tutorny
Posts:
39
Registered:
1/28/06


Re: Help with probability&stat problem
Posted:
Jun 11, 2007 1:27 PM


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.
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!!

