Question

The probability that a specific hydraulic actuator can be successfully repaired in the field, once it has failed, is estimated at 0.4. You are asked to optimize the shipment of a limited supply of spares and maintenance personnel. If 15 actuators have failed today, what is the probability that A) at least 10 are repairable? B) from 3 to 8 are repairable? C) exactly 5 are repairable?

Answer 1

ANSWER:

Given that:

The probability that a specific hydraulic actuator can be successfully repaired in the field, once it has failed, is estimated at 0.4.

You are asked to optimize the shipment of a limited supply of spares and maintenance personnel.

P = 0.4

N = 15

It is a binomial distribution with two outcomes.

1. Repairable

2. Not Repairable

P = probability of success be that the actuator is repairable

P = 0.4

Now

perform a binomial to normal approximation as

np = 0.4 * 15

= 6 > 5

n ( 1 - P ) = 15 * ( 1-0.4 )

= 9 > 5

mu = nP

= 0.4 * 15

mu = 6

A)

at least 10 are repairable

B)

from 3 to 8 are repairable

P(3<=x<=8)

= P( X < = 8 ) -P( X < =3 )

C)

exactly 5 are repairable