## 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

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 μ = 40 and standard deviation σ = 14. 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 602)Thickness measurements of ancient prehistoric Native American pot shards discovered in a Hopi village are approximately normally distributed, with a mean of 4.5 millimeters (mm) and a standard deviation of 1.0 mm. For a randomly found shard, find the following probabilities. (Round your answers to four decimal places.)a) the thickness is less than 3.0 mm b) the thickness is more than 7.0 mm c) the thickness is between 3.0 mm and 7.0 mm.

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1 month 2021-11-08T13:35:09+00:00 1 Answer 0 views 0

Question 1:

(a) P(x<60) = 0.9236

(b) P(x>16) = 0.9564

(c) P(16<x<60) = 0.88

(d) P (x>60) = 0.0764

Question 2:

(a) P(x<3) = 0.0668

(b) P(x>7) = 0.0062

(c) P(3<x<7) = 0.927

Step-by-step explanation:

Question 1:

x = no. of mg of porphyrin per deciliter of blood.

μ = 40

σ = 14

(a) We need to compute P(x<60). We need to find the z-score using the normal distribution formula:

z = (x – μ)/σ

P(x<60) = P((x – μ)/σ < (60 – 40)/14)

= P(z < 20/14)

= P(z<1.43)

Using the normal distribution probability table we can find the value of p at z=1.43.

P(z<1.43) = 0.9236

so, P(x<60) = 0.9236

(b) P(x>16) = P(z>(16-40)/14)

= P(z>-1.71)

= 1 – P(z<-1.71)

= 1 – 0.0436

P(x>16) = 0.9564

(c) P(16<x<60) = P((16-40)/14) < x < (60-40)/14)

= P(-1.71 < z < 1.43)

This probability can be calculated as: P(z<1.43) – P(z<-1.71)

P(16<x<60) = 0.9236 – 0.0436

P(16<x<60) = 0.88

(d) P(x>60) = 1 – P(x<60)

we have calculated P(x<60) in part (a) so,

P(x>60) = 1 – 0.9236

P (x>60) = 0.0764

Question 2:

μ = 4.5 mm

σ = 1.0 mm

In this question, we will again compute the z-scores and then find the probability from the normal distribution table.

(a) P(x<3) = P(z<(3-4.5)/1)

= P(z<-1.5)

P(x<3) = 0.0668

(b) P(x>7) = 1 – P(x<7)

= 1 – P(z<(7-4.5)/1)

= 1 – P(z<2.5)

= 1 – 0.9938

P(x>7) = 0.0062

(c) P(3<x<7) = P(x<7) – P(x<3)

we have computed both of these probabilities in parts (a) and (b) so,

P(3<x<7) = 0.9938 – 0.0668

P(3<x<7) = 0.927