glucose level in patients free of diabetes are assumed to follow a normal distribution with a mean of 120 and a standard deviation of 16. (20 pts) What proportion of patients has glucose levels exceeding 115?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
To find the proportion of patients with glucose levels exceeding 115, we need to calculate the area under the normal distribution curve that is to the right of 115.
We know that the glucose levels of patients are normally distributed with a mean (μ) of 120 and a standard deviation (σ) of 16.
To calculate the proportion of patients with glucose levels exceeding 115, we can use a standard normal distribution table or a statistical calculator. However, I will demonstrate how to calculate it manually using the standard normal distribution.
The first step is to standardize the value of 115 using the z-score formula:
z = (x - μ) / σ
Where:
x = 115 (value we want to standardize)
μ = 120 (mean)
σ = 16 (standard deviation)
Substituting these values into the formula:
z = (115 - 120) / 16 = -0.3125
Now, we need to find the area to the right of this z-score on the standard normal distribution table.
Using the table or a calculator, we can find that the area to the left of -0.3125 is approximately 0.3775.
To calculate the area to the right of -0.3125, we subtract the value from 1:
Area to the right = 1 - 0.3775 = 0.6225
Therefore, approximately 0.6225 (or 62.25%) of patients have glucose levels exceeding 115.
To summarize, the proportion of patients with glucose levels exceeding 115 is approximately 0.6225, or 62.25%.