Question

q1 ).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?

q2). At a Supermarket 45% of customers pay by credit card. Find the probability that
in a randomly selected sample of fifteen customers.
a. Exactly four pay by credit card
b. More than ten pay by cash
(give answers to 3 decimal places

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

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