# Solved – Calculate Expected Values in Poisson Distribution What I have done so far:

I have found that $$Pr(X=3)= 0.0934$$ because the sum of probabilities equals $$1.$$
The mean I found in the previous question which was $$1.34615$$.
When I plug in mean $$= 1.34615$$ and $$x=0$$, get that $$P(X=0) = 0.26024.$$

What is it that I'm doing wrong?

Thank you for any help.

Contents

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*}