Answer:
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:
[tex]Z= \frac{X-\mu}{\sigma}[/tex]
For X = 120, the z-score is:
[tex]Z= \frac{120-93.5}{19.8}\\ Z=1.3384[/tex]
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:
[tex]P(X>120) = 1-0.9096=0.0904 = 9.04\%[/tex]
Answer:
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.; [tex]\mu[/tex] = 93.5 and [tex]\sigma[/tex] = 19.8 .
Let X = amount of glucose i.e. X ~ N([tex]\mu = 93.5 , \sigma^{2} = 19.8^{2}[/tex])
Now, the Z score probability is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
So,Probability that glucose exceeds 120 in this population =P(X>120)
P(X > 120) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{120-93.5}{19.8}[/tex] ) = P(Z > 1.34) = 1 - P(Z <= 1.34)
= 1 - 0.90988 = 0.09012 .
