Problem 94 - Applied Statistics and Probability for Engineers

Question

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.

Step-by-Step Solution

Answer

(a) EF for PH exceeded with 5% probability: 55.74. (b) Probability for a control EF < 55.74 is 48.68%. (c) PH and control subjects are distinguishable by EF due to different means.

Step 1: Define the Normal Distribution for PH Subjects

The right ventricle ejection fraction (EF) for PH subjects follows a normal distribution with a mean (\( \mu \)) of 36 and a standard deviation (\( \sigma \)) of 12. We denote EF as a random variable \( X \sim N(36, 12^2) \).

Step 2: Find the 95th Percentile for PH Subjects

To find the EF value exceeded with 5% probability for PH subjects, we need to find the 95th percentile of the normal distribution \( N(36, 12^2) \). We use the z-table to find the z-score that corresponds to 0.95 probability. The closest z-score is 1.645. Use this z-score to find the EF value:\[ EF_{5\%} = \mu + z \cdot \sigma = 36 + 1.645 \times 12 = 55.74 \]

Step 3: Define the Normal Distribution for Control Subjects

For control subjects, the right ventricle ejection fraction (EF) is also normally distributed with a mean of 56 and a standard deviation of 8. Thus, let \( Y \sim N(56, 8^2) \).

Step 4: Calculate Probability for Control Subjects

Using the EF value from part (a), which is 55.74, find the probability that the EF of a control subject is less than this value. This probability corresponds to the cumulative density function (CDF) of \( Y \) at 55.74. We find the standardized z-score:\[ z = \frac{55.74 - 56}{8} = -0.0325 \]From the z-table, the probability corresponding to \( z = -0.0325 \) is approximately 0.4868.

Step 5: Analyze Distinguishability Based on EF

To determine how well control and PH subjects can be distinguished by EF measurements, consider their mean EFs and standard deviations. The mean EF of control subjects is higher at 56 compared to 36 for PH subjects, but there is some overlap due to their standard deviations. However, the means are sufficiently different that they could be distinguished in many cases, especially towards the tails of the distributions.

Key Concepts

Understanding Standard Deviation

Standard deviation is a measure that tells us how spread out the values in a data set are around the mean. In simpler terms, it shows us the typical distance of the values from the average. This concept is crucial because it helps us understand the variability within the data.

For example, in the context of the right ventricle ejection fraction (EF) for PH subjects, the standard deviation is 12. This means that the EF values for this group tend to vary by 12 units from their mean, which is 36. Similarly, for control subjects, the EF standard deviation is 8, indicating less variability around their mean of 56.

The smaller the standard deviation, the more tightly clustered around the mean the values are. When interpreting data like heart ejection fractions, a smaller standard deviation implies more consistency in measurements. A larger standard deviation would suggest greater variability.

The Role of Z-Score

A z-score allows us to understand how far away a particular value is from the mean of the data, in terms of standard deviations. It’s a vital tool for calculating probabilities in a normal distribution.

The z-score is calculated using the formula: \[ z = \frac{(X - \mu)}{\sigma} \]where \(X\) is the value of interest, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

For PH subjects, when we need to find an EF value that only 5% of the subjects exceed, we identify the z-score corresponding to 95% since that's the complement. By referring to a z-table or standard normal distribution table, we find that this z-score is approximately 1.645. This z-score helps in figuring out the exact EF value by converting it back using the given mean and standard deviation.

Interpreting Percentiles

Percentiles are a useful way to understand where a particular value stands in a distribution. The 95th percentile, for example, is the value below which 95% of the data falls. This is significant in many real-world contexts because it tells us about the relative position of data within the whole distribution.

In the case of PH subjects, finding the 95th percentile means we are looking for an EF level that 95% of individuals have or fall below. Conversely, it’s the EF level that only 5% exceed. This was calculated to be approximately 55.74, using the mean and standard deviation provided for the PH subjects.

Percentiles give us an intuitive grasp of where data values fall, allowing for easy comparison between different data sets or groups.

Exploring the Cumulative Density Function (CDF)

The cumulative density function (CDF) is an essential statistical tool used to determine the probability that a random variable will be less than or equal to a certain value. It's a function derived from the normal distribution and provides a simple way to calculate probabilities.

The CDF is what we used when solving for the probability of a control subject’s EF being below 55.74. After finding the z-score (using the formula to standardize), we applied the CDF from the normal distribution table. This calculation showed that the probability was approximately 0.4868, indicating there’s about a 48.68% chance of a control subject having an EF less than 55.74.

Using the CDF helps in understanding data distributions more comprehensively, providing insights into probabilities across different ranges or scenarios.