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 = 6150 and estimated standard deviation σ = 2300. 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 = 6150 and σ_x = 1150.00.

- The probability distribution of x is approximately normal with μ_x = 6150 and σ_x = 1626.35.    

- The probability distribution of x is not normal.

- The probability distribution of x is approximately normal with μ_x = 6150 and σ_x = 2300.

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 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 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.

Answer 1

for normal distribution z score =(X-μ)/σx
here mean=       μ=	6150
std deviation   =σ=	2300

a)

probability =P(X<3500)=(Z<(3500-6150)/2300)=P(Z<-1.15)=0.1251

b)

- The probability distribution of x is approximately normal with μ_x = 6150 and σ_x = 1626.35.  

probability =P(X<3500)=(Z<(3500-6150)/1626.346)=P(Z<-1.63)=0.0516

c)

probability =P(X<3500)=(Z<(3500-6150)/1327.906)=P(Z<-2)=0.0228

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.