1) 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 = 43 and standard deviation σ = 9. Find the following probabilities.
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
2)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. 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.

1)

a) 0.9706 = 97.06% probability that x is less than 60.

b) 0.9987 = 99.87% probability that x is greater than 16.

c) 0.9693 = 96.93% probability that x is between 16 and 60.

d) 0.0294 = 2.94% probability that x is more than 60.

2)

a) 0.0668 = 6.68% probability that the thickness is less than 3.0 mm.

b) 0.0062 = 0.62% probability that the thickness is more than 7.0 mm

c) 0.9270 = 92.70% probability that the thickness is between 3.0 mm and 7.0 mm.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

1)

We have that

a) x is less than 60

This is the pvalue of Z when X = 60.

has a pvalue of 0.9706

0.9706 = 97.06% probability that x is less than 60.

b) x is greater than 16

This is 1 subtracted by the pvalue of Z when X = 16.

has a pvalue of 0.0013.

1 - 0.0013 = 0.9987

0.9987 = 99.87% probability that x is greater than 16.

c) x is between 16 and 60

This is the pvalue of Z when X = 60 subtracted by the pvalue of Z when X = 16.

From a), Z when X = 60 has a pvalue of 0.9706.

From b), Z when X = 16 has a pvalue of 0.0013

0.9706 - 0.0013 = 0.9693

0.9693 = 96.93% probability that x is between 16 and 60.

d) x is more than 60

This is 1 subtracted by the pvalue of Z when X = 60.

From a), Z when X = 60 has a pvalue of 0.9706.

1 - 0.9706 = 0.0294

0.0294 = 2.94% probability that x is more than 60.

2)

Now

a) the thickness is less than 3.0 mm

This is the pvalue of Z when X = 3.

has a pvalue of 0.0668

0.0668 = 6.68% probability that the thickness is less than 3.0 mm.

b) the thickness is more than 7.0 mm

This is 1 subtracted by the pvalue of Z when X = 7.

has a pvalue of 0.9938.

1 - 0.9938 = 0.0062

0.0062 = 0.62% probability that the thickness is more than 7.0 mm

c) the thickness is between 3.0 mm and 7.0 mm.

This is the pvalue of Z when X = 7 subtracted by the pvalue of Z when X = 3.

From b), Z when X = 7 has a pvalue of 0.9938.

From a), Z when X = 3 has a pvalue of 0.0668

0.9938 - 0.0668 = 0.9270

0.9270 = 92.70% probability that the thickness is between 3.0 mm and 7.0 mm.

x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x

step-by-step explanation:

last month maria hiked the 5-mile mountain trail a number of times and she hiked the 10-mile canal trail several times. let x represent the number of times she hiked the 5-mile trail, and let y represent the number of times she hiked the 10-mile trail. if she hiked a total of 90 miles, which equation can be used to find the number of times maria hiked each trail? x + y = 90 5x – 10y = 90 90 – 10y = 5x 90 + 10y = 5x