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sandra2223

New member
Joined
Oct 14, 2009
Messages
8
A manufacturer claims that their manufacturing process makes widgets with at most 5 percent defective rate. assume that the defective rate is 5%. Let x denote the number of defective widgets in a box of 50 widgets.

What is the probabilty of finding more than 2 defective widgets in the box?

what is the variance of X?
 

soroban

Elite Member
Joined
Jan 28, 2005
Messages
5,588
Hello, Sandra!

The first part is a binomial probability.


A manufacturer claims that their manufacturing process makes widgets with at most 5% defective rate.
Let x denote the number of defective widgets in a box of 50 widgets.

What is the probabilty of finding more than 2 defective widgets in the box?

The opposite of "more than two" is "two or less".

. . \(\displaystyle \begin{array}{ccccccc}P(\text{0 def}) &=& (0.95)^{50} &=& 0.076944975 \\ P(\text{1 def}) &=& {50\choose1}(0.05)(0.95)^{49} &=& 0.202486777 \\ \\[-3mm] P(\text{2 def}) &=& {50\choose2}(0.05)^2(0.95)^{48} &=& 0.261101370 \\ \hline & &\text{Total:} && 0.540533122\end{array}\)


\(\displaystyle \text{Hence: }\:p(\text{2 or less def}) \:\approx\:0.54 \:=\:54\%\)


\(\displaystyle \text{Therefore: }\:p(\text{more than 2 def}) \;=\;100\% - 54\% \;=\;46\%\)

 

sandra2223

New member
Joined
Oct 14, 2009
Messages
8
Thank you so much. I really appreciate it. I still need help with finding the variance of X, i cant figure it out.
 
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