Answer to Question #259171 in Statistics and Probability for idkman

Question

1.     The number of white corpuscles on a slide has a Poisson distribution with mean 4.5.

a.     Find the most likely number of white corpuscles om a slide.

b.     Calculate correct to three decima places the probability of obtaining this number

c.     If such two slides are prepared, what is the probability, correct to three decimals places of obtaining at least two white corpuscles in total on the two slides?

Answer 1

Let X be the number of white corpuscles on a slide, then X~Pois(4.5)

a) We have to find the expected value of Pois(4.5)

The expected value of Poisson distrubution with mean $\lambda$ is equal to $\lambda$. So, in the given case, expected value is 4.5. But the number of corpuscles cannot be non-integer, so, if we want to find the most likely number in one experiment, we should determine it between 4 and 5.

In the common case P(X=k) when X~Pois($\lambda$) is equal to ${\frac {\lambda^{k}} {k!}}e^{-\lambda}$ . $e^{-\lambda}$ is a constant in terms of k, so the point is to find out what is greater: ${\frac {4.5^4} {4!}}$ or ${\frac {4.5^5} {5!}}$ . ${\frac {4.5^5} {5!}}={\frac {4.5^4} {4!}} * {\frac {4.5} 5}$ .${\frac {4.5} 5}<1$ , So, ${\frac {4.5^4} {4!}}$ is greater

So, the most expected value in any given experiment is 4.

b) $P(X = 4) ={\frac {4.5^4} {4!}}e^{-4.5}=0.190$

c) Let Y be the total number of corpuscles on two screens. Then $Y=X{\scriptscriptstyle 1}+X{\scriptscriptstyle 2}$ , where $X{\scriptscriptstyle 1}, X{\scriptscriptstyle 2}$ - number os corpuscles on the first and secon slide respectively. Then Y = Pois(4.5 + 4.5) = Pois(9)

The goal is to find P(Y ≥ 2)

$P(Y ≥ 2)=1-P(Y=0)-P(Y=1)=1-{\frac {9^0} {0!}}e^{-9}-{\frac {9^1} {1!}}e^{-9}=1-10e^{-9}=0.999$