Question

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?

Short answer

a. The proportion of healthy children with left atrial diameters less than 24 mm is the cumulative probability associated with the calculated Z score for 24 mm. \nb. Similarly, the proportion of healthy children with left atrial diameters greater than 32 mm is calculated as 1 minus the cumulative probability associated with the Z score for 32 mm. \nc. Proportion of healthy children with diameters between 25 and 30 mm is obtained by subtracting the cumulative probabilities associated with Z scores for 25 mm and 30 mm. \nd. The left atrial diameter for which only about 20% of healthy children exceed is calculated by transforming the Z score for 80% back to the value of left atrial diameter.

Step-by-step solution

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 \( Z_4 = \frac{30 - 26.4}{4.2} \).

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).

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.

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. \( X = Z * \sigma + \mu \)).