The lifespan of an electric bulb is normally distributed with a mean 2020 hours and standard deviation 420 hours. The manufacturer of the electric bulbs guarantees that the ligh bulb will last for 2000 hours.
a) calculate the percentage of light bulb which fail within the guantee period

Probability(X < 2000), where X~N(2020,420)


The first step is to normalize X so that the transformed variable is distributed N(0,1) (referred to as standard normal, random variable is typically denoted Z). You must perform the same transformation on 2000.

P(X < 2000) = P((X-2020)/420 < (2000-2020)/420) = P(Z < -.0476), which is the same as saying the standard normal cdf of -.0476 (it is the cdf because we are looking at the probability a RV is LESS THAN some threshhold).

Now you can look up -.0476 in a Z-table to find the probability, which turns out to be approximately 48%.

Make sure to check my numbers as I make errors often

This post was edited by zackill4 on Aug 30 2013 02:29pm

hmm since manufacturer guarantees 2000 but the mean is greater than 2000, i would have thought it would fail slightly over 50% and not under?

No - the guarantee is less than the expected life of the lightbulb so that it will be easier for the guarantee to succeed most of the time.

derp. dunno what i was thinking.