Let x be a random variable that represents the level of glucose in the blood (mi
ID: 3392391 • 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, x has a distribution that is approximately normal, with mean = 50 and estimated standard deviation = 35. 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.
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.)
Explanation / Answer
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 40
u = mean = 50
s = standard deviation = 35
Thus,
z = (x - u) / s = -0.285714286
Thus, using a table/technology, the left tailed area of this is
P(z < -0.285714286 ) = 0.387548481 [ANSWER]
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b)
It will have a new standard deviation, which is 1/sqrt(2) of the original.
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 40
u = mean = 50
n = sample size = 2
s = standard deviation = 35
Thus,
z = (x - u) * sqrt(n) / s = -0.404061018
Thus, using a table/technology, the left tailed area of this is
P(z < -0.404061018 ) = 0.343083925 [ANSWER]
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c)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 40
u = mean = 50
n = sample size = 3
s = standard deviation = 35
Thus,
z = (x - u) * sqrt(n) / s = -0.494871659
Thus, using a table/technology, the left tailed area of this is
P(z < -0.494871659 ) = 0.310345359 [ANSWER]
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d)
We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as
x = critical value = 40
u = mean = 50
n = sample size = 5
s = standard deviation = 35
Thus,
z = (x - u) * sqrt(n) / s = -0.638876565
Thus, using a table/technology, the left tailed area of this is
P(z < -0.638876565 ) = 0.261451617 [ANSWER]
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