Step 1 :The problem is asking for the probability of a score being less than 489 in a normal distribution with a mean of 500 and a standard deviation of 10.5.
Step 2 :This can be solved using the Z-score formula, which is \((X - μ) / σ\), where X is the score, μ is the mean, and σ is the standard deviation.
Step 3 :The Z-score gives us the number of standard deviations a score is from the mean.
Step 4 :Substituting the given values into the Z-score formula, we get \((-1.0476190476190477)\).
Step 5 :We can then use a Z-table to find the probability associated with this Z-score, which is approximately 0.1474.
Step 6 :Final Answer: The probability that a randomly selected medical student who took the test had a total score that was less than 489 is approximately \(\boxed{0.1474}\).