Divide. 7/$7.42 =$___
1 answer:
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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.
<h3>Normal Probability Distribution</h3>In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- 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
.
- The standard deviation is of
.
Item 1:
Considering his ratio, we have that X = 0.73, hence:
His z-score was of Z = -1.88.
Item 2:
The probability is the <u>p-value of Z = -1.88</u>, 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:
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