. It is known that the glucose level in blood of diabetic persons follows a normal distribution model with mean 106 mg/100 ml and standard deviation 8 mg/100 ml. a. Calculate the 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?

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 and standard deviation , the z-score of a measure X is given by:

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

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

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

has a p-value of 0.9599

X = 90

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.

i think it is c, not positive though

step-by-step explanation: