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 = 7300 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 μx = 7300 and σx = 2121.32. The probability distribution of x is not normal. The probability distribution of x is approximately normal with μx = 7300 and σx = 3000. The probability distribution of x is approximately normal with μx = 7300 and σx = 1500.00. 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 decreased as n increased. The probabilities stayed the same as n increased. The probabilities increased 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 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 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 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

Answer 1

a) P(X < 3500)

= P((X - )/ < (3500 - )/)

= P(Z < (3500 - 7300)/3000)

= P(Z < -1.27)

= 0.1020

b) = 7300

=

        = 3000/ = 2121.32

The probability distribution of is approximately normal with = 7300 and = 2121.32.

P( < 3500)

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

= P(Z < (3500 - 7300)/2121.32)

= P(Z < -1.79)

= 0.0367

c) P( < 3500)

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

= P(Z < (3500 - 7300)/(3000/))

= P(Z < -2.19)

= 0.0143

d) The probabilities decreased as n increased.

It would be an extremely rare event for a person to have two or three tests below 3500 purely by chance.The person probably does not have leukopenia.