Calculate the proportion of days on which (i) neither car is used (ii) some demand is refused.

Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 6M

Year: May 2014

Let X denote number of cars which are hired out per day.

For Poisson distribution mean = m = 1.5

$$P(X=x) = \frac{e^{-m}.m^x}{x!}=\mathrm{\ }\frac{e^{-1.5}.{1.5}^x}{x!}\ $$

1) P(neither car is used ) $$P(X=0) = \frac{e^{-1.5}.{1.5}^0}{0.2231}$$

2) P(Some demand is refused ) = P(Demand is more than 2 cars per days) $$P(x \gt 2)$$

$$= 1 - P(x\mathrm{\le}2)$$

$$= 1 - [P(x=0) + P(x=1) + P(x=2)]$$

$$= 1 - [\frac{e^{1.5}\ {1.5}^0}{0!}+\ \frac{e^{1.5}\ {1.5}^1}{1!}+\frac{e^{1.5}\ {1.5}^2}{2!}]$$

$$= 1 - e^{1.5}\ \left[\ 1+1.5+\frac{2.25}{2}\right] = 0.1912$$

$\\ \therefore$ Proportion of days on which neither car is used = 0.2231 = 22.31 %

And Proportion of days on which some demand is refused = 0.1912 = 19.12 %