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 μ = 57 and estimated standard deviation σ = 34. 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.)
What is the probability that 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 7.1.
The probability distribution of x is not normal.The probability distribution of x is approximately normal with μx = 57 and σx = 17.00. The probability distribution of x is approximately normal with μx = 57 and σx = 24.04.The probability distribution of x is approximately normal with μx = 57 and σx = 34.
(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.)
a)
mean = 57 , sigam = 34
P(x< 40)
= P(z< (x -mean)/sigma)
= P(z< (40 -57)/34)
= P(z< -0.50)
= 0.3085
b)
n = 2
mean = 57 , std.dev = sigma/sqrt(n)
= 34/sqrt(2) = 24.04
The probability distribution of x is approximately normal with μx = 57 and σx = 24.04
P(x< 40)
= P(z< (x -mean)/sigma)
= P(z< (40 -57)/24.04)
= P(z< -0.71)
= 0.2397
c)
n = 3
mean = 57 , std.dev = sigma/sqrt(n)
= 34/sqrt(3) = 19.6299
The probability distribution of x is approximately normal with μx = 57 and σx = 19.63
P(x< 40)
= P(z< (x -mean)/sigma)
= P(z< (40 -57)/19.63)
= P(z< -0.87)
= 0.1932
d)
n= 5
mean = 57 , std.dev = sigma/sqrt(n)
= 34/sqrt(5) = 15.21
The probability distribution of x is approximately normal with μx =
57 and σx = 15.21
P(x< 40)
= P(z< (x -mean)/sigma)
= P(z< (40 -57)/15.21)
= P(z< -1,12)
= 0.1319
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