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Given that x is a random variable for which a Poisson probability distribution provides a good approximation, use statistical software to find the following:

a.P(x2) when =1

b.P(x2) when =2

c.P(x2) when =3

d. What happens to the probability of the event {x2} as it increases from 1 to 3? Is this intuitively reasonable?

Short Answer

Expert verified

a.The probabilityPx2is obtained from the R-software is 0.6766764.

b.The probabilityPx2is obtained from the R-software is 0.6766764.

c.The probability Px2 is obtained from the R-software is 0.4231901.

d.The average probability value decreases when the value increases are reasonable.

Step by step solution

01

Given Information

The x is a Poisson random variable.

02

Sate the statistical software used to compute the probability

The R-software is used to compute the Poisson probabilities.

The general form for R-command used for computing the Poisson probability is given as follows:

dpois(x,lambda,log=FALSE)ppois(x,lambda,lower.tail=TRUE)

03

State the Poisson probability distribution

The Poisson probability distribution is,

Px=xe-x!x=0,1,2,...

The parameter is the mean of the Poisson distribution.

04

(a)Compute the probability  P(x⩽2)  when λ=1

The value of the Poisson parameter,=1 .

ThePx2when =1 can be obtained by,

Px2=Px=0+Px=1+Px=2

The R-code used to compute thePx2is,

ppois2,lambda=1,lower.tail=TRUE

Therefore,

The probability Px2 is obtained from the R-software is 0.9196986.

05

(b) Compute the probability P(x⩽2) when λ=2

The value of the Poisson parameter,=2.

ThePx2when =2 can be obtained by,

Px2=Px=0+Px=1+Px=2

The R-code used to compute thePx2is,

ppois2,lambda=2,lower.tail=TRUE

Therefore,

The probability Px2 is obtained from the R-software is 0.6766764.

06

(c) Compute the probability P(x⩽2) when λ=3

The value of the Poisson parameter,=3.

ThePx2when =3 can be obtained by,

Px2=Px=0+Px=1+Px=2

The R-code used to compute thePx2is,

ppois2,lambda=3,lower.tail=TRUE

Therefore,

The probability Px2 is obtained from the R-software is 0.4231901.

07

(d) State the reason what happens to the probability of an event when the λvalues increase from 1 to 3

The probability of the event {x2} when =1 is 0.9196986.

The probability of the event {x2} when =2 is 0.6766764.

The probability of the event {x2} when =3 is 0.4231901.

Here, the event's probability decreases when the value of is increases. The reason is that there are a lot more events that could happen during the interval.

Therefore, The average probability value decreases when the value increases are reasonable.

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91.28 92.83 89.35 91.90 82.85 94.83 89.83 89.00 84.62

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90.96 92.85 89.39 89.82 89.91 92.16 88.67 89.35 86.51

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Mean

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Standard deviation(n-1)

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