Diabetic patients have normally distributed cholesterol with mean 200 and standard deviation=10. Find the percentage of patients whose cholesterol is between 198 mg/dL and

207 mg/dL ?

The percentage of patients whose cholesterol is between 198 mg/dL and 207 mg/dL is 33.73%

Step-by-step explanation:

To calculate this proportion, we follow the probability route, using the z-score statistics

Mathematically;

z-score = (x-mean)/SD

from the question, mean = 200 and SD = 10

So for 198

z-score = (198-200)/10 = -2/10 = -0.2

For 207

z-score = (207-200)/10 = 7/10 = 0.7

So the probability we want to calculate is;

P(-0.2<z<0.7)

Mathematically this can be calculated as;

P(z<0.7) - P(z<-0.2)

We can calculate the required probability using the standard normal distribution table

P(-0.2<x<0.7) = 0.3373 from the standard distribution table

So it is this 0.3373 that we now convert to percentage and that is 33.73%

step-by-step explanation:

i have done this all on a piece of paper but idk how to put it on here cuz it needs to be drawn