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 = 6300 and estimated standard deviation σ = 2700. 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 μ_x = 6300 and σ_x = 2700.   

-) The probability distribution of x is approximately normal with μ_x = 6300 and σ_x = 1909.19.

-) The probability distribution of x is approximately normal with μ_x = 6300 and σ_x = 1350.00.

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 decreased 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 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 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 has leukopenia.

Answer 1

SOLUTION:-

Given that

A) To find the probability of test X<3500

P(X<3500) = P(Z<((3500-6300)/2700))

= P(Z<-1.04)

From standard normal table

=0.1492

Required probability is 0.149

B)   Probability distribution of X for two tests taken about a week apart

_x = E(X) = 6300

_x =  /(n^1/2)

= 1909/(2^1/2)

  _x = 1349.87

Hence the probability distribution of X is approximately normal with _x = 6300 and  _x = 1349.87

Hence the correct option is D

To find the probability of (X<3500)

P(X<3500)=P(Z<(((3500-6300)/1349.87))

=P(Z<-2.07)

From the table P(X<3500) = 0.0192

Required probability is 0.0192

C) Probability distribution of X for three tests taken about a week apart

_x = E(X) = 6300

_x =  /(n^1/2)

= 1909/(3^1/2)

  _x = 1102.16

Hence the probability distribution of X is approximately normal with _x = 6300 and  _x = 1102.16

To find the probability of (X<3500)

P(X<3500)=P(Z<(((3500-6300)/1102.16))

=P(Z<-2.54)

From the table P(X<3500) = 0.0055

Required probability is 0.0055

D) The correct option is b , we ca see that A and B questions n value is increased then the probability is decreased.

  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