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I have this problem:

A professor of statistics noticed that the marks in his course are normally distributed. He also noticed that his morning classes average 70% with a standard deviation of 13% on their final exams. His afternoon classes average 79% with a standard deviation of 9%.

What is the probability that a randomly selected student in the morning class has a higher final exam mark than a randomly selected student from an afternoon class?

I've been trying to solve this problem myself for quite awhile and not really sure what is wrong with my approach.

I let the morning class = X1 and the afternoon class = X2. So E(X1) = 70, σ(X1) = 13, E(X2) = 79, and σ(X2) = 9. Then I let the variable Y = X1-X2, and since X1 and X2 are both normally distributed, then Y will be as well. So I have the following for Y:

$$E(Y) = E(X_1)-E(X_2) = 70-79 = -9$$ $$σ(Y) = \sqrt{σ(X_1)^2 + σ(X_2)^2} = \sqrt{13^2+9^2} = \sqrt{250}$$

In oder to find the probability that a randmly selected student in the morning class has a higher final exam mark than one from the afternoon class, we want P(X1 > X2) = P(X1 - X2 > 0) = P(Y > 0). So I converted this to z-score...

$$Z = \frac{X-E(Y)}σ = \frac{0-(-9)}{\sqrt{250}} = \frac 9{\sqrt{250}} \approx 0.57$$

So that we are looking for P(Z > 0.57) or 1-P(Z$\leq$0.57).

I looked at my z-score table and found my answer to be 0.2843, which is marked as wrong. I've tried doing different things for rounding etc., and still can't seem to find the right answer. Any insight is greatly appreciated - maybe I made some dumb error that I can't seem to see. Thank you! :)

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  • $\begingroup$ DO NOT ROUND MORE THAN YOU HAVE TO UNTIL THE LAST STEP! $\endgroup$
    – Michael Hardy
    Commented Apr 11, 2017 at 16:59

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Your approach is perfectly fine, but you've rounded too much precision off your $z$-score, which is why your answer is off. The $z$-score is $0.56921$, thus the probability is $0.284607$.

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  • $\begingroup$ For some reason we are only allowed to use tables that take a z-score to 2 decimal pts only, and give a probability to 4 decimal points, even though the program is looking for an exact answer :| Thank you very much! $\endgroup$
    – Emma G.
    Commented Apr 8, 2017 at 22:00
  • $\begingroup$ @EmmaG. In that case, you will need to use linear interpolation of the table to achieve the necessary precision. Since $0.56921$ is $0.00921$ away from $0.56$ and $0.00079$ away from $0.57$, you would look up $\Pr[Z \le 0.56] = 0.2877$ and $\Pr[Z \le 0.57] = 0.2843$. Then you would calculate $$\Pr[Z \le 0.56921] \approx (0.079)(0.2877) + (0.921)(02843) = 0.2846.$$ This gets you the precision you need. $\endgroup$
    – heropup
    Commented Apr 9, 2017 at 0:52

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