# ahh help please (normal approx. to binomial distribution)

• Nov 25th 2008, 06:01 PM
perron
ahh help please (normal approx. to binomial distribution)
The human resources manager at a company knows that 34% of the workforce belong to a union. If she randomly surveys 50 employees, what is the probability that exactly 30 of them do not belong to a union?

p=0.34 q=0.66 n=50 np=17

p(x<30)=p(x<29)=p(x<29.5)

6=((square root)(npq))
=3.33

z1=x1 - x /6
=29.5-17/3.33
=3.73?????????
• Nov 25th 2008, 07:48 PM
perron
do I need to subtract 3.73 by one?
• Nov 26th 2008, 02:09 AM
Quote:

p=0.34 q=0.66 n=50 np=17

p(x<30)=p(x<29)=p(x<29.5)

6=((square root)(npq))
=3.33

z1=x1 - x /6
=29.5-17/3.33
=3.73?????????
What you have established with this working is that P(X<30) = P(Z<3.73) where Z has the standard normal distribution. When you use this approximation, it is assumed that you have a table to look up P(Z<3.73) in, so feel free to get a probability from your calculator, then find P(X>30) in the same way.

I would like to point out though that this question is much easier done exactly: $P(X=x) = {n \choose x} p^xq^{n-x}$. I never liked the normal approximation, it never seemed logical to me that I should approximate something calculable by hand with something that isn't.

also
Quote:

p(x<30)=p(x<29)=p(x<29.5)
should read $P(X<30 = P(X\leq29) = P(X<29.5)$, but I suspect this was just an issue with typing the symbols