Question

1. An aircraft emergency locator transmitter (ELT) is a device designed to transmit a signal in...

1. An aircraft emergency locator transmitter (ELT) is a device designed to transmit a signal in the case of a crash. The Altigauge Manufacturing Company makes 70% of the ELTs, the Bryant Company makes 20% of them, and the Chartair Company makes the rest. The ELTs made by Altigauge have a 2% rate of defects, the Bryant ELTs have a 4% rate of defects, and the Chartair ELTs have a 6% rate of defects (which helps to explain why Chartair has the lowest market share).  

(a) How likely is a randomly selected ELT from the general population of all ELTs is defective?

(b) If a randomly selected ELT is tested and found to be good (not defective), find the probability that it was made by the Altigauge Company. Final answer in 4 decimals.

Homework Answers

Answer #1

answer:

  • For Simplicity reason for existing, how about we expect add up to ELT's are 100.
  • At that point,
  • Produced by Alpha Manufacturing Company = 70
  • Produced by Beta Manufacturing Company = 20
  • Made by Gamma Manufacturing Company = 10
  • Rate of imperfections of ELT's:
  • Alpha Manufacturing organization = 4% = 4% of 70 = 2.8
  • Beta Manufacturing organization = 6% = 6% of 20 = 1.2
  • Gamma Manufacturing organization = 2% = 2% of 10 = 0.2

a)

  • Likelihood that a haphazardly chose ELT from all inclusive community is deserted =

= (Total number of inadequate ELT)/Total ELT

= (2.8 + 1.2 + 0.2)/(100)

=(4.2)/(100)

= 4.2/100

0.042

  • Hence, there's a 4.2% likelihood of a haphazardly chose ELT from the overall public of all ELTs to be deficient

b)

  • This is an instance of restrictive likelihood.
  • Let Event A = ELT is deficient = 4.2% and
  • Occasion B = ELT was made by Beta Manufacturing Company
  • Likelihood (B | A) = P(A and B)/P(A)
  • Presently, P(A and B) = P(A | B) * P(B)
  • P(A | B) = Probability of an ELT being imperfect given it was made by Beta Manufacturing organization = 6%
  • P(B) = 20%
  • P(A and B) = 6% * 20%
  • = 0.06 * 0.20 = 0.0120
  • In this way,
  • Likelihood (B | A) = P(A and B)/P(A) = (0.0120)/(0.042)
  • 0.285714285

= 0.2857

  • Therefor, there's a 28.57 likelihood that it was made by Beta Manufacturing Company
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