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 = 71 and estimated standard deviation σ = 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 7.1.

The probability distribution of x is not normal.The probability distribution of x is approximately normal with μ_x = 71 and σ_x = 35.    The probability distribution of x is approximately normal with μ_x = 71 and σ_x = 17.50.The probability distribution of x is approximately normal with μ_x = 71 and σ_x = 24.75.

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.

Answer 1

Solution :

Given that ,

mean = = 71

standard deviation = = 35

a) P(x < 40) = P[(x - ) / < (40 - 71) / 35]

= P(z < -0.89)

Using z table,

= 0.1867

b) n = 2

= = 71

= / n = 35/ 2 = 24.75

The probability distribution of x is approximately normal with μx = 71 and σx = 24.75.

P( < 40) = P(( - ) / < (40 - 71) / 24.75)

= P(z < -1.25)

Using z table

= 0.1056

c) n = 3

= = 71

= / n = 35/ 3 = 20.21

The probability distribution of x is approximately normal with μx = 71 and σx = 20.21.

P( < 40) = P(( - ) / < (40 - 71) / 20.21)

= P(z < -1.53)

Using z table

= 0.0630

d) n = 5

= = 71

= / n = 35/ 5 = 15.65

The probability distribution of x is approximately normal with μx = 71 and σx = 15.65

P( < 40) = P(( - ) / < (40 - 71) / 15.65)

= P(z < -1.98)

Using z table

= 0.0239