Answer:
- Preventive Maintenance
Explanation:
Given that,
No. of breakdowns 0 1 2 3 4
Frequency 9 2 4 4 1
So,
No. of breakdowns Monthly probability
0 0.45
1 0.10
2 0.20
3 0.20
4 0.05
_______________________________________
Total 1.00
________________________________________
As per the above data,
The breakdowns expected every month = (0 * 0.45) + (1 * 0.10) + (2 * 0.20) + (3 * 0.20) + (4 * 0.05)
= 1.30
∵ The cost expected for the breakdowns every month = 1.30 * $400
= $ 520
Now,
Per month cost of preventive maintenance = (0.5 * $400) + $200
= $ 400
Thus, <u>preventive maintenance</u> would be more economical.
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