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 = 6450 and estimated standard deviation σ = 2400. 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?

(b) What is the probability of x < 3500?

(c) Repeat part (b) for n = 3 tests taken a week apart. (Round your answer to four decimal places.)

Answer 1

Solution :

Given that ,

mean = = 6450

standard deviation = = 2400

a)

P(x < 3500) = P((x - ) / < (3500 - 6450) / 2400)

= P(z < -1.23)

= 0.1093 Using standard normal table,

Probability = 0.1093

b)

P(x < 3500) = P((x - ) / < (3500 - 6450) / 2400)

= P(z < -1.23)

= 0.1093 Using standard normal table,

Probability = 0.1093

​​​​​​c)

n = 3

= = 6450 and

= / n = 2400 / 3= 1385.6406

P( < 3500) = P(( - ) / < (3500-6450) / 1385.6406)

= P(z < -2.13)

= 0.0166 Using standard normal table,   

Probability = 0.0166