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=
2 answers:
Answer:
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
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