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

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

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.

true !

step-by-step explanation:

the 8 in 8,420 is in the thousands place so it's value is 8 thousand.

the 8 in 6,870 is in the hundreds place so it's value is 8 hundred.

divide 8000 by 800 to find the answer:

8000 / 800 = 10

the 8 in 8,420 is 10 times greater than the 8 in 6,870