Normal Distribution Problem. Porphyrin is a pigment in blood protoplasm and other body fluids that is significant in body energy and storage. Assume porphyrin is normally distributed with mean (mu) = 38 milligrams per deciliter and standard deviation (sigma) = 12 milligrams per deciliter. a) What is the probability that x is less than 56? (56-38)/12= 18/12= 1.5 → 0.4332 (?) b) What is the probability that x is greater than 23? (23-38)/12= -1.25 → Z=

A) P(X < 56) = 0.9332

B) P(X > 56) = 0.8944

Step-by-step explanation:

From the 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.

x is approximately normally distributed with;

mean μ = 38

standard deviation σ = 15.

So the distribution of X is written as:

X ~ N (μ = 38, σ = 15)

A) The required probability that x is less than 56 is given as;

P(X < 56) = P((x - μ)/σ) < (56-38)/15)

Thus, we have;

P(z ≤ 1.5)

From the table attached, it gives a value of; 0.93319 ≈ 0.9332

B)The required probability that x is greater than 23 is given as;

P(X > 56) = P((x - μ)/σ) < (23-38)/12)

Thus, we have;

P(z ≥ -1.25) = 1 - P(z ≤ -1.25)

From the second table attached,

P(z ≥ -1) = 1 - 0.10565 = 0.89435 ≈ 0.8944

; )

step-by-step explanation: ; )

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