At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
Answer:
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 106, \sigma = 8[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{120 - 106}{8}[/tex]
[tex]Z = 1.75[/tex]
[tex]Z = 1.75[/tex] 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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{120 - 106}{8}[/tex]
[tex]Z = 1.75[/tex]
[tex]Z = 1.75[/tex] has a p-value of 0.9599
X = 90
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{120 - 106}{8}[/tex]
[tex]Z = -2[/tex]
[tex]Z = -2[/tex] 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.
We hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
