Answer to Question #330048 in Statistics and Probability for Mae

Question

According to Chemical Engineering Progress (Nov. 1990), approximately 30% of all pipework failures in chemical plants are caused by operator error.

(a) What is the probability that out of the next 20 pipework failures at least 10 are due to operator error?

(b) What is the probability that no more than 4 out of 20 such failures are due to operator error?

(c) Suppose, for a particular plant, that, out of the random sample of 20 such failures, exactly 5 are operational errors.

Do you feel that the 30% figure stated above applies to this plant? Comment.

Answer 1

$p=0.3\\Normal\,\,approximation:\\a:\\P\left( \hat{p}\geqslant 0.5 \right) =P\left( \sqrt{n}\frac{\hat{p}-p}{\sqrt{p\left( 1-p \right)}}\geqslant \sqrt{n}\frac{0.5-p}{\sqrt{p\left( 1-p \right)}} \right) =\\=P\left( Z\geqslant \sqrt{20}\frac{0.5-0.3}{\sqrt{0.3\left( 1-0.3 \right)}} \right) =1-\varPhi \left( 1.9518 \right) =\\=1-0.9745=0.0255\\b:\\P\left( \hat{p}\leqslant \frac{4}{20} \right) =P\left( \sqrt{n}\frac{\hat{p}-p}{\sqrt{p\left( 1-p \right)}}\leqslant \sqrt{n}\frac{0.2-p}{\sqrt{p\left( 1-p \right)}} \right) =\\=P\left( Z\leqslant \sqrt{20}\frac{0.2-0.3}{\sqrt{0.3\left( 1-0.3 \right)}} \right) =\varPhi \left( -0.9759 \right) =0.1646\\c:\\H_0:p=0.3\\H_1:p\ne 0.3\\Z=\sqrt{n}\frac{\hat{p}-p}{\sqrt{p\left( 1-p \right)}}=\sqrt{20}\frac{\frac{5}{20}-0.3}{\sqrt{0.3\left( 1-0.3 \right)}}=-0.48795\\P\left( \left| Z \right|>0.48795 \right) =2\varPhi \left( -0.48795 \right) =2\cdot 0.313=0.626>0.05\\Since\,\,the\,\,P-value\,\,is\,\,large, the\,\,null\,\,hyoithesis\,\,is\,\,not\,\,rejected.\\The\,\,30\% figure\,\,applies.$