Solved – Calculate Expected Values in Poisson Distribution

You know that $P(X=x)=frac{lambda^xe^{-lambda}}{x!}$, so $P(X=0)=e^{-lambda}=0.2865$ implies $lambda=-log(0.2865)=1.25$. Indeed: begin{align*}P(X=0)&=e^{-lambda}=e^{-1.25}=0.2865\P(X=1)&=lambda e^{-lambda}=1.25e^{-1.25}=0.3581\P(X=2)&=frac{lambda^2 e^{-lambda}}{2!}=frac{1.25^2e^{-1.25}}{2}=0.2238 end{align*} and $P(X=3)=frac{lambda^3 e^{-lambda}}{3!}=frac{1.25^3 e^{-1.25}}{6}=0.0933$.

$lambda$ is the average number of events in an interval, i.e.: $$frac{0times 77+1times 90+2times 55+3times 30+4times y+5times 3}{77+90+55+30+y+5}=1.25$$ therefore $y=5$ and $lambda=frac{325}{260}$.

As to the expected values, you have 325 events in 260 days, and $lambda=1.25$ is the expected value (the mean) of this Poisson distribution (the average number of events in one day), but the expected values (plural) are the number of days with $X=0,1,2,3,dots$ events out of 260: begin{align*} X=0: &quad P(X=0)times 260=0.2865times 260=74.5\X=2: &quad P(X=2)times 260=0.2238times 260=58.2\X=4: &quad P(X=4)times 260=0.0291times 260=7.6\X=5: &quad P(X=5)times 260=0.0073times 260=1.9end{align*} Furthermore, $P(Xge 6)=1-P(X=0)-cdots-P(X=5)=0.0018$ and $0.0018times 260=0.5$.

Finally, begin{align*}E_1&=P(X=1)times 260=0.3581times 260=93.1\E_3&=P(X=3)times 260=0.0933times 260=24.3end{align*}