Question

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=

Answer 1

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

Answer 2

Answer: a) 0.9332, b) 0.8944

Step-by-step explanation: the probability value attached to a z score is gotten by using a z distribution table.

The z score is gotten by making use of the formulae below

Z = x - u / σ

Where x = sample mean, u = population mean and σ = population standard deviation.

a)

For our question, u = 38, σ = 12, we are to look for the z score at z ≤ 56, that's x = 56

By substituting the parameters, we have that

z ≤ 56 = 56 - 38/ 12

z ≤ 56 = 18/12

z ≤ 56 = 1.5

To get the probabilistic value, we check the normal distribution table.

The table I'm using will be giving me probabilistic value towards the left of the area.

From the table, p ( z ≤ 56) = 0.9332.

b)

Z >23 = 23 - 38/ 12

Z >23 = - 15/ 12

Z >23 = - 1.25

The probability value of this z score is towards the right of the distribution but the table I'm using is only giving probability values towards the left.

Hence Z >23 = 1 - Z<23

From the table, Z<23 = 0.1056.

Z >23 = 1 - 0.1056

Z >23 = 0.8944