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Test: Continuous Time Signals


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10 Questions MCQ Test Analog Electronics | Test: Continuous Time Signals

Test: Continuous Time Signals for Electrical Engineering (EE) 2023 is part of Analog Electronics preparation. The Test: Continuous Time Signals questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Continuous Time Signals MCQs are made for Electrical Engineering (EE) 2023 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Continuous Time Signals below.
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Test: Continuous Time Signals - Question 1

(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=2.1.

Q. The probability that more than 3 cars will arrive during any 10 min period is

Detailed Solution for Test: Continuous Time Signals - Question 1

Evaluate 1 – P(x = 0) – P(x = 1) – P(x = 2) – P(x = 3).

Test: Continuous Time Signals - Question 2

The probability that no car will arrive is

Detailed Solution for Test: Continuous Time Signals - Question 2

Evaluate P(x = 0).

Test: Continuous Time Signals - Question 3

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

Q. The probability that there will be five or more murder in a given week is

Detailed Solution for Test: Continuous Time Signals - Question 3

P(5 or more) = 1 – P(0) – P(1) – P(2) – P(3) – P(4) = 0.1847.

Test: Continuous Time Signals - Question 4

On the average, how many weeks a year can Delhi expect to have no murders ?

Detailed Solution for Test: Continuous Time Signals - Question 4

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

Test: Continuous Time Signals - Question 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?

Detailed Solution for Test: Continuous Time Signals - Question 5

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.

Test: Continuous Time Signals - Question 6

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

Q. The expect value of g(x) = X3 is

Detailed Solution for Test: Continuous Time Signals - Question 6

Solve E[g(x)] = E[X3].

Test: Continuous Time Signals - Question 7

The mean of the random variable x is

Detailed Solution for Test: Continuous Time Signals - Question 7

Solve integral (x f(x) dx) from negative infinity to x.

Test: Continuous Time Signals - Question 8

The variance of the random variable x is

Detailed Solution for Test: Continuous Time Signals - Question 8

Variance is given by E[X230] – 1/16.

Test: Continuous Time Signals - Question 9

(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.

Q. The probability of the event{X  2.5, Y  6} is

Detailed Solution for Test: Continuous Time Signals - Question 9

 The required answer is given by Fxy(2.5, 6.0) = 0.1 + 0.35 = 0.45.

Test: Continuous Time Signals - Question 10

The probability of the event that X is less than three is

Detailed Solution for Test: Continuous Time Signals - Question 10

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

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