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Find $P(3 < X < 6)$, where $X$ is a Poisson variable with parameter $\lambda = 3.5$.
Hint: For a Poisson distribution, $P(X = r) = \mathrm{e}^{-\lambda} \dfrac{\lambda^r}{r!}$.
$P(X = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c0}}\hspace{35px}}~}) = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c2}}\hspace{100px}}~}}{\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c3}}\hspace{100px}}~}} = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c4R}}\hspace{100px}}~}$ $P(X = [4]) = [0.03019738] \times \dfrac{[150.0625]}{[24]} = [0.18881229]$
$P(X = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c0}}\hspace{35px}}~}) = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c2}}\hspace{100px}}~}}{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c3}}\hspace{100px}}~}} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c4R}}\hspace{100px}}~}$ $P(X = [5]) = [0.03019738] \times \dfrac{[525.21875]}{[120]} = [0.1321686]$
Therefore, Hint: Add together the individual probabilities.
$P(3 < X < 6) = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BR}}\hspace{100px}}~}$ $P(3 < X < 6) = [0.32098089]$