Question

A DC battery has a time to failure that is normally distributed with a mean of 30 hours and a standard deviation of 4 hours.

(a) What is the 25 hour reliability?

(b) When should a battery be replaced to ensure that there is not more than a 5% change of failure prior to replacement?

(c) Two batteries are connected in parallel to power a light. Assuming that the light does not fail, what is the 35 hour reliability for the power source?

Answer 1

P(X < A) = P(Z < (A - mean)/standard deviation)

a) 25 hour reliability = P(X > 25)

= 1 - P(X < 25)

= 1 - P(Z < (25 - 30)/4)

= 1 - P(Z < -1.25)

= 1 - 0.1056

= 0.8944

= 89.44%

b) Let the battery be replaced at R hours to ensure that there is no more that 5% chance of failure

P(X < R) = 0.05

P(Z < (R - 30)/4) = 0.05 (take Z score corresponding to 0.05 from standard normal distribution table)

(R - 30)/4 = -1.645

R = 23.424 hours

c) P(a battery lasting 35 hours) = P(X > 35)

= 1 - P(X < 35)

= 1 - P(Z < (35 - 30)/4)

= 1 - P(Z < 1.25)

= 1 - 0.8944

= 0.1056

P( a battery not lasting 35 hours = 1 - 0.1056 = 0.8944

35 hour power reliability = P(at least one battery not failing)

= 1 - P(both batteries failing)

= 1 - 0.8944²

= 0.2000