Question

Company XYZ know that replacement times for the quartz time pieces it produces are normally distributed with a mean of 11 years and a standard deviation of 1.8 years.

a) Find the probability that a randomly selected quartz time piece will have a replacement time less than 7.58 years?


b) If the company wants to provide a warranty so that only 0.8% of the quartz time pieces will be replaced before the warranty expires, what is the time length of the warranty?
warranty =  years

Answer 1

Given

= 11 , = 1.8

We convert this to standard normal as

P(X < x) = P(Z < (X - ) / )

a

P(X < 7.58) = P(Z < ( 7.58 - 11) / 1.8)

= P(Z < -1.9)

= 0.0287

b)

We have to calculate x such that P(X < x) = 0.008

P(Z < (X - ) / ) = 0.008

From Z table, z-score for the probability of 0.008 is -2.409

(X - ) / = -2.409

( x - 11) / 1.8 = -2.409

Solve for x

x = 6.6638