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 = 6250 and estimated standard deviation σ = 2750. 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 for two tests taken about a week apart. What can we say about the probability distribution of x?

- The probability distribution of x is approximately normal with μ_x = 6250 and σ_x = 1944.54.
- The probability distribution of x is approximately normal with μ_x = 6250 and σ_x = 2750.
- The probability distribution of x is approximately normal with μ_x = 6250 and σ_x = 1375.00.
- The probability distribution of x is not normal.

What is the probability of x < 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 stayed the same as n increased.
- The probabilities increased as n increased.
- The probabilities decreased as n increased.

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

- 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.
- 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 leukopeni.
- 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.

Answer 1

Solution :

Given that ,

mean = = 6250

standard deviation = = 2750

a) P(x < 3500) = P[(x - ) / < (3500 - 6250) / 2750]

= P(z < -1.00)

Using z table,

= 0.1587

b) n = 2

= = 6250

= / n = 2750/ 2 = 1944.54

The probability distribution of x is approximately normal with μx = 6250 and σx = 1944.54

P( < 3500) = P(( - ) / < (3500 - 6250) / 1944.54)

= P(z < -1.41)

Using z table

= 0.0793

c) n = 3

= = 3500

= / n = 2750/ 3 = 1587.71

The probability distribution of x is approximately normal with μx = 6250 and σx = 1587.71

P( < 3500) = P(( - ) / < (3500 - 6250) / 1587.71)

= P(z < -1.73)

Using z table

= 0.0.0418

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.