Let x be a random variable that represents the level of glucose in the blood (mi

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, x has a distribution that is approximately normal, with mean ? = 72 and estimated standard deviation ? = 23. 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 approximately normal with ?x = 72 and ?x = 16.26.The probability distribution of x is approximately normal with ?x = 72 and ?x = 23.     The probability distribution of x is approximately normal with ?x = 72 and ?x = 11.50.The probability distribution of x is not normal.

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

Explanation / Answer

the PDF of normal distribution is = 1/? * ?2? * e ^ -(x-u)^2/ 2?^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 72
standard Deviation ( sd )= 23
a.
LESS THAN
P(X < 40) = (40-72)/23
= -32/23= -1.3913
= P ( Z <-1.3913) From Standard Normal Table
= 0.0821
b.
mean of the sampling distribution ( x ) = 72
standard Deviation ( sd )= 23/ Sqrt ( 2 ) =16.2635
sample size (n) = 2
The probability distribution of x is approximately normal with ?x = 72 and ?x = 16.26
LESS THAN
P(X < 40) = (40-72)/23/ Sqrt ( 2 )
= -32/16.2635= -1.9676
= P ( Z <-1.9676) From Standard NOrmal Table
= 0.02456
c.
mean of the sampling distribution ( x ) = 72
standard Deviation ( sd )= 23/ Sqrt ( 3 ) =13.2791
sample size (n) = 3
LESS THAN
P(X < 40) = (40-72)/23/ Sqrt ( 3 )
= -32/13.2791= -2.4098
= P ( Z <-2.4098) From Standard NOrmal Table
= 0.00798
d.
mean of the sampling distribution ( x ) = 72
standard Deviation ( sd )= 23/ Sqrt ( 5 ) =10.2859
sample size (n) = 5
LESS THAN
P(X < 40) = (40-72)/23/ Sqrt ( 5 )
= -32/10.2859= -3.1111
= P ( Z <-3.1111) From Standard NOrmal Table
= 0.00093

e
probability is decreased

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