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Sagot :
Answer:
The answer is "[tex]\bold{r=9, \lambda=\frac{1}{2}\ and \ \ \text{exponential, 9 steps}}[/tex]"
Step-by-step explanation:
In point a:
We are aware of the random gamma variable X:
[tex]\to \mathbb{E} =\frac{r}{\lambda}\\\\\to Std(X) =\frac{\sqrt{r}}{\lambda}\\\\[/tex]
It is given:
[tex]\to \mathbb{E} = 18 \\\\\to std(X)=6[/tex]
[tex]\to 18=\frac{r}{\lambda}\\\\ \to 6=\frac{\sqrt{r}}{\lambda}\\\\[/tex]
Substituting the value:
[tex]\to \frac{\sqrt{r}}{6}=\frac{r}{18}\\\\\to r=9\\\\\to \lambda=\frac{1}{2}\\\\[/tex]
In point b:
When building the Erlang/Gammas distribution, these could reasonably be assumed to become an exponential distribution only with \lambda = 1/2 parameter with one step but to be r = 9 for one step.
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