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Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 2 - Problem 106e
Get Full Access to Probability And Statistics For Engineers And The Scientists - 9 Edition - Chapter 2 - Problem 106e

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# One method used to distinguish between granitic (G) and ISBN: 9780321629111 32

## Solution for problem 106E Chapter 2

Probability and Statistics for Engineers and the Scientists | 9th Edition

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Problem 106E

One method used to distinguish between granitic (?G?) and basaltic (?B?) rocks is to examine a portion of the infrared spectrum of the sun’s energy reflected from the rock surface. Let ?R?1, ?R?2, and ?R?3 denote measured spectrum intensities at three different wavelengths; typically, for graniteR123 , whereas for basalt R123. When measurements are made remotely (using aircraft), various orderings of the ?R?is may arise whether the rock is basalt or granite. Flights over regions of known composition have yielded the following information: Suppose that for a randomly selected rock in a certain region P(granite) =.25 , and P(basalt) = .75.. a. Show thatP(granite| R123) >P(basalt| R123). If measurements yielded R123 , would you classify the rock as granite or basalt? b. If measurements yielded R123, how would you classify the rock? Answer the same question for R123. c. Using the classification rules indicated in parts (a) and (b), when selecting a rock from this region, what is the probability of an erroneous classification? [?Hint: ?Either ?G ?could be classified as ?B ?or ?B ?as ?G, ?and ?P?(?B?) and ?P?(?G?) are known.] d. If P(granite) = p rather than .25, are there values of ?p ?(other than 1) for which one would always classify a rock as granite?

Step-by-Step Solution:

Answer Step 1 of 6 G = {rock is granite} P(G) = 0.25 B = {rock is basalt} P(B) = 0.75 Where R1, R2, and R3 denote measured spectrum intensities at three different wavelengths R123 = {R1 < R2 < R3} R132 = {R1 < R3 < R2} R312 = {R3 < R1 < R2} P(R123 | G) = 0.60 P(R123 | B) = 0.10 P(R132 | G) = 0.25 P(R132 | B) = 0.20 P(R312 | G) = 0.15 P(R312 | G) =0.7 Step 2 of 6 a) The probability that the rock reflecting the light is granite given the wavelength reflected. P(G/R123)= P(G R123)= P(G) P(R123 | G) R123 R123 = P(G) P(R123 | G) P(G)P(R123 | G)+P(B)(R123 | B) = 0.25(0.6) 0.25(0.6)+0.75(0.1) =0.15/0.625=0.667 Step 3 of 6 the probability that the rock reflecting the light is basalt given the wavelength reflected. P(B/R123)= P(B R123)= P(B) P(R123 | B) R123 R123 = 0.75(0.1) 0.625 =0.33 P(granite R1 < R2 < R3) > P(basalt R1 < R2 < R3) I would choose granite because P(G|R123) > P(B|R123).

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##### ISBN: 9780321629111

The answer to “One method used to distinguish between granitic (?G?) and basaltic (?B?) rocks is to examine a portion of the infrared spectrum of the sun’s energy reflected from the rock surface. Let ?R?1, ?R?2, and ?R?3 denote measured spectrum intensities at three different wavelengths; typically, for graniteR123 , whereas for basalt R123. When measurements are made remotely (using aircraft), various orderings of the ?R?is may arise whether the rock is basalt or granite. Flights over regions of known composition have yielded the following information: Suppose that for a randomly selected rock in a certain region P(granite) =.25 , and P(basalt) = .75.. a. Show thatP(granite| R123) >P(basalt| R123). If measurements yielded R123 , would you classify the rock as granite or basalt? b. If measurements yielded R123, how would you classify the rock? Answer the same question for R123. c. Using the classification rules indicated in parts (a) and (b), when selecting a rock from this region, what is the probability of an erroneous classification? [?Hint: ?Either ?G ?could be classified as ?B ?or ?B ?as ?G, ?and ?P?(?B?) and ?P?(?G?) are known.] d. If P(granite) = p rather than .25, are there values of ?p ?(other than 1) for which one would always classify a rock as granite?” is broken down into a number of easy to follow steps, and 208 words. This full solution covers the following key subjects: Rock, granite, basalt, classify, measurements. This expansive textbook survival guide covers 18 chapters, and 1582 solutions. Since the solution to 106E from 2 chapter was answered, more than 577 students have viewed the full step-by-step answer. Probability and Statistics for Engineers and the Scientists was written by and is associated to the ISBN: 9780321629111. This textbook survival guide was created for the textbook: Probability and Statistics for Engineers and the Scientists, edition: 9. The full step-by-step solution to problem: 106E from chapter: 2 was answered by , our top Statistics solution expert on 05/06/17, 06:21PM.

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