Company XYZ know that replacement times for the quartz time pieces it produces are normally distributed with a mean of 17.7 years and a standard deviation of 1.8 years.

Question

the probability that a randomly selected quartz time piece will have a replacement time less than 14.1 years?

P(X < 14.1 years)

if the company wants to provide a warranty so that only 1% of the quartz time pieces will be replaced before the warranty expires, what is the time length of the warranty? 

Answer 1

Sounds kind of weird when it is known that typical failure rates follows a Weibull distribution NOT a normal distribution.

Assuming some weird quartz time piece with built in failure we want x where P(x) <= .01

given Ν(17.7,1.8)

Part 1. Using the cumulative normal distribution from our TI calculator nmcdf(-10e-3,14.1,17.7,1.8) =0.829 or 82.9%

Part 2. Using the inverse cumulative normal distribution from our calculator invnm(.01,17.8, 1.8) = 13.51 years

or use (https://onlinestatbook.com/2/calculators/inverse_normal_dist.html)

Alternate: Use Z table where Z for P(Z) <.01. = -2.326 , then use algebra to solve for x where Z = (x-μ)/σ

remembering that μ=17.7 and σ = 1.8 The answer is the same (13.51 years)