This question is incomplete, the complete question is;
Suppose that for a particular piece of machinery, the frequency distribution of monthly breakdowns is as follows;
Number of breakdowns prob.
0 0.50
1 0.25
2 0.10
3 0.10
4 0.05
The cost of a breakdown is $2,000, and the cost of a preventive maintenance program is $2,000 per month. If the preventive maintenance program is adopted, the probability of a machine breakdown is negligible. How much better off per month would the firm be if it adopted preventive maintenance
Answer:
the firm will be $100 worse off per month if it adopted preventive maintenance.
Step-by-step explanation:
Given the data in the question;
the probability of breakdown iwill be;
= (0×0.5) + (1×0.25) + (2×0.10) + (3×0.10) +(4×0.05)
= 0 + 0.25 + 0.2 + 0.3 + 0.2 = 0.95 or 95%
Now the cost of machine failure will be;
⇒ 2000 × 0.95= $1900
if they implemented preventive measures, the cost is $2000,
if the preventive measure taken the cost is $2000, the benefit is for $1900, and the output is additional cost of $100.
Therefore the firm will be $100 worse off per month if it adopted preventive maintenance.
