Answer to Question #87028 in Statistics and Probability for Swati malik

Question

The number of cars arriving at a railway station follows the poisson distribution. If the average number of cars arriving during a specified period of half an hour is 2. Find the probabilities that during a give half an hour.
i. No car will arrive
ii. At least two cars will arrive
iii. At the most 3 cars will arrive
iv. Between 1 and 3 cars will arrive

Answer 1

The probability distribution of the Poisson random variable X, representing the number of outcomes occurring

in a given time interval or specified region denoted by t, is

$$p(x; \lambda t)={e^{-\lambda t}(\lambda t)^x \over {x!}}$$

Use the Poisson distribution with 

$$\lambda t=2$$

i. No car will arrive 

$$P(X=0)={e^{-2}(2)^0 \over {0!}}=e^{-2}\approx0.135335$$

ii. At least two cars will arrive 

$$P(X\ge2)=1-(P(X=0)+P(X=1))=$$

$$=1-({e^{-2}(2)^0 \over {0!}}+{e^{-2}(2)^1 \over {1!}})=1-e^{-2}-2e^{-2}=1-3e^{-2}\approx0.593994$$

iii. At the most 3 cars will arrive 

$$P(X\le3)=P(X=0)+P(X=1)+P(X=2)+P(X=3)=$$

$$={e^{-2}(2)^0 \over {0!}}+{e^{-2}(2)^1 \over {1!}}+{e^{-2}(2)^2 \over {2!}}+{e^{-2}(2)^3 \over {3!}}=$$

$$={19 \over 3}e^{-2}\approx0.857123$$

iv. Between 1 and 3 cars will arrive

$$P(1\le X\le3)=P(X=1)+P(X=2)+P(X=3)=$$

$$={e^{-2}(2)^1 \over {1!}}+{e^{-2}(2)^2 \over {2!}}+{e^{-2}(2)^3 \over {3!}}={16 \over 3}e^{-2}\approx0.721788$$