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?????????
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.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?????????
I would like to point out though that this question is much easier done exactly:. 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.
alsoshould readp(x<30)=p(x<29)=p(x<29.5), but I suspect this was just an issue with typing the symbols
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