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 μ = 87 and standard deviation σ = 28. 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?
o the probability distribution of x is not normal.
o the probability distribution of x is approximately normal with μx = 87 and σx = 14.
o the probability distribution of x is approximately normal with μx = 87 and σx = 19.7990.
o the probability distribution of x is approximately normal with μx = 87 and σx = 28.
what is the probability that x < 40? (round your answer to four decimal places.)
(c) repeat part (b) for a = 3 tests taken a week apart. (round your answer to four decimal places.)
(d) repeat part (b) for a = 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?
o yes
o no
(f) explain what this might imply if you were a doctor or a nurse.
o the more tests a patient completes, the stronger is the evidence for excess insulin.
o the more tests a patient completes, the stronger is the evidence for lack of insulin.
o the more tests a patient completes, the weaker is the evidence for lack of insulin.
o the more tests a patient completes, the weaker is the evidence for excess insulin.

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{\bar{x}-\mu_{\bar{x}}}{\sigma_{\bar{x}}}\). Given \(\bar{x} = 40\), \(\mu_{\bar{x}}=87\), and \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\). For part (a), \(n = 1\), so \(\sigma_{\bar{x}}=\sigma=28\).

$$z=\frac{40 - 87}{28}=\frac{- 47}{28}\approx - 1.68$$

Step2: Find the probability using the standard normal table

We want to find \(P(\bar{X}<40)\). Using the standard normal table (or a calculator with a normal - distribution function), for \(z=-1.68\), \(P(Z < - 1.68)=0.0465\)

Step3: For \(n = 3\)

First, calculate \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{28}{\sqrt{3}}\approx16.17\)

$$z=\frac{40 - 87}{16.17}=\frac{-47}{16.17}\approx - 2.91$$

Then, using the standard normal table, \(P(Z < - 2.91)=0.0019\)

Step4: For \(n = 5\)

Calculate \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}=\frac{28}{\sqrt{5}}\approx12.52\)

$$z=\frac{40 - 87}{12.52}=\frac{-47}{12.52}\approx - 3.75$$

Using the standard normal table, \(P(Z < - 3.75)\approx0.0001\)

Answer:

(a) \(0.0465\)
(b) For \(n = 3\): \(0.0019\); For \(n = 5\): \(0.0001\)