1. Which of the following is NOT true about the Taguchi DOE approach?

I. It includes the concept of loss function in factorial experiments.

II. It assumes losses occur when a process fails to meet a target value.

III. It assumes losses are due to variability within the process.

IV. It assumes that the loss function is a step function relative to the specification limits.

A. I only

B. II only

C. III only

**D. IV only**

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2. In general, reliability testing is performed for which of the following reasons?

I. To detect unanticipated failure modes.

II. To compare estimated failure rates to actual failure rates.

III. To monitor reliability growth over time.

IV. To meet or exceed customer expectations.

A. I and III only

B. II and IV only

C. I, II and III only

**D. I, II, III and IV**

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3. The time to failure of a component is Weibull distributed, β=0.5, η=10,000 hours.

What is the probability that the component will still be operational after one year?**A. 0.3922**

B. 0.3214

C. 0.3680

D. 0.6320

4. Seventy-two bearings are available for life testing. To save test time

they are divided into six groups of twelve bearings and tested using the

sudden-death technique. The Resultant Data is :

Group 1: Bearing #3 fails at | 110 hrs |

Group 2: Bearing #4 fails at | 75 hrs |

Group 3: Bearing #11 fails at | 165 hrs |

Group 4: Bearing #5 fails at | 310 hrs |

Group 5: Bearing #7 fails at | 210 hrs |

Group 6: Bearing #12 fails at | 270 hrs |

What is the 0.90 Reliability of the bearing population?

A. 69 hrs

B. 79 hrs

**C. 241 hrs**

D. 120 hrs

Time1(hrs) | Number | Censor(F=1,C=0) |

110 | 1 | 1 |

75 | 1 | 1 |

165 | 1 | 1 |

310 | 1 | 1 |

210 | 1 | 1 |

270 | 1 | 1 |

110 | 11 | 0 |

75 | 11 | 0 |

165 | 11 | 0 |

310 | 11 | 0 |

210 | 11 | 0 |

270 | 11 | 0 |

5. The ratio of the Bill of Material (BoM) turbopump field failures to the BoM turbopump bench lab test failures is:

Both Field & lab data had a Weibull β=2.59

The turbopump was redesigned to fix the problem and two units were tested in the lab for 500 hours without failure under the same accelerated conditions. What service experience should be expected in terms of characteristic life?

Hint: Use the equation for MLE η

A. 252 hrs

B. 707 hrs

C. 1000 hrs

**D. 653 hrs**

**A. 0.34**

What is the probability of at least one system functioning properly?

**A. 0.9999**

8. While the AMSAA-Duane model is N(t)=λtβ ; the model can be rearranged in a number of ways to satisfy a customer’s demands for reporting. Which of the following functions could be used for plotting cumulative Failure rate and Cumulative MTBF:

A. I. & II

**B. I. & III.**

C. II. & IV.

D. II. & III.

9. In constructing programs for environmental stress screening(ESS), the selection of the stress levels and the choice of exposure times is a challenging task. Which of the following are true:

I. ESS is not a test. ESS is a screening process

II. ESS is not Burn-in, ESS stresses a product to operational extremes

III. ESS is used in manufacturing/production to catch latent failures.

IV. All items in a product line should be exposed to ESS.

A. I., II., IV.

B.I.,III., IV.

C. II., III., IV.

**D. I.,II.,III.,& IV.**

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10. A test was done on ultrasonic inspection kits to determine how effective they are in discovering microscopic cracks in aircraft parts. When a crack was present, the equipment signaled a crack 98% of the time. There was a “false alarm” on 3% of the parts which had no cracks. Suppose five percent of all the parts have a crack in them. If the percentages in the test can be assumed to be the true probabilities, find the probability that a part is really bad when a kit signals a crack.

A. 0.9215

B.0.9225.

**C. 0.6322.**

D. 0.7533.

An engineer just ran 50 compressor start tests with NO failures.

The engineer needs to pass the customer’s requirement of 80% confidence of at least 0.90 Reliability.

Has the engineer achieved that goal?

**A. Yes**

B. No

C. More data required

D. Question not worded correctly

We want 80% of the intervals we would estimate from repeated

Testing to contain the true value.

Therefore,

1 – Confidence =

Therefore, 0.2=

So, R = 0.968

For N=50, no failures, we can state with 80% confidence that

the true unknown reliability lies between 0.968 and 1.0.

The answer is A.