Poisson and Binomial Distribution

Question

I need help with a simple question.

Suppose that a hairdresser with a single store has 5 customers arrive on average every hour while they are open. Assume that customers’ arrivals don’t depend on time of day or on whether other customers have arrived. We wish to model the arrival of customers in the morning (9am to 12 midday). Suppose we define the random variable X as the number of customers arriving in the morning. What would be a suitable distribution for X, along with its mean and variance?

a) Poisson, lambda = 15; mean = 15; variance = 15;
b) Poisson, lambda = 15; mean = 15; variance = 225
c) Binomial, n = 15, p = 0.5; mean = 15; variance = 15  
d) Binomial, n = 15, p = 0.5; mean = 15; variance = 225

I know that poisson is used with discreet events. And binomial with continuous events. In this case I would choose a because the mean is equal to the variance. Did I choose correctly? or is the case in this question is a special case?

Answer 1

Since you are given an average rate you should use the Poisson distribution. You are correct that the answer is a since the mean is equal to the variance for the Poisson distribution.

Note that the binomial distribution is also discrete, not continuous.

Answer 2

Let $\{N(t):t\geqslant0\}$ be a Poisson process with rate $\lambda=5$. Then $Y=N(3)$ which has Poisson distribution with mean $3\cdot5=15$. Since the variance of a Poisson distribution is equal to its mean, the correct choice is a).