Electronic Devices and Circuits Questions and Answers – Continous Time Signals

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This set of Electronic Devices and Circuits Multiple Choice Questions & Answers (MCQs) focuses on “Continous Time Signals”.

(Q.1-Q.2) The number of cars arriving at ICICI bank drive-in window during 10-min period is Poisson random variable X with b=2.

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1. The probability that more than 3 cars will arrive during any 10 min period is
a) 0.249
b) 0.143
c) 0.346
d) 0.543
View Answer

Answer: b
Explanation: Evaluate 1 – P(x = 0) – P(x = 1) – P(x = 2) – P(x = 3).

2. The probability that no car will arrive is
a) 0.516
b) 0.459
c) 0.246
d) 0.135
View Answer

Answer: d
Explanation: Evaluate P(x = 0).

(Q.3-Q.5) Delhi averages three murder per week and their occurrences follow a Poisson distribution.

3. The probability that there will be five or more murder in a given week is
a) 0.1847
b) 0.2461
c) 0.3927
d) 0.4167
View Answer

Answer: a
Explanation: P(5 or more) = 1 – P(0) – P(1) – P(2) – P(3) – P(4) = 0.1847.
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4. On the average, how many weeks a year can Delhi expect to have no murders ?
a) 1.4
b) 1.9
c) 2.6
d) 3.4
View Answer

Answer: c
Explanation: P(0) = 0.0498. Hence average number of weeks per year with no murder is 52 x P(0) = 2.5889 week.

5. How many weeds per year (average) can the Delhi expect the number of murders per week to equal or exceed the average number per week?
a) 15
b) 20
c) 25
d) 30
View Answer

Answer: d
Explanation: P(3 or more) = 1 – P(0) – P (1) – P(2) = 0.5768. Therefore average number of weeks per year = 52 x 0.5768 or 29.994 weeks.

(Q.6-Q.8) The random variable X is defined by the density f(x) = 0.5u(x) e(0.5x)

6. The expect value of g(x) = X3 is
a) 48
b) 192
c) 36
d) 72
View Answer

Answer: a
Explanation: Solve E[g(x)] = E[X3].
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7. The mean of the random variable x is
a) 1/4
b) 1/6
c) 1/3
d) 1/5
View Answer

Answer: a
Explanation: Solve integral (x f(x) dx) from negative infinity to x.

8. The variance of the random variable x is
a) 1/10
b) 3/80
c) 5/16
d) 3/16
View Answer

Answer: b
Explanation: Variance is given by E[X230] – 1/16.

(Q.9-Q.10) A joint sample space for two random variable X and Y has four elements (1,1), (2,2), (3,3) and (4,4). Probabilities of these elements are 0.1, 0.35, 0.05 and 0.5 respectively.

9. The probability of the event{X  2.5, Y  6} is
a) 0.45
b) 0.50
c) 0.55
d) 0.60
View Answer

Answer: a
Explanation: The required answer is given by Fxy(2.5, 6.0) = 0.1 + 0.35 = 0.45.
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10. The probability of the event that X is less than three is
a) 0.45
b) 0.50
c) 0.55
d) 0.60
View Answer

Answer: b
Explanation: The required answer is given by Fx(3.0) = Fxy(3.0, infinity) = 0.1 + 0.35 + 0.05 = 0.50.

Sanfoundry Global Education & Learning Series – Electronic Devices and Circuits.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn