Suppose that in a population of adults between the ages of 18 and 49, glucose follows a normal distribution with a mean of 93.5 and standard deviation of 19.8. What is the probability that glucose exceeds 120 in this population

0.0904 or 9.04%

Step-by-step explanation:

Mean glucose (μ) = 93.5

Standard deviation (σ) = 19.8

In a normal distribution, the z-score for any glucose value, X, is given by:

For X = 120, the z-score is:

A z-score of 1.3384 corresponds to the 90.96th percentile of a normal distribution. Therefore, the probability that glucose exceeds 120 in this population is:

Probability that glucose exceeds 120 in this population is 0.09012.

Step-by-step explanation:

We are given that in a population of adults between the ages of 18 and 49, glucose follows a normal distribution with a mean of 93.5 and standard deviation of 19.8, i.e.; = 93.5  and   = 19.8 .

Let X = amount of glucose i.e. X ~ N()

Now, the Z score probability is given by;

           Z = ~ N(0,1)

So,Probability that glucose exceeds 120 in this population =P(X>120)

P(X > 120) = P( > ) = P(Z > 1.34) = 1 - P(Z <= 1.34)

                                                   = 1 - 0.90988 = 0.09012 .

4a. is 3 miles every hour

4b. is 1/3 of an hour per mile

step-by-step explanation: