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 = 6600 and estimated standard deviation σ = 3000. 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 = 6600 and σ_x = 3000.The probability distribution of x is approximately normal with μ_x = 6600 and σ_x = 1500.00.    The probability distribution of x is not normal.The probability distribution of x is approximately normal with μ_x = 6600 and σ_x = 2121.32.

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 increased as n increased.The probabilities decreased as n increased.    The probabilities stayed the same 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 does not have 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 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.

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	6600
std deviation   =σ=	3000.0000

probability =

P(X<3500)

=

P(Z<-1.03)=

0.1515

b)

The probability distribution of x is approximately normal with μ_x = 6600 and σ_x = 2121.32.

probability =

P(X<3500)

=

P(Z<-1.46)=

0.0721

c)

probability =

P(X<3500)

=

P(Z<-1.79)=

0.0367

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.