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Statistics 9 Online
OpenStudy (anonymous):

A population of 100 contains 12 nonconforming items. a) What is the probability of selecting a random sample of 25 items containing 3 to 5 nonconforming items? Sampling is without replacement. b) Calculate the parameters mean and variance of the distribution of nonconforming items in samples of 25. c) Explain on the meaning of the quantities mean and variance in the context of this problem.

ganeshie8 (ganeshie8):

*

OpenStudy (kropot72):

a) You need to find the probabilities of 3, 4 and 5 nonconforming items and add the three resulting values of probability: \[P(3\ nonconforming)=\frac{\left(\begin{matrix}12 \\ 3\end{matrix}\right)\left(\begin{matrix}88 \\ 22\end{matrix}\right)}{\left(\begin{matrix}100 \\ 25\end{matrix}\right)}\] \[P(4\ nonconforming)=\frac{\left(\begin{matrix}12 \\ 4\end{matrix}\right)\left(\begin{matrix}88 \\ 21\end{matrix}\right)}{\left(\begin{matrix}100 \\ 25\end{matrix}\right)}\] \[P(5\ nonconforming)=\frac{\left(\begin{matrix}12 \\ 5\end{matrix}\right)\left(\begin{matrix}88 \\ 20\end{matrix}\right)}{\left(\begin{matrix}100 \\ 25\end{matrix}\right)}\] The required probability is found as follows: \[P(3\ -\ 5\ nonconforming)=P(3\ nonconforming)+P(4\ nonconforming)+P(5\ nonconforming)\]

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