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Question Number 105670 by Ar Brandon last updated on 30/Jul/20

A machine manufactures washers, and 20% of the production is substandard. A random sample o f 10 washers is selected. Find the mean and standard deviation of the number of substandard washers in the sample.

$$\mathrm{A}\:\mathrm{machine}\:\mathrm{manufactures}\:\mathrm{washers},\:\mathrm{and}\:\mathrm{20\%}\:\mathrm{of}\:\mathrm{the}\:\mathrm{production}\:\mathrm{is} \\ $$$$\mathrm{substandard}.\:\mathrm{A}\:\mathrm{random}\:\mathrm{sample}\:\mathrm{o}\:\mathrm{f}\:\mathrm{10}\:\mathrm{washers}\:\mathrm{is}\:\mathrm{selected}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{mean}\:\mathrm{and}\:\mathrm{standard}\:\mathrm{deviation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{number} \\ $$$$\mathrm{of}\:\mathrm{substandard}\:\mathrm{washers}\:\mathrm{in}\:\mathrm{the}\:\mathrm{sample}. \\ $$

Answered by Ar Brandon last updated on 30/Jul/20

This is a binomial probability distribution. Therefore; mean, E(X)=np. Where n is the number of trials and p is the probability of X on a trial. In this case n=10, p=20% ⇒E(X)=10×0.2=2 For a binomial probability distribution, variance, var(X)=np(1−p)=2(1−0.2)=1.6 standard deviation, σ=(√(var(X)))=(√(1.6))=1.26 (to 3 S∙F)

$$\mathrm{This}\:\mathrm{is}\:\mathrm{a}\:\mathrm{binomial}\:\mathrm{probability}\:\mathrm{distribution}.\:\mathrm{Therefore}; \\ $$$$\mathrm{mean},\:\mathrm{E}\left(\mathrm{X}\right)=\mathrm{np}.\:\mathrm{Where}\:\mathrm{n}\:\mathrm{is}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{trials}\:\mathrm{and} \\ $$$$\mathrm{p}\:\mathrm{is}\:\mathrm{the}\:\mathrm{probability}\:\mathrm{of}\:\mathrm{X}\:\mathrm{on}\:\mathrm{a}\:\mathrm{trial}.\:\mathrm{In}\:\mathrm{this}\:\mathrm{case}\:\mathrm{n}=\mathrm{10},\:\mathrm{p}=\mathrm{20\%} \\ $$$$\Rightarrow\mathrm{E}\left(\mathrm{X}\right)=\mathrm{10}×\mathrm{0}.\mathrm{2}=\mathrm{2} \\ $$$$\mathrm{For}\:\mathrm{a}\:\mathrm{binomial}\:\mathrm{probability}\:\mathrm{distribution}, \\ $$$$\mathrm{variance},\:\mathrm{var}\left(\mathrm{X}\right)=\mathrm{np}\left(\mathrm{1}−\mathrm{p}\right)=\mathrm{2}\left(\mathrm{1}−\mathrm{0}.\mathrm{2}\right)=\mathrm{1}.\mathrm{6} \\ $$$$\mathrm{standard}\:\mathrm{deviation},\:\sigma=\sqrt{\mathrm{var}\left(\mathrm{X}\right)}=\sqrt{\mathrm{1}.\mathrm{6}}=\mathrm{1}.\mathrm{26}\:\:\left(\mathrm{to}\:\mathrm{3}\:\mathrm{S}\centerdot\mathrm{F}\right) \\ $$

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