Question

Question: Researchers in Pakistan wanted to better understand the effects of anthracycline (a chemotherapeutic drug) on the hearts of young children. More information on the details of their study can be found here: https://datadryad.org/stash/dataset/doi:10.5061/dryad.cg849
((The data set can be downloaded. After going to the website, the download is to the right)).

One variable the researchers calculated in these children was the E/A ratio. The E/A ratio is a measurement that reflects the health of the heart. The (E) velocity is greater than the late (A) velocity in a healthy heart. Particularly low or particularly high elevated E/A ratios indicate signs of heart issues.


The mean E/A ratio of the 110 patients in the study was 1.35. The standard deviation was 0.33.


1. One patient had an E/A ratio of 0.73. What is the z-score for this patient?


2. Assuming the population of E/A ratio scores is approximately normally distributed, what's the probability that a randomly selected person has a smaller E/A ratio than the patient in question 1?


3. Another patient had an E/A ratio of 1.96. What's the z-score for this patient? What's the probability that a randomly selected patient has a higher E/A ratio?


4. Which patient has a more extraordinary E/A ratio? Why? Be sure to refer to the z-scores to support your answer.

Answer 1

Using the normal distribution, it is found that:

1. His z-score was of Z = -1.88.

2. There is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

3. Z-score of z = 1.85, there is a 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

4. Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

Normal Probability Distribution

In a normal distribution with mean $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

• It measures how many standard deviations the measure is from the mean.

• After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

In this problem:

• The mean is of $\mu = 1.35$.

• The standard deviation is of $\sigma = 0.33$.

Item 1:

Considering his ratio, we have that X = 0.73, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.73 - 1.35}{0.33}$

$Z = -1.88$

His z-score was of Z = -1.88.

Item 2:

The probability is the p-value of Z = -1.88, hence, there is a 0.0301 = 3.01% probability that a randomly selected person has a smaller E/A ratio than the patient in question 1.

Item 3:

Score of X = 1.96, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1.96 - 1.35}{0.33}$

$Z = 1.85$

The probability is 1 subtracted by the p-value of Z = 1.85, hence, 1 - 0.9678 = 0.0322 = 3.22% probability that a randomly selected patient has a higher E/A ratio.

Item 4:

Due to the higher absolute value of the z-score, the first patient had a more extraordinary result.

More can be learned about the normal distribution at brainly.com/question/24663213