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 = 2950. 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 = 6100 and σ_x = 1475.00. The probability distribution of x is approximately normal with μ_x = 6100 and σ_x = 2085.97.     The probability distribution of x is approximately normal with μ_x = 6100 and σ_x = 2950. 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.

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 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 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

Given that

mean (mu) = 6100

standard deviation (sigma) = 2950

(A) using normalcdf

setting lower = -99999

upper = 3500

mean (mu) = 6100

standard deviation (sigma) = 2950

P(X<3500)= nomalcdf(-99999,3500,6100,2950)

= 0.1891

(B)

mean will remain same, so mu = 6100

standard deviation = sd/sqrt{n}

= 2950/sqrt(2)

= 2085.97

The probability distribution of x is approximately normal with μ_x = 6100 and σ_x = 2085.97.

using normalcdf

setting lower = -99999

upper = 3500

mean (mu) = 6100

standard deviation (sigma) = 2085.97

P(Xbar<3500)= nomalcdf(-99999,3500,6100,2085.97)

= 0.1063

(C)

using normalcdf

setting lower = -99999

upper = 3500

mean (mu) = 6100

standard deviation (sigma) = 2950/sqrt(3)

P(Xbar<3500)= nomalcdf(-99999,3500,6100,2950/sqrt(3))

= 0.0634

(D) We can see that as n increases from 1 to 3, probability gets decreased from 0.1891 to 0.0634

so, The probabilities decreased as n increased

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.