Using the normal distribution and the central limit theorem, it is found that:
- There is a 0.0359 = 3.59% probability that a randomly-selected person will find an indication of severe excess insulin.
- Considering the mean of two tests, there is a 0.0054 = 0.54% probability that the person will find an indication of severe excess insulin.
- Three tests: 0.0009 = 0.09%.
- Five tests: 0% probability.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable that has mean represented by and standard deviation represented by is given by the following rule:
- The z-score measures how many standard deviations the measure X is above or below the mean, depending if the z-score score is positive or negative.
- From the z-score table, the p-value associated with the z-score is found, which represents the percentile of the measure X in the distribution of interest.
- By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation .
The mean and the standard deviation of the glucose levels are given, respectively, by:
The probability of a reading of less than 40 mg/dl(severe excess insulin) is the p-value of Z when X = 40, hence:
Z = (40 - 85)/25
Z = -1.8.
Z = -1.8 has a p-value of 0.0359.
For the mean of two tests, the standard error is:
s = 25/sqrt(2) = 17.68.
Hence, by the Central Limit Theorem:
Z = (40 - 85)/17.68
Z = -2.55.
Z = -2.55 has a p-value of 0.0054.
For 3 tests, we have that:
s = 25/sqrt(3) = 14.43.
Z = (40 - 85)/14.43
Z = -3.12.
Z = -3.12 has a p-value of 0.0009.
For 5 tests, we have that:
s = 25/sqrt(5) = 11.18.
Z = (40 - 85)/11.18
Z = -4.03
Z = -4.03 has a p-value of 0.
More can be learned about the normal distribution at brainly.com/question/4079902
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