In: Statistics and Probability
Maintenance of Production Facility
A production facility has several 3-D printing machines which need maintenance for two reasons. Maintenance type A: a sensor may have detected a fault, e.g. an oil leak. Maintenance type B: the quality of the printed product is unacceptable. Each maintenance requirement occurs at random and independently of other requirements for maintenance, with characteristics summarized in the following table:
Maintenance Type |
Average rate of occurrence per week totalled over all 3-D printing machines in the production facility |
Cost of maintenance |
A |
6.7 (maintenance events per week) |
$430 (per maintenance event) |
B |
2.1 (maintenance events per week) |
$2600 (per maintenance event) |
What is the probability of exactly 6 maintenance events of type A in half a week?
Question 45 options:
<0.02 |
|
Between 0.02 and 0.04 |
|
Between 0.04 and 0.06 |
|
Between 0.06 and 0.08 |
|
>0.08 |
Question 46 (1 point)
Maintenance of Production Facility
A production facility has several 3-D printing machines which need maintenance for two reasons. Maintenance type A: a sensor may have detected a fault, e.g. an oil leak. Maintenance type B: the quality of the printed product is unacceptable. Each maintenance requirement occurs at random and independently of other requirements for maintenance, with characteristics summarized in the following table:
Maintenance Type |
Average rate of occurrence per week totalled over all 3-D printing machines in the production facility |
Cost of maintenance |
A |
6.7 (maintenance events per week) |
$430 (per maintenance event) |
B |
2.1 (maintenance events per week) |
$2600 (per maintenance event) |
What is the standard deviation of the total cost of maintenance of both type A and type B events for one week?
Question 46 options:
<$2000 |
|
Between $2000 and $4000 |
|
Between $4000 and $6000 |
|
Between $6000 and $8000 |
|
>$8000 |
Question 47 (1 point)
Maintenance of Production Facility
A production facility has several 3-D printing machines which need maintenance for two reasons. Maintenance type A: a sensor may have detected a fault, e.g. an oil leak. Maintenance type B: the quality of the printed product is unacceptable. Each maintenance requirement occurs at random and independently of other requirements for maintenance, with characteristics summarised in the following table:
Maintenance Type |
Average rate of occurrence per week totalled over all 3-D printing machines in the production facility |
Cost of maintenance |
A |
6.7 (maintenance events per week) |
$430 (per maintenance event) |
B |
2.1 (maintenance events per week) |
$2600 (per maintenance event) |
What is the probability of more than 2 maintenance events of type B in one week?
Question 47 options:
<0.2 |
|
Between 0.2 and 0.4 |
|
Between 0.4 and 0.6 |
|
Between 0.6 and 0.8 |
|
>0.8 |
Question 48 (1 point)
Maintenance of Production Facility
A production facility has several 3-D printing machines which need maintenance for two reasons. Maintenance type A: a sensor may have detected a fault, e.g. an oil leak. Maintenance type B: the quality of the printed product is unacceptable. Each maintenance requirement occurs at random and independently of other requirements for maintenance, with characteristics summarized in the following table:
Maintenance Type |
Average rate of occurrence per week totalled over all 3-D printing machines in the production facility |
Cost of maintenance |
A |
6.7 (maintenance events per week) |
$430 (per maintenance event) |
B |
2.1 (maintenance events per week) |
$2600 (per maintenance event) |
What is the standard deviation of the number of maintenance events of type A in one week?
Question 48 options:
<2 |
|
Between 2 and 4 |
|
Between 4 and 6 |
|
Between 6 an 8 |
|
>8 |
45.
For half a week, Average rate of occurrence maintenance events of type A = 6.7/2 = 3.35
Assuming Poisson distribution, X ~ Poisson(3.35)
probability of exactly 6 maintenance events of type A in half a week
= P(X = 6) = exp(-3.35) * 3.35^6 / 6! = 0.0689
Between 0.06 and 0.08
46.
Total Cost = 430A + 2600B
Variance of Total Cost = Var(430A + 2600B)
For Poisson distribution, Var(X) = Average rate
Standard deviation of Total Cost = = 3928.719
Between $2000 and $4000 |
47.
Probability of more than 2 maintenance events of type B in one week
= P(Y > 2) where Y ~ Poisson(2.1)
= 1 - (P(Y = 0) + P(Y = 1) + P(Y = 2))
= 1 - [exp(-2.1) * 2.1^0 / 0! + exp(-2.1) * 2.1^1 / 1! + exp(-2.1) * 2.1^2/ 2! ]
= 1 - (0.1224564 + 0.2571585 + 0.2700164)
= 0.3504
Between 0.2 and 0.4
48.
Standard deviation of the number of maintenance events of type A in one week
= = 2.59
Between 2 and 4