document.write( "Question 1118652: A car firm has two cars, which it hires out day by day. The number of demand for a car on each day is distributed as poison variate with mean 1.5 .\r
\n" ); document.write( "\n" ); document.write( "1.calculate the proportion of days on which neither car is used
\n" ); document.write( "2. Calculate the proportion of days on which some demand will be refused \n" ); document.write( "

Algebra.Com's Answer #734141 by addingup(3677) $\"\"$ $\"About$

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Let X denote number of cars hired out per day
\n" ); document.write( "Poisson distribution mean = m = 1.5
\n" ); document.write( "P(X=x)=(((e^−m)(m^x))/(x!))= (((e^−1.5)(1.5^x))/(x!))
\n" ); document.write( "1) P(neither car is used):
\n" ); document.write( "P(X=0)=(e^−1.5)(1.5^0)/0.2231
\n" ); document.write( "2) P(Some demand is refused ) = P(Demand is more than 2 cars per days)
\n" ); document.write( "P(x>2)
\n" ); document.write( "=1−P(x≤2)
\n" ); document.write( "=1−[P(x=0)+P(x=1)+P(x=2)]
\n" ); document.write( "=1−[((e^1.5)(1.5^0)/0!)+ ((e^1.5)(1.5^1)/1!)+((e^1.5)(1.5^2)/2!)]
\n" ); document.write( "=1−e^1.5[1+1.5+(2.25/2)]=0.1912
\n" ); document.write( "Proportion of days on which neither car is used = 0.2231 = 22.31 %
\n" ); document.write( "Proportion of days on which some demand is refused = 0.1912 = 19.12 % \n" ); document.write( "

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