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For the following homework question, I haven't been able to get an answer, and I haven't been able to figure out my mistake. Was hoping someone could help.

My first thought was that X-Y will be normal with mean- 5-5=0 and variance- (.2)^2 + (.2)^2= 0.08. Then we can simply compute the value (which I got to be .2763). This is not correct according to the answer.

I then realized that we should have 25 sets (for the students). So I ended up using a distribution 25(X-Y) w/ mean- 0 * 25=0 and variance- .08 * 25= 2. We also adjust the bounds, getting P(-2.5<25(X-Y)<2.5). This got me .9229

For part b, I'm guessing we use the central limit theorem (we have >30 iid samples). Am I correct there? Can't verify since my approach for part 1 is wrong.

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  • $\begingroup$ Your general approach is definitely sound. But you computed the answer if there is just one student each in the morning and afternoon. How will your numbers change if there are 25 students in each? (What is the difference between the mean and variance of a single $X$ versus $\overline X$?) $\endgroup$
    – Greg Martin
    Apr 6, 2021 at 6:56
  • $\begingroup$ I just made that change haha. $\endgroup$
    – pasha
    Apr 6, 2021 at 6:58

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