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 = 2100. 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 = 2100. The probability distribution of x is approximately normal with μ_x = 6600 and σ_x = 1050.00.     The probability distribution of x is approximately normal with μ_x = 6600 and σ_x = 1484.92. 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 decreased as n increased. The probabilities increased as n increased.     The probabilities stayed the same as n increased.

(c) 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 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 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.

Answer 1

Solution :

Given that ,

mean = = 6600

standard deviation = = 2100

a) P(x < 3500) = P[(x - ) / < (3500 - 6600) / 2100]

= P(z < -1.48)

Using z table,

= 0.0694

b) n = 2

= = 6600

= / n = 2100/ 2 = 1484.92

The probability distribution of x is approximately normal with μx = 6600 and σx = 1484.92

P( < 3500) = P(( - ) / < (3500 - 6600) / 1484.92)

= P(z < -2.09)

Using z table

= 0.0183

c) n = 3

= = 6600

= / n = 2100/ 3 = 1212.44

The probability distribution of x is approximately normal with μx = 6600 and σx = 1212.44

P( < 3500) = P(( - ) / < (3500 - 6600) / 1212.44)

= P(z < -2.56)

Using z table

= 0.0052

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