In: Statistics and Probability
A specialist electronics company carries out its own repairs.
The manager of the repair center is becoming concerned that the
cost of repairs is continually increasing. She has observed the
operations of the repair center and collected data relating to the
number of items being repaired each day, and also the cost of
repairing these items over a 50-day period. This information is
presented in the following tables:
Repairs per day |
Probability |
1 |
0.08 |
2 |
0.22 |
3 |
0.38 |
4 |
0.26 |
5 |
0.06 |
Cost of repair ($) |
Probability |
5 |
0.28 |
10 |
0.42 |
15 |
0.22 |
20 |
0.06 |
25 |
0.02 |
Use the following series of random numbers to simulate the number
and total daily cost of repairs carried out by the department over
a 5-day period. Use the following random numbers in order (from
left to right) for the simulation of repairs per day:
0.05 |
0.33 |
0.25 |
0.57 |
0.90 |
Use the following random numbers in order (from left to right,
first row first - as you need them) for the simulation of the cost
of repairs:
0.01 |
0.46 |
0.21 |
0.78 |
0.99 |
0.06 |
0.20 |
0.96 |
0.81 |
0.18 |
0.03 |
0.67 |
0.26 |
0.19 |
0.90 |
0.32 |
0.86 |
0.10 |
0.84 |
0.76 |
Please note you may find that you do not need all of the last
sequence of 24 random numbers.
What is the total cost of repairs after 5 days of simulation?
Solution
Preparatory Work
Random Number Assignment |
|||||||||
Repairs per day |
Probability |
Cumulative Probability |
Random numbers assigned |
||||||
1 |
0.08 |
0.08 |
0.01 - 0.08 |
||||||
2 |
0.22 |
0.3 |
0.09 - 0.30 |
||||||
3 |
0.38 |
0.68 |
0.31 - 0.68 |
||||||
4 |
0.26 |
0.94 |
0.69 - 0.94 |
||||||
5 |
0.06 |
1 |
0.95 - 0.00 |
||||||
Cost of repair ($) |
Probability |
Cumulative Probability |
Random numbers assigned |
||||||
5 |
0.28 |
0.28 |
0.01 - 0.28 |
||||||
10 |
0.42 |
0.7 |
0.29 - 0.70 |
||||||
15 |
0.22 |
0.92 |
0.71 - 0.92 |
||||||
20 |
0.06 |
0.98 |
0.93 - 0.98 |
||||||
25 |
0.02 |
1 |
0.99, 0.00 |
||||||
Back-up Theory
Total Cost of Repair per Day = (Number of Repairs per Day) x (Repair Cost per Repair)
Now to work out the solution,
Following codes are employed for ease in presentation.
RN: Random Number; NR: Number of Repairs per Day; CR: Repair Cost per Repair
TC: Total Cost of Repair per Day
Five-day Simulation
Day |
Repairs per Day |
Cost per Repair |
TC |
||
RN |
NR |
RN |
CR |
||
1 |
0.05 |
1 |
0.01 |
5 |
5 |
2 |
0.33 |
3 |
0.46 |
10 |
30 |
3 |
0.25 |
2 |
0.21 |
5 |
10 |
4 |
0.57 |
3 |
0.78 |
15 |
45 |
5 |
0.90 |
4 |
0.99 |
25 |
100 |
Total |
190 |
Thus, total cost of repair for the 5-day period = $190 ANSWER
DONE