Paleomagnetic studies of Canadian volcanic rock known as the Carmacks Group have recently been completed. The studies revealed that the northward displacement of the rock units has an approximately normal distribution with a standard deviation of 500 kilometers (Canadian Journal of Earth Sciences, Vol. 27, 1990). One group of researchers estimated the mean displacement at 1,200 kilometers, whereas a second group estimated the mean at 1,000 kilometers.

Required:
a. Assuming the mean is 1500 kilometers, what is the probability of a northward displacement of less than 500 kilometers?
b. Assuming the mean is 1200 kilometers, what is the probability of a northward displacement of less than 500 kilometers?
c. If, in fact, the northward displacement is less than 500 kilometers, which is the more plausible mean, 1200 or 1500 kilometers?

Answer:

a) 0.0228 = 2.28% probability of a northward displacement of less than 500 kilometers

b) 0.0808 = 8.08% probability of a northward displacement of less than 500 kilometers

c) 1200 kilometers, because in this case, the will be a higher probability of a northward displacement that is less than 500 kilometers.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

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.

Standard deviation of 500 kilometers

This means that [tex]\sigma = 500[/tex]

a. Assuming the mean is 1500 kilometers, what is the probability of a northward displacement of less than 500 kilometers?

Mean of 1500 km means that [tex]\mu = 500[/tex]

The probability is the pvalue of Z when X = 500. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{500 - 1500}{500}[/tex]

[tex]Z = -2[/tex]

[tex]Z = -2[/tex] has a pvalue of 0.0228

0.0228 = 2.28% probability of a northward displacement of less than 500 kilometers.

b. Assuming the mean is 1200 kilometers, what is the probability of a northward displacement of less than 500 kilometers?

Now [tex]\mu = 1200[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{500 - 1200}{500}[/tex]

[tex]Z = -1.4[/tex]

[tex]Z = -1.4[/tex] has a pvalue of 0.0808

0.0808 = 8.08% probability of a northward displacement of less than 500 kilometers.

c. If, in fact, the northward displacement is less than 500 kilometers, which is the more plausible mean, 1200 or 1500 kilometers?

1200 kilometers, because in this case, the will be a higher probability of a northward displacement that is less than 500 kilometers.

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