Question

A factory manufactures lightbulbs where the lifespan can be approximated by a normal distribution having an average of 1000 hours and a standard deviation of 200 hours.

a) what is the probability that a lightbulb's lifespan exceeds 1330 hours?

b) The factory would like to offer a replacement warranty for all lightbulb that does not last longer than "Y" hours. Determine what "Y" should be if the company wants to replace at MOST 2.5% of its lightbulbs with this warranty.

Answer 1

Solution :

(a)

P(x > 1330) = 1 - P(x < 1330)

= 1 - P[(x - ) / < (1330 - 1000) /200 ]  

= 1 - P(z < 1.65)

= 0.0495

(b)

P(Z > 1.96) = 0.025

z = 1.96

Using z-score formula,

x = z * +

x = 1.96 * 200 + 1000 = 1392

Y = 1392