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Porphyrin is a pigment in blood protoplasm and other body fluids that is significant in body energy and storage. Let x be a random variable that represents the number of milligrams of porphyrin per deciliter of blood. In healthy circles, x is approximately normally distributed with mean = 35 and standard deviation = 9. Find the following probabilities. (Round your answers to four decimal places.)(a) x is less than 60(b) x is greater than 16(c) x is between 16 and 60(d) x is more than 60 (This may indicate an infection, anemia, or another type of illness.)

User Vrijdenker
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We are given a random variable with a normal distribution and we are asked the following:

Part A.


P(x<60)

We need to determine the z-score for this value. To do that we will use the following formula:


z=(x-\mu)/(\sigma)

Now we substitute the values:


z=(60-35)/(9)

Solving the operations:


z=2.78

Therefore, we have:


P(x<60)=P(z<2.78)

From the table of probabilities for normal distribution we get:


P(z<2.78)=0.9973

Therefore, the probability that x is less than 60 is 0.9973

Part B. We are asked the following:


P(x>16)

We need to determine the z-score for this value. We use the same formula as in part A:


z=(16-35)/(9)

Solving the operations we get:


z=-2.11

Therefore, we have:


P(x>16)=P(z>-2.11)

Since the table will give us only the values that are less than -2.11 we need to use the following relationship:


P(z>-2.11)=1-P(z<-2.11)

From the table we get:


P(z<-2.11)=0.0174

Substituting we get:


P(z>-2.11)=1-0.0174

Solving the operations we get:


P(z>-2.11)=0.9826

Therefore, the probability that x is greater than 16 is 0.8257

Part C. We are asked the following:


P(16To determine that, we will use the following formula:<p></p>[tex]P(16We have already determined the formula for these probabilities in parts A and B. Therefore, we substitute and we get:<p></p>[tex]P(16Solving the operations we get:<p></p>[tex]P(16Therefore, the probability that x is between 16 and 60 is 0.9799.<p>Part C. We are asked the following:</p><p></p>[tex]P(x>60)

We use the same relationship as is part B:


P(x>60)=1-P(x<60)

We already determine the probability that x is less than 60 in part A, therefore, we substitute and we get:


P(x>60)=1-0.9973

Solving the operations we get:


P(x>60)=0.0027

Therefore, the probability that x is more than 60 is 0.027.

User Ulvi
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