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 = 8100 and estimated standard deviation σ = 2800. 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 not normal. The probability distribution of x is approximately normal with μx = 8100 and σx = 2800. The probability distribution of x is approximately normal with μx = 8100 and σx = 1400.00. The probability distribution of x is approximately normal with μx = 8100 and σx = 1979.90. 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 stayed the same as n increased. The probabilities decreased 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 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. Need Help? Read It

Answer 1

a)

Here, μ = 8100, σ = 2800 and x = 3500. We need to compute P(X <= 3500). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (3500 - 8100)/2800 = -1.64

Therefore,
P(X <= 3500) = P(z <= (3500 - 8100)/2800)
= P(z <= -1.64)
= 0.0505

b)
sigma = 2800/sqrt(2) = 1979.90

The probability distribution of x is approximately normal with μx = 8100 and σx = 1979.90.

Here, μ = 8100, σ = 1979.9 and x = 3500. We need to compute P(X <= 3500). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (3500 - 8100)/1979.9 = -2.32

Therefore,
P(X <= 3500) = P(z <= (3500 - 8100)/1979.9)
= P(z <= -2.32)
= 0.0102

c)

sigma = 2800/sqrt(3) = 1616.58

The probability distribution of x is approximately normal with μx = 8100 and σx = 1616.58

Here, μ = 8100, σ = 1616.58 and x = 3500. We need to compute P(X <= 3500). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (3500 - 8100)/1616.58 = -2.85

Therefore,
P(X <= 3500) = P(z <= (3500 - 8100)/1616.58)
= P(z <= -2.85)
= 0.0022

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