random sampling question - I need help

[anthonys]

A journalist wanted to learn the views of the chief executive officers of the 500 largest United States corporations on program trading of stocks. In the time available, it was only possible to contact a random sample of eighty-one of these chief executive officers. If 55% of all the population members believe that program trading should be banned, what is the probability that less than half the sample members hold this view?

Ans. 0.8159 (but I think it might be wrong)

I’ve been unable to attain this answer despite numerous attempts, I would really appreciate some help.

[JohnM]

You can use the normal approximation to the binomial for this problem. The mean is n*p, and the standard deviation is the square root of n*p*q

q = 1-p
n = 500
p = .55
q = .45
x = .5 * 500 = 250

z = ( x - np) / sqrt(npq)
= (250 - (500*.55)) / sqrt(500*.55*.45)
= -0.277

Find the probability of z = -0.277 or less.

[anthonys]

thank you for your help