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 = 6100 and estimated standard deviation σ = 2000. 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? Pick one

-The probability distribution of x is approximately normal with μ_x = 6100 and σ_x = 1000.00.

-The probability distribution of x is approximately normal with μ_x = 6100 and σ_x = 2000.    

-The probability distribution of x is not normal.

-The probability distribution of x is approximately normal with μ_x = 6100 and σ_x = 1414.21.

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? Pick one

-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? Pick one

-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 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.

Answer 1

Solution :

Given that ,

mean = = 6100

standard deviation = = 2000

a) P(x < 3500) = P[(x - ) / < (3500 - 6100) / 2000]

= P(z < -1.30)

Using z table,

= 0.0968

b) n = 2

= = 6100

= / n = 2000/ 2 = 1414.21

The probability distribution of x is approximately normal with μx = 6100 and σx = 1414.21

P( < 3500) = P(( - ) / < (3500 - 6100) / 1414.21)

= P(z < -1.84)

Using z table

= 0.0329

c) n = 3

= = 6100

= / n = 2000/ 3 = 1154.70

The probability distribution of x is approximately normal with μx = 6100 and σx = 1154.70

P( < 3500) = P(( - ) / < (3500 - 6100) / 1154.70)

= P(z < -2.25)

Using z table

= 0.0122

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