91Ó°ÊÓ

The size of the left upper chamber of the heart is one measure of cardiovascular health. When the upper left chamber is enlarged, the risk of heart problems is increased. The paper "Left Atrial Size Increases with Body Mass Index in Children" (International Journal of Cardiology [2009]: 1-7) described a study in which the left atrial size was measured for a large number of children ages 5 to 15 years. Based on these data, the authors concluded that for healthy children, left atrial diameter was approximately normally distributed with a mean of \(26.4 \mathrm{~mm}\) and a standard deviation of \(4.2 \mathrm{~mm}\). a. Approximately what proportion of healthy children have left atrial diameters less than \(24 \mathrm{~mm}\) ? b. Approximately what proportion of healthy children have left atrial diameters greater than \(32 \mathrm{~mm}\) ? c. Approximately what proportion of healthy children have left atrial diameters between 25 and \(30 \mathrm{~mm}\) ? d. For healthy children, what is the value for which only about \(20 \%\) have a larger left atrial diameter?

Step by step solution

01

Calculate the Z-score for each value

The Z score is calculated using the formula \( Z = \frac{X - \mu}{\sigma} \), where X is the value, \( \mu \) is the mean and \( \sigma \) is the standard deviation. So, for the diameter less than 24 mm, the Z score would be \( Z_1 = \frac{24 - 26.4}{4.2} \), for the diameter greater than 32 mm, the Z score would be \( Z_2 = \frac{32 - 26.4}{4.2} \), and for diameters between 25 and 30 mm we calculate two Z scores \( Z_3 = \frac{25 - 26.4}{4.2} \) and \( Z_4 = \frac{30 - 26.4}{4.2} \).

02

Look for the probability associated with the Z scores in the Z table

After calculating the values for Z, find the corresponding probabilities in the Z table (they'll describe the proportion of healthy children).

03

Subtract probabilities for values between an interval

for the part c, you have two Z scores and therefore two probabilities. To find the proportion of children with diameters between 25 and 30 mm, subtract the two probabilities.

04

Find the Z score for the value below which 20% of the distribution lies

To find the left atrial diameter that only about 20% of the healthy children exceed, look in your Z-table for the Z score that corresponds to 80% (since we want 20% to be above this value). Then, use the equation for the Z score to solve for X (i.e. \( X = Z * \sigma + \mu \)).

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Key Concepts

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

Z-score

In the realm of statistics, particularly when dealing with normal distributions, the Z-score plays a vital role. The Z-score helps us understand how far a particular data point is from the mean, in terms of standard deviations. For instance, if you have a left atrial diameter of 24 mm, you can calculate its Z-score using the formula: \[ Z = \frac{X - \mu}{\sigma} \]where:

• \( X \) is the value of interest—in our example, the diameter of 24 mm.
• \( \mu \) is the mean, which was found to be 26.4 mm.
• \( \sigma \) is the standard deviation, which is 4.2 mm in this case.

This Z-score helps in finding the probability of a child having a left atrial diameter less than 24 mm by referencing a standard normal distribution table. It tells you the proportion of the data that falls to the left of this score on a normal distribution curve. A lower Z-score indicates that a value is below the mean, while a higher Z-score shows it is above the mean.

Standard Deviation

Standard deviation is a statistical measure that quantifies the amount of variation or dispersion in a set of values. When addressing normal distributions, it tells us how spread out the values in a dataset are around the mean. In the context of cardiovascular health, such as measuring the diameter of the left atrial chamber, the standard deviation helps to understand how individual measurements might differ from the average value. In our exercise, the standard deviation is given as 4.2 mm. A smaller standard deviation means the data points tend to be closer to the mean, implying less variability among the measurements of the left atrial diameter. Conversely, a larger standard deviation indicates that the values are more spread out. Understanding standard deviation is crucial when interpreting Z-scores because it provides the scale for the distances calculated. Thus, the variance captured by the standard deviation is directly used for calculating and interpreting Z-scores.

Cardiovascular Health

Cardiovascular health refers to the well-being of the heart and blood vessels. In the exercise, the focus is on the left atrial diameter of children, which is a significant indicator of heart health. The left atrium is one of four chambers of the heart. Its size can indicate potential health risks. If the chamber is enlarged, there could be an increased risk of cardiovascular complications. Therefore, measuring and comparing the diameters of the left atrium is important in pediatric health studies. This exercise utilized measurements to determine a normal range of diameters for the left atrial chamber in healthy children. By analyzing these measurements, researchers can distinguish between what's considered a healthy size and what's not, potentially indicating a need for further medical evaluation.

Probability Distribution

Probability distribution is a fundamental concept in statistics that describes how the probabilities are distributed over the different possible outcomes. In the context of this exercise, we are dealing with a normal distribution. A normal distribution is characterized by its bell-shaped curve, where most of the data points cluster around the mean. Within a normal distribution, about 68% of the values lie within one standard deviation of the mean, 95% within two, and 99.7% within three. This is particularly useful when predicting the proportion of children likely to have a left atrial diameter within a specific range. For example, to find the probability of a diameter between 25 mm and 30 mm, we use the Z-score of these boundaries and reference a standard normal distribution table to find the corresponding probabilities. This method helps in making informed predictions about cardiovascular health by providing insights into where most healthy children's measurements fall in comparison to established medical research data.