The average age of doctors in a certain hospital is 50 years old . Suppose the distribution of ages is normal and has a standard deviation of 4 years. If 16 doctors are chosen at random for a committee find the probability that the average age of those doctors is less than 50.3 years. Solution The standard deviation of sample means is 4 years/(16) = 1 year 50.3 years corresponds to a z-score of (50.3 - 50)/1 = 0.3 The probability of a z-score less than 0.3 is 0.618.