Question

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

Answer 1

Probability that glucose exceeds 120 in this population is 0.09012.

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.;
`\mu` = 93.5 and
`\sigma` = 19.8 .

Let X = amount of glucose i.e. X ~ N(
`\mu = 93.5 , \sigma^(2) = 19.8^(2)`)

Now, the Z score probability is given by;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

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

P(X > 120) = P(
`(X-\mu)/(\sigma)` >
`(120-93.5)/(19.8)` ) = P(Z > 1.34) = 1 - P(Z <= 1.34)

= 1 - 0.90988 = 0.09012 .

Answer 2

0.0904 or 9.04%

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:

`Z= (X-\mu)/(\sigma)`

For X = 120, the z-score is:

`Z= (120-93.5)/(19.8)\\ Z=1.3384`

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:

`P(X>120) = 1-0.9096=0.0904 = 9.04\%`