needhelp83
- 193
- 0
On a production line, only 45% of items produced meet quality standards. A random sample of 500 items will be taken. Using the normal approximation to the binomial distribution, approximate the probability that less than half of the sampled items meet quality standards.
500*.5 = 250
P(Y<250) = P(X<249.5) where Bin (500, .45) and N(225, 123.75)
E(Y)=225 and Var(Y)=500*(.45)*(.55)=123.75
[tex]P( \frac{\overline{X}-225}{\sqrt{123.75}}<\frac{249.5-225}{\sqrt{123.75}})=P(Z<2.20)=\Phi(2.20)=0.9861[/tex]
Wanted to make sure this is done correctly. Thanks
500*.5 = 250
P(Y<250) = P(X<249.5) where Bin (500, .45) and N(225, 123.75)
E(Y)=225 and Var(Y)=500*(.45)*(.55)=123.75
[tex]P( \frac{\overline{X}-225}{\sqrt{123.75}}<\frac{249.5-225}{\sqrt{123.75}})=P(Z<2.20)=\Phi(2.20)=0.9861[/tex]
Wanted to make sure this is done correctly. Thanks