A work cell is scheduled to build 120 digital light processor (dlp) assemblies each week. these assemblies are later installed i
1 answer:
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Answer:
x= 30/7
Step-by-step explanation:
See image below:)
Answer:
a) (1,2), (1,3), (1,4), (1,5), (2,3), (2,4),(2,5),(3,4), (3,5) and (4,5)
b)
PX(0) = 0.3
PX(1) = 0.6
PX(2) = 0.1
c)
- F(x) = 0 if x<0
- F(x) = 0.3 if 0 ≤ x < 1
- F(x) = 0.9 if 1 ≤ x < 2
- F(x) = 1 if x ≥2
Step-by-step explanation:
Lets call the boards 1,2,3,4 and 5
a) There are possible outcomes: (1,2), (1,3), (1,4), (1,5), (2,3), (2,4),(2,5),(3,4), (3,5) and (4,5).
b) X is 0 if the selected boards doesnt contain 1 or 2. There are possibilities for this, in concrete, (3,4), (3,5) and (4,5).
PX(0) = 3/10 = 0.3
There is only 1 possibility so that X = 2, which is (1,2), therefore
PX(2) = 1/10 = 0.1
In any other case (there are 6 remaining), X will be equal to 1, so
PX(1) = 6/10 = 0.6
Therefore, X has the following distribution
PX(0) = 0.3
PX(1) = 0.6
PX(2) = 0.1
c) F(0) = 0.3
F(1) = 0.3+0.6 = 0.9
F(2) = 1
We have, as a consequence
- F(x) = 0 if x<0
- F(x) = 0.3 if 0 ≤ x < 1
- F(x) = 0.9 if 1 ≤ x < 2
- F(x) = 1 if x ≥2