Question

The average number of people arriving at a fast-food drive-thru in any given 10-minute interval is 4 customers. Find the probability that at least two customers will arrive in the fast-food drive-thru in any given 15-minute interval. 

Answer 1

Let X = the number of people arriving at a fast-food drive-thru in any given 15-minute interval (The interval of interest is 15 minutes)

x = 0, 1, 2, 3, …

If, the average number of people arriving at a fast-food drive-thru in any given 10-minute interval is 4 customers and there are 2/3 15 minute intervals in 10 minutes, then we have;

$\frac{1}{2/3}$(4) = 6 customers in 15 minutes, on average. So, μ = 6 for this problem. Then, the random variable X follows a Poisson distribution;

X ~ P(6)

Where;

P(x) = $\frac{\mu^x e^{-\mu}}{x!}$

We need to find P(x ≥ 2) = 1- (P(x=0) + P(x=1))

Where P(X=0) = $\frac{6^0 e^{-6}}{0!}$ = 0.00248

and P(X=1) = $\frac{6^1 e^{-6}}{1!}$ = 0.01487

Therefore,

 P(x ≥ 2) = 1- (0.00248+0.01487)

= 0.9827