Suppose that for a particular piece of machinery, the frequency distribution of monthly breakdowns is as follows: Number of Breakdowns Prob. 0 .50 1 .25 2 .10 3 .10 4 .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

$100 worse off

Step-by-step explanation:

Number of Breakdowns : 0, 1, 2, 3, 4

Prob.: 0.50, 0.25, 0.10, 0.10, 0.05

Cost of a breakdown = $2,000

cost of a preventive maintenance program is $2,000 per month

Expected probability of breakdown:

E(x) = Σ(X * p(x)) = (0*0.5) + (1*0.25) + (2*0.10) + (3*0.10) + (4*0.05) = 0.95

E(x) * cost of breakdown = (0.95 * 2000) = $1900

Hence, difference between breakdown and preventive maintenance

1900 - 2000 = - $100

$100 worse off

it is a. 4.2 because 7.6-3.4 gets the remaining length of 4.2. i might be wrong but i am pretty sure this is the answer.

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step-by-step explanation: