1. The average life expectancy of computers produced by Ahmadi, Inc. is 5 years with a standard deviation of 11 months. Assume that the lives of computers are normally distributed. (Suggestion: For this problem, convert ALL of the units to months. ) a. What is the probability that a randomly selected computer will have a life expectancy of at least 8 years? b. Computers that fail in less than 4 years will be replaced free of charge. What percentage of computers are expected to be replaced free of charge? Answer: 5years x 12months = 60 months (mean value) Standard deviation = 11 months (SD) Normal distribution 8 years x 12 = 96 months (target value) 96-60 /11 = 3.2727 for the z-score and then find the value and subtract it from 1 for probability A. Probability is .5005 = 50.05% Probability of fail in 4 years = 48 months 48-60 /11 = -1.0909 B. Z-score -1.09 and P-value .1379 = 13.79%