A two-section exam total

Question

An exam has two independent sections. Section A's score is normally distributed with mean 45 and standard deviation 8. Section B's score has mean 55 and standard deviation 15.

What is the probability that a student's total score exceeds 108.5?

Accepted Answer

The total of the two independent section scores is normal. Add the means directly, and add the variances (the squares of the standard deviations). Mean of the total: 45 + 55 = 100. Variance of the total: 8^2 + 15^2 = 64 + 225 = 289, so the standard deviation is \sqrt{289} = 17. The total score is normal with mean 100 and standard deviation 17. Convert the cutoff 108.5 to a z-score: z = \frac{108.5 - 100}{17} = 0.5. The area beyond z = 0.5 is about 1 - 0.6915 = 0.3085, so P(	ext{total} 108.5) \approx 0.3085. Because the cutoff is only half a standard deviation above the average, roughly 31% o…

A two-section exam total

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