In: Statistics and Probability
One of the large photocopiers used by a printing company has a
number of special functions unique to that particular model. This
photocopier generally performs well but, because of the complexity
of its design and the frequency of usage, it occasionally breaks
down. The department has kept records of the number of breakdowns
per month over the last fifty months. The data is summarized in the
table below:
Number of Breakdowns |
Probability |
0 |
0.12 |
1 |
0.32 |
2 |
0.24 |
3 |
0.20 |
4 |
0.08 |
5 |
0.04 |
The cost of a repair depends mainly on the time taken, the level of
expertise required and the cost of any spare parts. There are four
levels of repair. The cost per repair for each level and
probabilities for different levels of repair are shown in table
below:
Repair Category |
Repair Cost |
Probability |
1 |
$35 |
0.50 |
2 |
$75 |
0.30 |
3 |
$150 |
0.16 |
4 |
$350 |
0.04 |
Based on the probabilities given in the two tables and using the
random number streams given below, simulate for each of 12
consecutive months the number of breakdowns and the repair cost of
each breakdown. Note that for each month you must compute both the
number of breakdowns, the repair cost for each breakdown (if any)
and the total monthly repair cost as well as the total annual
repair cost to answer the following questions.
Use the following random numbers in order (from left to right) for
the simulation of number of breakdowns per month:
Jan |
Feb |
Mar |
Apr |
May |
Jun |
Jul |
Aug |
Sep |
Oct |
Nov |
Dec |
0.13 |
0.21 |
0.08 |
0.09 |
0.89 |
0.26 |
0.65 |
0.28 |
0.97 |
0.24 |
0.10 |
0.90 |
Use the following random numbers in order (from left to right,
first row first - as you need them) for the simulation of repair
cost for each breakdown.
0.19 |
0.39 |
0.07 |
0.42 |
0.65 |
0.61 |
0.85 |
0.40 |
0.75 |
0.73 |
0.16 |
0.64 |
0.38 |
0.05 |
0.91 |
0.97 |
0.24 |
0.01 |
0.27 |
0.69 |
0.18 |
0.06 |
0.53 |
0.97 |
What was the total monthly repair cost in May?
A. |
$220 |
|
B. |
$35 |
|
C. |
$300 |
|
D. |
$75 |
|
E. |
$140 |
Answer:-
Given That:-
One of the large photocopiers used by a printing company has a number of special functions unique to that particular model. This photocopier generally performs well but, because of the complexity of its design and the frequency of usage, it occasionally breaks down. The department has kept records of the number of breakdowns per month over the last fifty months.
Given,
Random numbers assigned in number of breakdown.
No. of Breakdown | Probability | Cumulative Probability | Random Numbers |
0 | 0.12 | 0.12 | 0.00-0.11 |
1 | 0.32 | 0.44 | 0.12-0.43 |
2 | 0.24 | 0.68 | 0.44-0.67 |
3 | 0.20 | 0.88 | 0.68-0.87 |
4 | 0.08 | 0.96 | 0.88-0.95 |
5 | 0.04 | 1.00 | 0.93-0.99 |
Simulation of number of breakdown per month:-
Month | Random Numbers | No. of breakdown |
Jan | 0.13 | 1 |
Feb | 0.21 | 1 |
Mar | 0.08 | 0 |
Apr | 0.09 | 0 |
May | 0.89 | 4 |
Jun | 0.26 | 1 |
Jul | 0.65 | 2 |
Aug | 0.28 | 1 |
Sep | 0.97 | 5 |
Oct | 0.24 | 1 |
Nov | 0.10 | 0 |
Dec | 0.90 | 4 |
Random Numbers assigned in cost per Repairs
Repair category | Repair Cost($) | Probability | Cumulative Probability | Random Numbers |
1 | 35 | 0.50 | 0.50 | 0.00-0.49 |
2 | 75 | 0.30 | 0.80 | 0.50-0.79 |
3 | 150 | 0.16 | 0.96 | 0.80-0.95 |
4 | 350 | 0.04 | 1.00 | 0.96-0.99 |
simulation of monthly repair cost for each breakdown.
Month | No. of breakdown | Random Number | Repair Cost per breakdown | Total Repair cost |
Jan | 1 | 0.19 | $35 | $35 |
Feb | 1 | 0.39 | $35 | $35 |
Mar | 0 | 0.07 | $35 | 0 |
Apr | 0 | 0.42 | $35 | 0 |
May | 4 | 0.65 | $75 | $300 |
Jun | 1 | 0.61 | $75 | $75 |
Jul | 2 | 0.85 | $150 | $300 |
Aug | 1 | 0.40 | $35 | $35 |
Sep | 5 | 0.75 | $75 | $375 |
Oct | 1 | 0.73 | $75 | $75 |
Nov | 0 | 0.16 | $75 | 0 |
Dec | 4 | 0.64 | $75 | $300 |
Therefore The total monthly repair cost in May $ 300
The correct answer is option C.
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