Question

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.)

Answer 1

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:

$z=\frac{x-\mu}{\sigma}$

Now we substitute the values:

$z=\frac{60-35}{9}$

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:

$z=\frac{16-35}{9}$

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:

[tex]P(16We have already determined the formula for these probabilities in parts A and B. Therefore, we substitute and we get:

[tex]P(16Solving the operations we get:

[tex]P(16Therefore, the probability that x is between 16 and 60 is 0.9799.

Part C. We are asked the following:

[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.