The reliability of a piece of equipment is frequently defined to be the probability, P, that the equipment performs its intended function successfully for a given period of time under specific conditions. Because P varies from one point in time to another, some reliability analysts treat P as if it were a random variable. Suppose an analyst characterizes the uncertainty about the reliability of an articulating robot used in an automobile assembly line using the pdf. listed below. Then, after careful investigation, the analyst discovers that P is definitely between 0.82 and 0.97, but that there is complete uncertainty about where it lies between these values. Describe the pdf. the analyst should now use to describe P. f(p)

c. In the first instance, the probability that p > 0.95 would mean that we are looking for the area of the rectangle with a height of 1 between 0.95 and 1, which implies it has length 0.05.

Therefore, the area is length × height, which is 0.05 ×1 = 0.05.

Second case, the probability that p is less than 0.95 would then be one minus the probability that p is greater than 0.95, so 1 - 0.05 = 0.95

D. a horizontal line at 20 between 0.90 and 0.95, and will be zero outside of this range.

Step-by-step explanation:

With the use of probability distribution, the probability that the equipment performs has a uniform distribution with minimum 0 and maximum 1.

a) The graph of the probability distribution will be 0 outside of the range of 0 to 1, so therefore y = 0. Inside the interval from 0 to 1, it will be constant (which is a horizontal line) with height 1÷(1-0) = 1, therefore y = 1.

b) The mean of the uniform is (maximum+minimum)/2.

Therefore, (1+0)/2 = 1/2 or 0.5.

The variance of the uniform is

(maximum-minimum)^2/12,

so (1-0)^2/12 = 1/12.

c) Since the probability distribution is rectangular, you can find probabilities by recalling the area of a rectangle which is: area = length × breadth.

In the first instance, the probability that p > 0.95 would mean that we are looking for the area of the rectangle with a height of 1 between 0.95 and 1, which implies it has length 0.05.

Therefore, the area is length × breadth, which is 0.05 ×1 = 0.05.

In the second case, the probability that p is less than 0.95 would then be one minus the probability that p is greater than 0.95, so 1 - 0.05 = 0.95.

d) If it is known that p is between 0.90 and 0.95, without the value, then we would assume that p has a uniform distribution between 0.90 and 0.95 since p originally had a uniform distribution.

In this case,

f(p) =

1/(0.95-0.90) = 20, 0.90 < p < 0.95,

0, otherwise.

In a nutshell, this function will have a horizontal line at 20 between 0.90 and 0.95, and will be zero outside of this range.