Fast food customers

Let $X$ be the random variable that represents the number of customers approaching the register per minute. The average of $X$ is $\lambda=2.8.$ The problem is asking for $P(X=4)$ the probability that 4 customers approach the register in the next minute. Using the formula for the Poisson distribution,

$\begin{aligned} P(X=k) &= \dfrac{\lambda^k e^{-\lambda}}{k!} \\ \\ P(X=4) &= \frac{2.8^4 e^{-2.8}}{4!} \\ \\ &\approx 0.156 \end{aligned}$

The probability that 4 customers approach the register in the next minute is approximately $\boxed{0.156}.$

The values for the variables are average or Lamba is 2.8 K = 4 so we can calculate the Poission distribution for X = k to get the answer as 0.156