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Please show all steps and work for pts. A light fixture holds two lightbulbs. Bu

ID: 1859510 • Letter: P

Question

Please show all steps and work for pts.


A light fixture holds two lightbulbs. Bulb A is of a type whos lietime is nomally distributed with mean 800 hours and standard deviation 100 hours. Bulb B has a lifetime that is normally distributed with mean 900 hours and standard deviation 150 hours. Assume that the lifetimes of the bulbs are independent.


A: What is the probability that bulb B lasts longer than bulb A?


B: What is the probability that bulb B lasts more than 200 hours longer than bulb A?


C: Another light fixture holds only one bulb. A bulb of type A is installed, and when it burns out, a bulb of type B is installed. What is the probability that the total lifetime of the two bulbs is more than 2000 hours?



Explanation / Answer

a) What is the probability that bulb B lasts longer than bulb A?

Define a new Random Variable D. Let D = B - A

E(D) = E(B) - E(A) = 900 - 800 = 100

V(D) = V(B) + V(A) = 150^2 + 100^2 = 32,500
[NOTE: the "addition of variance is correct! Always add variances]

Std. Dev. of D = sqrt (32500) = 180.28

P(B > A) = P(D > 0) = P[z > (0 - 100) / 180.28] = P(z > -0.555) , now look up in a Normal table:

P(z > -0.555) = 0.7106

So, the probability that bulb B lasts longer than bulb A equals 0.7106



b) Another light fixture holds only one bulb. A bulb of type A is installed, and when it burns out, a bulb of type B is installed. What is the probability that the total lifetime of the two bulbs is more than 2000 hours?

Define a new Random Variable T. Let T = A + B

E(T) = E(A) + E(B) = 800 + 900 = 1700

V(T) = V(A) + V(B) = 32,500
[NOTE: same variance as Part (a) above]

Std. Dev. of T = sqrt (32500) = 180.28

P(A + B > 2000) = P(T > 2000) = P[z > (2000 - 1700) / 180.28] = P(z > 1.664) , now look up in a Normal table:

P(z > 1.664) = 0.0481

So, the probability that the total lifetime of the two bulbs is more than 2000 hours is 0.0481


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