Formulaic Maths Problems
- get another problem:
level 1
,
level 2
,
level 3
-
link to this problem
Loading, please wait...
Find $P(1 \le X \le 4)$, where $X$ is a Poisson variable with parameter $\lambda = 3.9$.
Hint: For a Poisson distribution, $P(X = r) = \mathrm{e}^{-\lambda} \dfrac{\lambda^r}{r!}$.
$P(X = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c0}}\hspace{35px}}~}) = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c2}}\hspace{100px}}~}}{\class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c3}}\hspace{100px}}~}} = \class{inputBox step1}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar0c4R}}\hspace{100px}}~}$
$P(X = [1]) = [0.02024191] \times \dfrac{[3.9]}{[1]} = [0.07894345]$
$P(X = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c0}}\hspace{35px}}~}) = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c2}}\hspace{100px}}~}}{\class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c3}}\hspace{100px}}~}} = \class{inputBox step2}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar1c4R}}\hspace{100px}}~}$
$P(X = [2]) = [0.02024191] \times \dfrac{[15.21]}{[2]} = [0.15393974]$
$P(X = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c0}}\hspace{35px}}~}) = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c2}}\hspace{100px}}~}}{\class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c3}}\hspace{100px}}~}} = \class{inputBox step3}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar2c4R}}\hspace{100px}}~}$
$P(X = [3]) = [0.02024191] \times \dfrac{[59.319]}{[6]} = [0.20012166]$
$P(X = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar3c0}}\hspace{35px}}~}) = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar3c1}}\hspace{100px}}~} \times \dfrac{\class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar3c2}}\hspace{100px}}~}}{\class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar3c3}}\hspace{100px}}~}} = \class{inputBox step4}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{Ar3c4R}}\hspace{100px}}~}$
$P(X = [4]) = [0.02024191] \times \dfrac{[231.3441]}{[24]} = [0.19511862]$
Therefore,
Hint: Add together the individual probabilities.
$P(1 \le X \le 4) = \class{inputBox step5}{~\bbox[border:2px solid blue]{\strut\rlap{\class{inputReplace}{BR}}\hspace{100px}}~}$
$P(1 \le X \le 4) = [0.62812346]$