Answer:
a. y=1
b. Mean= 0.5; variance=0.5
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.
