Let x be a random variable that represents white blood cell

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 = 6300 and estimated standard deviation = 2100. 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 _x = 6300 and _x = 1050.00.    The probability distribution of x is approximately normal with _x = 6300 and _x = 2100.The probability distribution of x is approximately normal with _x = 6300 and _x = 1484.92.

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 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.

Solution

A)

We first get the z score for the critical value. As z = (x - u) / s, then as          
          
x = critical value =    3500      
u = mean =    6300      
          
s = standard deviation =    2100      
          
Thus,          
          
z = (x - u) / s =    -1.333333333      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.333333333   ) =    0.09121122 [ANSWER]

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b)

The new standard deviation will be divided by sqrt(2).

The probability distribution of x is approximately normal with x = 6300 and x = 1484.92. [ANSWER]

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We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    3500      
u = mean =    6300      
n = sample size =    2      
s = standard deviation =    2100      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -1.885618083      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -1.885618083   ) =    0.029673219 [ANSWER]

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C)

We first get the z score for the critical value. As z = (x - u) sqrt(n) / s, then as          
          
x = critical value =    3500      
u = mean =    6300      
n = sample size =    3      
s = standard deviation =    2100      
          
Thus,          
          
z = (x - u) * sqrt(n) / s =    -2.309401077      
          
Thus, using a table/technology, the left tailed area of this is          
          
P(z <   -2.309401077   ) =    0.010460668 [ANSWER]

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d)

The probabilities decreased as n increased. [ANSWER]

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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]