# 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

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

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

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.

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

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

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

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

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.

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%

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

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.

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

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.

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