Understanding these probabilities offers valuable insight, especially in distinguishing between groups like PH patients and control subjects based solely on their characteristics.
Medical Data Interpretation
Medical data interpretation is about making sense of numerical data to foster better diagnostic, treatment, or preventive measures. In this scenario, we'd focused on how the EF values for PH and control subjects compare. Ultimately, the ability to distinguish PH patients from control subjects based solely on EF measurements is limited, as observed from the significant probability overlap. While PH subjects' EF values exceeded by only 5% mirror the median values for control subjects, this leads to substantial difficulty in categorizing them precisely.
It's essential for medical researchers and healthcare professionals to interpret such overlaps correctly. This can facilitate better decision-making when EF measurements alone can't clearly distinguish between patient groups, potentially guiding the search for additional biomarkers or diagnostic tools.

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