a. Theater No shows - A theater owner has found that5% of patrons do not show up

Question

a. Theater No shows - A theater owner has found that5% of patrons do not show up for the performance that theypurchased tickets for. If the theater has 100 seats, find theprobability that six or more patrons will not show up for the soldout performance.

b. Personal Computers - According to recent surveys,53% of households have personal computers. If a randomsample of 175 households is selected, what is the probability thatmore than 75 but fewer than 110 have a personal computer ?

Explanation / Answer

(a) Given X~Binomial(n=100, p=0.05) By using Normal approximation, X~Normal(=n*p=100*0.05=5,=n*p*(1-p)=sqrt(100*0.05*0.95)=2.18) So the probability is P(X>=6)=P(X>=5.5) =P((X-)/ >= (5.5-5)/2.18) =P(Z>=0.23) = 0.409 (check normal table) -------------------------------------------------------------------------------------------------------------- (b) Given X~Binomial(n=175, p=0.53) By using Normal approximation, X~Normal(=n*p=175*0.53=92.75,=n*p*(1-p)=sqrt(175*0.53*(1-0.53))=6.6) So the probability is P(75

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