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
Have a GREAT day!
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