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 = 56 and estimated standard deviation σ = 42. 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 three 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?

The probability distribution of x is not normal.

The probability distribution of x is approximately normal with μ_x = 56 and σ_x = 21.00.

The probability distribution of x is approximately normal with μ_x = 56 and σ_x = 29.70.

The probability distribution of x is approximately normal with μ_x = 56 and σ_x = 42.

What is the probability that x < 40? (Round your answer to three decimal places.)

C.) Repeat part (b) for n = 3 tests taken a week apart. (Round your answer to three decimal places.)

D.) Repeat part (b) for n = 5 tests taken a week apart. (Round your answer to three decimal places.)

Answer 1

Solution :

Given that ,

mean = = 56

standard deviation = = 42

a) P(x < 40) = P[(x - ) / < (40 - 56) / 42]

= P(z < -0.38)

Using z table,

= 0.352

b) n = 2

= = 56

= / n = 42/ 2 = 29.70

The probability distribution of x is approximately normal with μx = 56 and σx = 29.70

P( < 40) = P(( - ) / < (40 - 56) / 29.70)

= P(z < -0.54)

Using z table

= 0.295

c) n = 3

= = 56

= / n = 42/ 3 = 24.25

The probability distribution of x is approximately normal with μx = 56 and σx = 24.25.

P( < 40) = P(( - ) / < (40 - 56) / 24.25)

= P(z < -0.66)

Using z table

= 0.255

d) n = 5

= = 56

= / n = 42/ 5 = 18.78

The probability distribution of x is approximately normal with μx = 56 and σx = 18.78

P( < 40) = P(( - ) / < (40 - 56) / 18.78)

= P(z < -0.85)

Using z table

= 0.198