Assume that X is a random variable having a Poisson probability distribution with a mean of 1.5. Use statistical software to find the following probabilities:

1. $\mathbf{P}\mathbf{(}\mathbf{X}\mathbf{\le }\mathbf{3}\mathbf{)}$
   
2. role="math" localid="1664181123938" $\mathbf{P}\mathbf{(}\mathbf{X}\mathbf{\ge }\mathbf{3}\mathbf{)}$
   
3. $\mathbf{P}\mathbf{(}\mathbf{X}\mathbf{=}\mathbf{3}\mathbf{)}$
   
4. $\mathbf{P}\mathbf{(}\mathbf{X}\mathbf{=}\mathbf{0}\mathbf{)}$
   
5. $\mathbf{P}\mathbf{(}\mathbf{X}\mathbf{>}\mathbf{0}\mathbf{)}$
   
6. $\mathbf{P}\mathbf{(}\mathbf{X}\mathbf{>}\mathbf{6}\mathbf{)}$
   

The x is a Poisson random variable.

The mean of the Poisson Probability distribution is 1.5.

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:

$\mathrm{dpois}\left(\mathrm{x},\mathrm{lambda},\log =\mathrm{FALSE}\right)$

$\mathrm{ppois}\left(\mathrm{x},\mathrm{lambda},\mathrm{lower}.\mathrm{tail}=\mathrm{TRUE}\right)$

The probability $\mathrm{P}\left(\mathrm{X}\le 3\right)$can be obtained

localid="1664184047402" $\mathrm{P}\left(\mathrm{X}\le 3\right)=\mathrm{P}\left(\mathrm{X}=0\right)+\mathrm{P}\left(\mathrm{X}=1\right)+\mathrm{P}\left(\mathrm{x}=2\right)+\mathrm{P}\left(\mathrm{x}=3\right)$

The R-code used to compute the localid="1664184032966" $\mathrm{P}\left(\mathrm{x}\le 3\right)$is,

localid="1664184019959" $\mathrm{ppois}\left(3,\mathrm{lambda}1.5,\mathrm{lower}.\mathrm{tail}=\mathrm{TRUE}\right)$

Therefore,

The probability$\mathrm{P}\left(\mathrm{X}\le 3\right)$is obtained from the R-software is 0.9343575

The probability$\mathrm{P}(\mathrm{X}\ge 3)$can be obtained

$$\begin{gathered} \mathrm{P}\left(\mathrm{X}\ge 3\right)=1-\mathrm{P}\left(\mathrm{X}<3\right) \\ =1-\mathrm{P}\left(\mathrm{X}\le 2\right) \end{gathered}$$

The R-code used to compute the$\mathrm{P}\left(\mathrm{X}\ge 3\right)$is

$1-\mathrm{ppois}\left(2,\mathrm{lambda}1.5,\mathrm{lower}.\mathrm{tail}=\mathrm{TRUE}\right)$

Therefore,

The probability $\mathrm{P}(\mathrm{X}\ge 3)$is obtained from the R-software is 0.1911532.

The R-code used to compute the P(X=3) is,

$\mathrm{dpois}\left(\mathrm{x}=3,\mathrm{lambda}=1.5,\log =\mathrm{FALSE}\right)$

Therefore,

The probability P(X=3) is obtained from the R-software is 0.1255107.

The R-code used to compute the P(X=0) is,

$\mathrm{dpois}\left(\mathrm{x}=0,\mathrm{lambda}=1.5,\log =\mathrm{FALSE}\right)$

Therefore,

The probability P(X=0) is obtained from the R-software is 0.2231302.

The probability$\mathrm{P}(\mathrm{X}>0)$can be obtained,

$\mathrm{P}\left(\mathrm{X}>0\right)=1-\mathrm{P}\left(\mathrm{X}\le 0\right)$

The R-code used to compute thelocalid="1664184073912" $\mathrm{P}\left(\mathrm{X}>0\right)$is,

$1-\mathrm{ppois}\left(0,\mathrm{lambda}1.5,\mathrm{lower}.\mathrm{tail}=\mathrm{TRUE}\right)$

Therefore,

The probability localid="1664184067768" $\mathrm{P}\left(\mathrm{X}>0\right)$is obtained from the R-software is 0.7768698.

The probability$\mathrm{P}(\mathrm{X}>6)$is obtained,

$\mathrm{P}(\mathrm{X}>6)=1-\mathrm{P}\left(\mathrm{X}\le 6\right)$

The R-code used to compute the$\mathrm{P}(\mathrm{X}>6)$ is,

$1-\mathrm{ppois}\left(6,\mathrm{lambda}1.5,\mathrm{lower}.\mathrm{tail}=\mathrm{TRUE}\right)$

Therefore, the probability $\mathrm{P}(\mathrm{X}>6)$isobtained from the R-software is 0.0009259919.