Total Quality Management Questions and Answers – Numericals on Reliability

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This set of Total Quality Management Multiple Choice Questions & Answers (MCQs) focuses on “Numericals on Reliability”.

1. A system consists of four components in series with two components having reliability of 0.9 and two others having reliability of 0.8 at the end of one year. What is the system reliability at the end of one year?
a) 0.9000
b) 0.8000
c) 0.7000
d) 0.5184
View Answer

Answer: d
Explanation: The reliability of the system at the end of one year is given by R(s)=R1*R2*R3*R4. So, R(s)=0.9*0.9*0.8*0.8=0.5184.

2. A production line is using five robots for its daily operations. If each robot has a reliability of 0.95, what is the total system reliability?
a) 0.7737
b) 0.9500
c) 0.9000
d) 1.0000
View Answer

Answer: a
Explanation: The reliability of the system is given by R(s)=R1*R2*R3*R4*R5. So, R(s)=0.95*0.95*0.95*0.95*0.95, or R(s)=0.7737.

3. If the system reliability of five robots in a production line is 0.95, what is the individual reliability?
a) 0.9354
b) 0.6535
c) 0.9897
d) 0.8143
View Answer

Answer: c
Explanation: The system reliability of five robots in a production line is 0.95. The reliability of the system is given by R(s)=R1*R2*R3*R4*R5. Therefore, R(x)5=0.95, or R(x)=0.9897.
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4. A system consists of four components in parallel with two components having reliability of 0.9 and two others having reliability of 0.8 at the end of one year. What is the system reliability at the end of one year?
a) 0.9996
b) 0.8000
c) 0.7000
d) 0.5180
View Answer

Answer: a
Explanation: System reliability for parallel components is given by R(s)=1-[(1-R1)2(1-R2)2] or, R(s)=1-[(1-0.9)2(1-0.8)2] = 0.9996.

5. For a 2-out-of-3 system, each component has reliability of 0.9. What is the reliability of the system?
a) 0.942
b) 0.854
c) 0.972
d) 0.999
View Answer

Answer: c
Explanation: The reliability of the system is given by R(s)=P(X=2)+P(X=3), or R(s)=(3C2)(0.9)2(0.1)1+(3C3)(0.9)3(0.1)0=0.243+0.729=0.972.

6. For a 3-out-of-4 system, each component has reliability of 0.97. What is the reliability of the system?
a) 0.9148
b) 0.9248
c) 0.9448
d) 0.9947
View Answer

Answer: d
Explanation: The reliability of the system is given by R(s)=P(X=3)+P(X=4), or R(s)=(4C3)(0.97)3(0.03)1+(4C4)(0.97)4(0.03)0=0.1095+0.8852=0.9947.

7. If the failure rate of the system is 0.000128 units, what is the mean time to fail?
a) 7812.5 units
b) 7830.5 units
c) 4530.5 units
d) 9821.5 units
View Answer

Answer: a
Explanation: If the failure rate of the system is given by MTTF = 1/λs=1/0.000128, or MTTF=7812.5 units. MTTF refers to the mean time to fail.
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8. If the failure rate is 6.67 x 10-6 failures/hr, calculate reliability for a period of 1000 hours.
a) 98.45%
b) 99.34%
c) 97.56%
d) 94.67%
View Answer

Answer: b
Explanation: The reliability is given by R(t)=e-λt. Here, t=1000 hours and λ=6.67×10-6 failures/hr. So, reliability=99.34%.

9. If the time to fail for 3 components are 807, 820, and 810, calculate mean time to failure.
a) 808.34
b) 810.34
c) 812.34
d) 814.34
View Answer

Answer: c
Explanation: If the time to fail for 3 components are 807, 820, and 810, then
MTTF=(807+820+810)/3=812.34 hrs. MTTF refers to the mean time to fail.
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10. If the reliability is 0.3012 and the period of 1000 hours, calculate failure rate per hour.
a) 0.0010
b) 0.0012
c) 0.0014
d) 0.0016
View Answer

Answer: b
Explanation: The reliability is given by R(t)=e-λt. If t=1000 hours, R=0.3012, then, λ=0.0012 failures/hr.

Sanfoundry Global Education & Learning Series – Total Quality Management.

To practice all areas of Total Quality Management, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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