Question

The weekly repair cost Y for a machine has a probability density function given by

f(y) = 3(1-y)², 0 < y < 1 and 0 elsewhere
with measurements given in hundreds of dollars. How much money should be budgeted each week for repair costs so that the actual cost will exceed the budgeted amount only by 10% of the time?

Answer 1

To budget for weekly machine repair costs given a probability density function, we need to calculate the 90th percentile of the distribution. This involves integrating the density function, setting the result equal to 0.90, and solving for the percentile value of y that yields this area under the curve.

Step-by-step explanation:

The student is asking about how to determine a budget for weekly machine repair costs given a probability density function, where the actual cost exceeds the budgeted amount only 10% of the time. This requires calculating the 90th percentile of the distribution represented by the probability density function (f(y) = 3(1-y)² for 0 < y < 1).

To find the 90th percentile, we need to solve for y in the equation ∫ f(y) dy = 0.90, where the integral is computed from 0 to y. This calculus-based problem involves finding the value of y such that the area under the probability density function from 0 to y is equal to 0.90. We begin by setting up the integral:

∫0y3(1-y)^2 dy = 0.90

Integrating 3(1-y)^2 with respect to y gives us [y - y^2 + y^3/3] evaluated from 0 to y. After performing the integration and simplifying, we can solve the resulting equation for y to find the value that corresponds to the 90th percentile of the cost distribution.

The solution to this equation will tell us the budget that should be allocated for repairs such that the actual costs exceed this budget only 10% of the time.