When red blood cells are counted using a certain electronic counter, the standar

Question

When red blood cells are counted using a certain electronic counter, the standard deviation of repeated counts of the same blood specimen is about 0.8% of the true value mu. Assume that the distribution of repeated counts is normal with mean equal to the true value mu: a. Suppose the true value of the red count for a certain specimen is 5,000,000 cells/mm^3, what is the probability that the counter will give a reading between 4, 900,000 and 5, 100000 b. If the true value is mu, what is the probability that the count will be between .98 mu and 1.02 mu? c. A certain lab performs a large number of counts of blood specimens daily. For what percentage of these specimens does the reported count differ from the true value by 2% or more?

Explanation / Answer

Part a

We have to find P(4900000<X<5100000)

We are given

µ = 5000000

= 5000000*0.8% = 40000

P(4900000<X<5100000) = P(X<5100000) – P(X<4900000)

Z = (X - µ)/

For X=5100000

Z = (5100000 – 5000000)/40000 = 2.5

P(X<5100000) = P(Z<2.5) = 0.99379

For X = 4900000

Z = (4900000 – 5000000)/40000 = -2.5

P(X<4900000) = P(Z<-2.5) = 0.00621

P(4900000<X<5100000) = P(X<5100000) – P(X<4900000)

P(4900000<X<5100000) = 0.99379 - 0.00621 = 0.98758

Required probability = 0.98758

Part b

Here, we have to find P(0.98µ<X<1.02µ)

0.98µ = 0.98*5000000 = 4900000

1.02µ = 1.02*5000000 = 5100000

P(4900000<X<5100000) = P(X<5100000) – P(X<4900000)

Z = (X - µ)/

For X=5100000

Z = (5100000 – 5000000)/40000 = 2.5

P(X<5100000) = P(Z<2.5) = 0.99379

For X = 4900000

Z = (4900000 – 5000000)/40000 = -2.5

P(X<4900000) = P(Z<-2.5) = 0.00621

P(4900000<X<5100000) = P(X<5100000) – P(X<4900000)

P(4900000<X<5100000) = 0.99379 - 0.00621 = 0.98758

Required probability = 0.98758

Part c

µ = 5000000

= 5000000*0.8% = 40000

0.02*µ = 0.02*5000000 = 100000

µ - 100000 = 5000000 – 100000 = 4900000

µ + 100000 = 5000000 + 100000 =5100000

We have to find P(4900000<X<51000)

P(4900000<X<5100000) = P(X<5100000) – P(X<4900000)

P(4900000<X<5100000) = 0.99379 - 0.00621 = 0.98758

Required probability = 0.98758

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