Let x be a random variable that represents the level of glucose in the blood (mi
ID: 3363984 • Letter: L
Question
Let x be a random variable that represents the level of glucose in the blood (milligrams per deciliter of blood) after a 12 hour fast. Assume that for people under 50 years old, xhas a distribution that is approximately normal, with mean = 53 and estimated standard deviation = 27. A test result x < 40 is an indication of severe excess insulin, and medication is usually prescribed.
(a) What is the probability that, on a single test, x < 40? (Round your answer to four decimal places.)
(b) Suppose a doctor uses the average x for two tests taken about a week apart. What can we say about the probability distribution of x? Hint: See Theorem 6.1.
The probability distribution of x is not normal.The probability distribution of x is approximately normal with x = 53 and x = 13.50. The probability distribution of x is approximately normal with x = 53 and x = 27.The probability distribution of x is approximately normal with x = 53 and x = 19.09.
What is the probability that x < 40? (Round your answer to four decimal places.)
(c) Repeat part (b) for n = 3 tests taken a week apart. (Round your answer to four decimal places.)
(d) Repeat part (b) for n = 5 tests taken a week apart. (Round your answer to four decimal places.)
(e) Compare your answers to parts (a), (b), (c), and (d). Did the probabilities decrease as n increased?
YesNo
Explain what this might imply if you were a doctor or a nurse.The more tests a patient completes, the stronger is the evidence for lack of insulin.The more tests a patient completes, the stronger is the evidence for excess insulin. The more tests a patient completes, the weaker is the evidence for lack of insulin.The more tests a patient completes, the weaker is the evidence for excess insulin.
Explanation / Answer
sOLUTIONa:
mean=53
sd=27
P(X<40)
z=x-mean/sd
=40-53/27
=-13/27
=-0.48
P(Z<-0.48)
=1-P(Z<0.48)
=1-0.6844
=0.3156
ANSWER:0.3156
Solutionb:
for two tests taken about a week
n=2
sample mean=pop mean=53
sd=sd/sqrt(27/sqrt(2)=19.09
The probability distribution of x is approximately normal with x = 53 and x = 19.09.
Solutionc:
P(X <40)
z=40-53/19.09
z=-0.68
P(Z<-0.68)
=1-P(Z<0.68)
=1-0.7517
=0.2483
answer:0.2483
SOLUTIONC:
P(X<40) FOR N=5
Z=40-53/27/SQRT(5)
=-1.08
P(Z<-1.08)
=1-P(Z<1.08)
=1-0.8599
=0.1401
ANSWER:0.1401
the probabilities decrease as n increased?
YesNo
YES
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