Question

At a cafeteria the customers arrive at an average of 0.3 per minute. The probability
that a)exactly 2 customers arrive in a 10 minute span b)2 or more customers arrive in a 10
minute span c) one customer arrives in a 5 minute span and one customer arrives in the next
minute span is

Answer 1

Answer on Question #44828 – Math - Statistics and Probability

At a cafeteria the customers arrive at an average of $\lambda = 0.3$ per minute. The probability that a) exactly 2 customers arrive in a 10 minute span b) 2 or more customers arrive in a 10 minute span c) one customer arrives in a 5 minute span and one customer arrives in the next minute span is

Solution

We use a Poisson process with a parameter $\lambda t$, $\lambda t = 0.3 \times 10 = 3$ in items a) and b).

a)

$$P(2 \text{ customers in 10 minute span}) = \frac{e^{-0.3 \cdot 10} (0.3 \cdot 10)^2}{2!} = 0.224.$$

b)

$$\begin{array}{l} P(2 \text{ or more in 10 minute span}) = 1 - P(0) - P(1) \\ = 1 - \frac{e^{-0.3 \cdot 10} (0.3 \cdot 10)^0}{0!} - \frac{e^{-0.3 \cdot 10} (0.3 \cdot 10)^1}{1!} = 0.8. \end{array}$$

c)

$$\begin{array}{l} P(\text{one customer in 5 minute and one customer in the next minute}) \\ = \frac{e^{-0.3 \cdot 5} (0.3 \cdot 5)^1}{1!} \cdot \frac{e^{-0.3 \cdot 1} (0.3 \cdot 1)^1}{1!} = 0.074. \end{array}$$

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