Question

1. A recent study showed that 50 percent of households in Syracuse have at least 2 cars. Let X denote a binomially distributed random variable that is the number of households that have at least 2 cars. Out of 16 randomly chosen households, what is the probability that

   1. exactly nine have at least 2 cars;

   2. at most six have at least 2 cars;

   3. anywhere from 8 to 12 have at least 2 cars?

Answer 1

here p=0.5=probability of households in Syracuse have at least 2 cars

sample size=n=16

here we use binomial distribution and for Binomial distribution ,P(X=r)=ⁿC_rp^r(1-p)^n-r  

(a) exactly nine have at least 2 cars;

P(X=9)=¹⁶C₉(0.5)⁹(1-0.5)^16-9=11440*0.001953*0.007813=0.1746

(or using ms-excel=BINOMDIST(9,16,0.5,0)

(b)at most six have at least 2 cars;

P(X<=6)=0.2272 (using ms-excel =BINOMDIST(6,16,0.5,1))

(c)any where from 8 to 12 have at least 2 cars?

P(8<=X<=12)=P(X<=12)-P(X<=8)=0.9894-0.2272=0.7622

P(X<=12)=0.9894( using ms-excel=BINOMDIST(12,16,0.5,1))

P(X<=8)=0.2272 ( using ms-excel=BINOMDIST(8,16,0.5,1))