Calculate the reliability of this series/parallel system, assuming that the prob

Question

Calculate the reliability of this series/parallel system, assuming that the probability that each component works is the same, p. The phone lines to an airline reservations system are occupied 45% of the time. Assume that the events that the lines are occupied on successive calls are independent. Assume that eight calls are placed to the airline. What is the probability that for exactly two calls the lines are occupied? What is the probability that for at least one call the lines are occupied? What is the expected number of calls in which the lines are occupied? A data set suggests that for a particular experimental setup involving a small bar of polonium, the number of collisions of a particle with a small screen placed near the bar during an 8-minute period can be modeled as a Poisson variable with mean lambda = 7.74. Let X be the number of collisions in the next 4-minute period. Evaluate the following: Scores on a standardized test are approximately normally distributed with a mean of 480 and a standard deviation of 90. What proportion of the scores are above 700? What proportion of the scores are between 420 and 520? 85% of the scores are below what value? Consider a continuous random variable X with density f(x) = {3x^2/56 if 2

Explanation / Answer

1) Series System This is a system in which all the components are in series and they all have to work for the system to work. If one component fails, the system fails.

Parallel System This is a system that will fail only if they all fail.

reliability = P(A) * P(atleast (B and C) OR (D and E))

P(A) = p

P(B and C) = p * p = p2

P(D and E) = p * p = p2

P(atleast (B and C) OR (D and E)) = 1 – P(both fail) = 1 – (1 – p2)2
= 1 – 1 - p4 + 2p2 = 2p2 – p4

Reliability = p * (2p2 – p4) = 2p3 – p5

Note: A is the first component from the left, B and C are the components in series in the top branch and D and E are the components in series of the 2nd branch

2)

Let X denote the number of times the line is occupied. Then, X has a binomial distribution with n = 8 and p = 0.45.

(a) P(X = 2) = (8 choose 2) * (0.45)2 (0.55)6 = 28 * 0.2025 * 0.02768 = 0.1569 = 15.69%

(b) P(X 1) = 1 P(X = 0) = 1 – (8 choose 0) * (0.45)0 * (0.55)8 = 0.9916
= 99.16%

(c) E(X) = np = 8(0.45) = 3.6

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