tutorny wrote: > Thanks so much for your reply! > >> I read that as the probability that 30 or more leave. You've apparently >> read it as the probability that 30 or fewer leave. The longer I look >> at the question, the less sure I am which of us is correct. > > A lot of us in the class had issue with the wording of the problem. > We emailed the professor, and she said that the correct understanding > is "30 or fewer leave" or P(x <=30), so that's what I've been working > with. > >> I personally agree that the normal approximation >> is a good fit for this kind of question. Especially since we're not >> way out on the tail of the curve. > > Exactly the reasoning I used. The book said that normal approximation > should work as long as p isn't too close to 0 or 1, and I figured that > 0.20 should be reasonable for this. > >>> We want P(x <=30) >>> When x = 30, z = (x - u)/s.d = (30 - 22)/4.1952 = 1.9069 >> Here, I think you've committed a fencepost error. If you're treating >> a normal distribution as if it were a discrete histogram then you >> want to put your cutoff points between the bars on the histogram, not >> in the middle of the bars. You want to look at x=29.5 or x=30.5. >> >> You decide whether to use the x=29.5 or the x=30.5 cutoff by considering >> whether the case when x=30 is included or excluded in the set of cases >> you are looking for. > > I totally missed that point, and looking over the chapter I see that > you are right. Since this is not continuous, I have to use either > 29.5 or 30.5. Since I've confirmed that the question asks for P(x <= > 30) I don't think that I can use 30.5, since that's > 30, so I have to > use 29.5.
Think again. How would you work out the probability that exactly 30 left? See if your textbook has any examples like that.
> > So, for x = 29.5: z = (x - u)/s.d = (29.5 - 22)/4.1952 = 1.7878 > P(z <= 1.7878) = 0.9631 and the answer is 96.31% which is also > reasonable. > > How is that? > > Thanks!! >
-- Bruce Weaver email@example.com www.angelfire.com/wv/bwhomedir