Question

A firm designs and manufactures automatic electronic control devices that are installed at customers' plant sites. The control devices are shipped by truck to customers' sites; while in transit, the devices sometimes get out of alignment. More specifically, a device has a prior probability of .10 of getting out of alignment during shipment. When a control device is delivered to the customer's plant site, the customer can install the device. If the customer installs the device, and if the device is in alignment, the manufacturer of the control device will realize a profit of $15,000. If the customer installs the device, and if the device is out of alignment, the manufacturer must dismantle, realign, and reinstall the device for the customer. This procedure costs $3,000, and therefore the manufacturer will realize a profit of $12,000. As an alternative to customer installation, the manufacturer can send two engineers to the customer's plant site to check the alignment of the control device, to realign the device if necessary before installation, and to supervise the installation. Since it is less costly to realign the device before it is installed, sending the engineers costs $400. Therefore, if the engineers are sent to assist with the installation, the manufacturer realizes a profit of $14,600 (this is true whether or not the engineers must realign the device at the site). Before a control device is installed, a piece of test equipment can be used by the customer to check the device's alignment. The test equipment has two readings, "in" or "out" of alignment. If the control device is in alignment, there is a 0.8 probability that the test equipment will read "in." If the control device is out of alignment, there is a 0.8 probability that the test equipment will read "out." Set up probability revision tables to: (a) Find the probability that the test equipment "reads in," and find the posterior probabilities of in alignment and out of alignment given that the test equipment "reads in." (Round intermediate calculations and final answers to 4 decimal places.) P("reads in") = Posterior Probabilities: (b) Find the probability that the test equipment "reads out," and find the posterior probabilities of in alignment and out of alignment given that the test equipment "reads out." (Round intermediate calculations and final answers to 4 decimal places.) P("reads out") = Posterior Probabilities:

Answer 1

The solution is as follows.

Ans Prior Date : Thus, prior internet lignment: ent-1-0.10 probability of out-alignment = 0.10 prior probability of = 0.90 equation i p Cout-aligiment) tp (in- aligment) According to Bayes Rule 0.90 PCH x PCA | B) probability of event A, given B has recused PCBA) P (B) where P(A) P (BIA) prior probability of A probability of event B. A has occurred prior probability of B. given P (B) Now, As per given information. the given the Probability that device 'reads -in' given device in- alignment 0.8 Thus, probability that device 'reads-out device is in-aligment 1 - 008 i.e. lP (reads-in in-alignment) P (reads -out / in-alignment) 0.8 equations

=0.8 Date Page No.1 given 3 Probability that device 'reads out given the device out - alignment Thus, probability that device reads-in the device out-alignment iep (reads-out, out-aligmmend) = 0.8 IP (reads in -alignment) a.) Probability of device is 'in-alignment given reading is read-in' H/prior probability of (in-aligment) (probability that device -aligment) out- quation reads in device is given the in- [from equations and equations 0.72 is lout- Probability of device I reading is alignment given 'read-in = (prior probability of out-alignment probability that device device given reads in is out-aligmen X 6. 2 [ from and equation 3 egation 0.02 in- Therefore Probality of the device that reads in irrespective of whether -alignment or out-aliyment ie. P (reads.in ) = (P (reads in) 0.74 0-72 + 0.02 equation y

Date Page No.: Thus according to Bayes rule Posterior probability of P ( in-alignment / readsin) = P (reads in in-align P Creads - in P (in-alignment) x 0.9 xo.8 0.74 0.72 from equation 2 84 - o. 973 P (out-alignment / Had -in) - P (out-aligment) pread in (out-align) P Creads in) from oolxo.2 0.74 equation 074 P Ereads in ) Postrinor probabilities 0.74 0.973 7 in-alignment 0.027 oot-alignment

6.) Probability of device in-alignment given reading is read out P (in-alignment *P (readout in-aligment) Date Paga No.: 0.2 0.90 18 . Tie plin Probability of of device dersice out-alignment given reading is 'read out Plout - alignment) + P (read-out / out- out-alignmen! xo.8 is in- Therefore P Cread out) isrespective wthet her the device aligment on out-alignment - P (reado out) 0-18 0:26 eyatims This according to Bayes rule Posterior probability of PC in-alignment I reads out). Plin. alignment P (reach out (in-alignment f(reads out) o do xo 2 0 6923

P Plout-alignment reads.out) = Perst-alignmenty x P (read out lost. alignmez P (read-out) 0-1xo.8 0.26 0.30769 P (reads out) Posterior probabilities 0 26 0.6923 0.30769

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