Check my though process-statistics?

Among U.S. households, 24% have telephone answering machines. When polling 2,500 households, find the probability that more than 650 have answering machines.

My answer: based on 650/2500=26%(p)

I used this methodology:

Sqrt of p(1-p) /n

.26(.74)/2500=.0000769, SQRT OF THIS =0087726 rounded up to .009

I then looked at z chart for .009 and found my final answer of .5359 or 53.59%

Thanks for you help. I am still tring to wrap my brain around your method.

using your steps, I entered the Z chart at 2.36 and it gave me a probability of .9909. This seems a high based on the original questions.

when,I did a search here(did not know I could do that,wow), there were several similar questions with several different answers

Sorry but that is not correct.

p not equal to 650 / 2500

We are already told that p = 0.24
For the size of the problem, you are right that we need to use the normal approximation

X ~ B(2500, 0.24)

As a normal approximation

mu = np = 0.24 * 2500 = 600
sigma^2 = npq = 456
sigma = 21.354
X ~ N(600, 456)

Remember we are dealing with the distribution of X not Xbar. So you don't need p(1-p) /n

P(X > 650) = P(X > 650.5)
This is a continuity correction.
Now proceed with the normal distribution
P (Z > [650.5 - 600]/21.354)
= P (Z > 2.365)
Now use the tables to find that.


**EDIT**
In your tables, you say you get 0.9909, but if you look at the shaded region just above the tables, you'll see that that represents P(Z < 2.365).
So P(Z > 2,365) = 1 - 0.9909 = 0.0091
which is much more realistic.

The only difference between my method and yours is the value of p. I used 0.24 rather than 0.26.