A major television manufacturer has determined that its 19-inch color tv picture tubes have a mean service life that can be modeled by a normal distribution with a mean of six years and a standard deviation of one-half year. What probability can you assign to service lives of atleast five years, six years, and seven years? If the manufacturer offers service contracts of four years on these picture tubes, what percentage can be expected to fail from wear-out during the service period?
Explanation / Answer
Normal Distribution mean = 6 yr standard deviation = 1.5 yr Probability to get service life at least 5 years = Area under standard normal distribution curve upto z value corresponding to 5 years z (X = 5 yr) = (5-6)/1.5 = -0.67 P(X>= 5yr) = 1 - P(z>= -0.67) = P(z= 6yr) = P(z>= 0) = 0.5 Probability to get service life at least 7 years = Area under standard normal distribution curve upto z value corresponding to 7 years z (X = 7 yr) = (7-6)/1.5 = 0.67 P(X>= 7yr) = P(z>= 0.67) = 1- P(z