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 = 28. a test result x < 40 is an indication of severe excess insulin, and medication is usually prescribed. details my notes previous answers ask your teacher practice another 14. 1/1 points use salt (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. 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. • the probability distribution of x is approximately normal with μx = 87 and σx = 28. 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 n = 3 tests taken a week apart. (round your answer to four decimal places.) (d) repeat part (b) for n = 8 tests taken a week apart. (round your answer to four decimal places.)

Explanation:

Step1: Calculate the z - score for part (a)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). Here, \(x = 40\), \(\mu=87\), and \(\sigma = 28\).

$$z=\frac{40 - 87}{28}=\frac{-47}{28}\approx - 1.6786$$

Using the standard normal table (or a calculator with a normal - distribution function), \(P(X\lt40)=\Phi(-1.6786)\)

$$P(X\lt40)=0.0465$$

Step2: Calculate the z - score for part (b)

For the sample mean \(\bar{x}\), the formula for the z - score is \(z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}\). Here, \(n = 2\), \(\bar{x}=40\), \(\mu = 87\), and \(\sigma=28\)
\(\frac{\sigma}{\sqrt{n}}=\frac{28}{\sqrt{2}}\approx19.799\)

$$z=\frac{40 - 87}{19.799}=\frac{-47}{19.799}\approx - 2.373$$

Using the standard normal table (or a calculator with a normal - distribution function), \(P(\bar{X}\lt40)=\Phi(-2.373)\)

$$P(\bar{X}\lt40)=0.0089$$

Answer:

(a) \(0.0465\)
(b) \(0.0089\)