Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Join our platform to get reliable answers to your questions from a knowledgeable community of experts. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
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.
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
