Step-by-step explanation:
since Poisson distribution parameter is not given so we have to estimate it from the sample data. The average number of of arrivals per minute at an ATM is
[tex]\hat{\lambda}=\bar{x}=\frac{\sum x}{n}=\frac{30}{30}=1[/tex]
So probabaility for [tex]\mathrm{X}=1[/tex] is
[tex]P(X=1)=\frac{e^{-\lambda} \lambda^{x}}{x !}=\frac{e^{-1} \cdot 1^{1}}{1 !}=0.3679[/tex]
So expected frequency for [tex]X=1[/tex] is [tex]0.3679^{*} 30=11.037[/tex] (or [tex]\left.11.04\right)[/tex] .
