Question

Q1 ).The number of white corpuscles on a slide has a Poisson distribution with mean 4.5.

Answer 1

The most likely number of white corpuscles om a slide is 4 ,the probability of obtaining this number 4 is 0.190 , probability of obtaining at least two white corpuscles in total on the two slides is 0.999.

Attending to the first question

What is Poison Distribution ?

Poisson Distribution is used for an independent event to find its  probability in a fixed interval of time .

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

a) We have to find the most likely number of white corpuscles on  a slide , which means expected value of Poise(4.5)

The expected value of Poisson distribution with mean λ is equal to λ.

Therefore the expected value is 4.5.

 the number of corpuscles cannot be non-integer,

To find the most likely number in one experiment,

determine it between 4 and 5.

In the common case P(X=k) when X~Poise(λ) is equal to $\frac {\lambda^{k}} {k!}}e^{-\lambda}$

,$\rm e^{-\lambda}$  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 likely number of white corpuscles om a slide is 4.

b)  the probability of obtaining this number 4 is

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.

$Y=X{\scriptscriptstyle 1}+X{\scriptscriptstyle 2}Y=X1+X2$ ,

where $X{\scriptscriptstyle 1}, X{\scriptscriptstyle 2}$  - number of corpuscles on the first and second slide respectively.

Then Y = Poise(4.5 + 4.5) = Poise(9)

the probability, correct to three decimals places of obtaining at least two white corpuscles in total on the two slides

The goal is to find P(Y ≥ 2)

Therefore the probability less than 2 will be subtracted from the total probability i.e. 1

P(Y ≥ 2 ) = 1 - P (Y=0) - P(Y=1)

= $1-{\frac {9^0} {0!}}e^{-9}-{\frac {9^1} {1!}}e^{-9}=1-10e^{-9}=0.999$

To know more about Poisson Distribution

brainly.com/question/17280826

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