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Answer:
a. 0.9599 = 95.99% probability of a random diabetic person having a glucose level less than 120 mg/100 ml.
b. 0.9371 = 93.71% of people have a glucose level between 90 and 120 mg/100 ml.
Step-by-step explanation:
Normal Probability Distribution
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
Mean 106 mg/100 ml and standard deviation 8 mg/100 ml
This means that [tex]\mu = 106, \sigma = 8[/tex]
a. Calculate the probability of a random diabetic person having a glucose level less than 120 mg/100 ml.
This is the p-value of Z when X = 120. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{120 - 106}{8}[/tex]
[tex]Z = 1.75[/tex]
[tex]Z = 1.75[/tex] has a p-value of 0.9599
0.9599 = 95.99% probability of a random diabetic person having a glucose level less than 120 mg/100 ml.
b. What percentage of persons have a glucose level between 90 and 120 mg/100 ml?
The proportion is the p-value of Z when X = 120 subtracted by the p-value of Z when X = 90. So
X = 120
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{120 - 106}{8}[/tex]
[tex]Z = 1.75[/tex]
[tex]Z = 1.75[/tex] has a p-value of 0.9599
X = 90
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{120 - 106}{8}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] has a p-value of 0.0228
0.9599 - 0.0228 = 0.9371
0.9371 = 93.71% of people have a glucose level between 90 and 120 mg/100 ml.
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