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Fasting glucose levels in patients free of diabetes are assumed to follow a normal distribution with...

Fasting glucose levels in patients free of diabetes are assumed to follow a normal distribution with a mean of 84.5 mg/dL and a standard deviation of 26. Indicate the 95th percentile of the mean fasting glucose for a sample of 30 patients.
a.

"X-bar" = 127.27
b.

"X-bar" = 92.3087
c.

"X-bar" = 135.46
d.

"X-bar" = 93.804
Math   Statistics-And-Probability   
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Homework Answers

Answer #1


= 84.5

= 26

n = 30

SE = /

= 26/ = 4.7469

95th percentile corresponds to 0.95 - 0.50 = 0.45 from mid value to Z on RHS.

Table of Area Under Standard Normal Curve gives Z = 1.645

So,

Z = 1.645 = (X - 84.5)/4.7469

So,

X = 84.5 + (1.645 X 4.7469) = 92.3087

So,

Correct option:

b.    = 92.3087

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