Question

Let x be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that x has a distribution that is approximately normal, with mean μ = 7500 and estimated standard deviation σ = 2850. A test result of x < 3500 is an indication of leukopenia. This indicates bone marrow depression that may be the result of a viral infection.

(a) What is the probability that, on a single test, x is less than 3500? (Round your answer to four decimal places.)


(b) Suppose a doctor uses the average x-bar for two tests taken about a week apart. What can we say about the probability distribution of x-bar?

The probability distribution of x-bar is approximately normal with μ_x-bar = 7500 and σ_x-bar = 2850.

The probability distribution of x-bar is not normal.     

The probability distribution of x-bar is approximately normal with μ_x-bar = 7500 and σ_x-bar = 1425.00.

The probability distribution of x-bar is approximately normal with μ_x-bar = 7500 and σ_x-bar = 2015.25.

What is the probability of x-bar < 3500? (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) Compare your answers to parts (a), (b), and (c). How did the probabilities change as n increased?

The probabilities decreased as n increased.

The probabilities increased as n increased.     

The probabilities stayed the same as n increased.

If a person had x-bar < 3500 based on three tests, what conclusion would you draw as a doctor or a nurse?

It would be an extremely rare event for a person to have two or three tests below 3,500 purely by chance. The person probably does not have leukopenia.

It would be an extremely rare event for a person to have two or three tests below 3,500 purely by chance. The person probably has leukopenia.     

It would be a common event for a person to have two or three tests below 3,500 purely by chance. The person probably has leukopenia.

It would be a common event for a person to have two or three tests below 3,500 purely by chance. The person probably does not have leukopenia.

Answer 1

Solution :

Given that ,

mean = = 7500

standard deviation = = 2850

a) P(x < 3500) = P[(x - ) / < (3500 - 7500) / 2850]

= P(z < -1.40)

Using z table,

= 0.0808

b) n = 2

= = 7500

= / n = 2850/ 2 = 2015.25

The probability distribution of x is approximately normal with μx = 7500 and σx = 2015.25

P( < 3500) = P(( - ) / < (3500 - 7500) / 2015.25)

= P(z < -1.98)

Using z table

= 0.0239

c) n = 3

= = 7500

= / n = 2850/ 3 = 1645.45

The probability distribution of x is approximately normal with μx = 7500 and σx = 1645.45

P( < 3500) = P(( - ) / < (3500 - 7500) / 1645.45)

= P(z < -2.43)

Using z table

= 0.0075

d) The probabilities decreased as n increased.

It would be an extremely rare event for a person to have two or three tests below 3,500 purely by chance. The person probably has leukopenia