In article <firstname.lastname@example.org>, tutorny <email@example.com> writes: > I am looking at solving the following problem: > > Past records show that at a given college 20% of the students who > began as psychology majors either changed their major or dropped out > the school. An incoming class has 110 > beginning psychology majors. What is the probability that as many as > 30 of these students leave the psychology program?
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.
> I think that I can solve it using the normal approximation to the > binomial probability distribution, as follows: > > n =110, p = 0.20 > mean = u = np = 110*0.20 = 22 > standard deviation = s.d. = (n*p*q)^.5 = (110*.20*.80)^.5 = 4.1952
Looks reasonable. And 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.
> 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.
Think about it this way. You're approximating p(x=30) in the discrete case by p(x<=30.5) - p(x<=29.5) in the continuous model.
Or think about it this way. If you were asked for the probability that x is 30 or more, do you want the answer to be the complement of the probability that x is 30 or less? It will be if you use p(x>=30) and p(x<=30) as your respective estimates. Or do you want the non-zero probability that x is 30 exactly to figure in somehow? That's where the 29.5 and 30.5 make themselves useful.