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Customers arrive at a service facility according to a Poisson process of rate λ customers/hour. Let X(t) be the number of custom

mash [69]

Answer:

Step-by-step explanation:

Given that:

X(t) = be the number of customers that have arrived up to time t.

$W_1,W_2$ ... = the successive arrival times of the customers.

(a)

Then; we can Determine the conditional mean E[W1|X(t)=2] as follows;

$E(W_!|X(t)=2) = \int\limits^t_0 {X} ( \dfrac{d}{dx}P(X(s) \geq 1 |X(t) =2))$

$= 1- P (X(s) \leq 0|X(t) = 2) \\ \\ = 1 - \dfrac{P(X(s) \leq 0 , X(t) =2) }{P(X(t) =2)}$

$= 1 - \dfrac{P(X(s) \leq 0 , 1 \leq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}$

$= 1 - \dfrac{P(X(s) \leq 0 ,P((3 \eq X(t)) - X(s) \leq 5 ) }{P(X(t) = 2)}$

Now $P(X(s) \leq 0) = P(X(s) = 0)$

(b) We can Determine the conditional mean E[W3|X(t)=5] as follows;

$E(W_1|X(t) =2 ) = \int\limits^t_0 X (\dfrac{d}{dx}P(X(s) \geq 3 |X(t) =5 )) \\ \\ = 1- P (X(s) \leq 2 | X (t) = 5 ) \\ \\ = 1 - \dfrac{P (X(s) \leq 2, X(t) = 5 }{P(X(t) = 5)} \\ \\ = 1 - \dfrac{P (X(s) \LEQ 2, 3 (t) - X(s) \leq 5 )}{P(X(t) = 2)}$

Now; $P (X(s) \leq 2 ) = P(X(s) = 0 ) + P(X(s) = 1) + P(X(s) = 2)$

(c) Determine the conditional probability density function for W2, given that X(t)=5.

So ; the conditional probability density function of $W_2$ given that X(t)=5 is:

$f_{W_2|X(t)=5}}= (W_2|X(t) = 5) \\ \\ =\dfrac{d}{ds}P(W_2 \leq s | X(t) =5 ) \\ \\ = \dfrac{d}{ds}P(X(s) \geq 2 | X(t) = 5)$

7 0

3 years ago

Convert 4/3 to an mixed number

wariber [46]

1 1/3. One whole number is 3/3 so 3/3=1 then u put 1/3

4 0

3 years ago

A number cube has sides labeled 1 to 6. Gia rolls the number cube. What is the probability that she rolls a 4?

Jlenok [28]

It would b C because there are 6 sides and out of the 6, you want one number, which is 4, so that’s why it would be C

3 0

3 years ago

Read 2 more answers

4. solve e) 4^x-10.2^x+16=0 I think the dot means multiply?

suter [353]

$4^x-10\cdot2^x+16=0\Rightarrow(2^2)^x-10\cdot2^x+16=0\Rightarrow(2^x)^2-10\cdot2^x+16=0$

$subtitute\ 2^x=t > 0\\\\t^2-10t+16=0\\t^2-2t-8t+16=0\\t(t-2)-8(t-2)=0\\(t-2)(t-8)=0\iff t-2=0\ or\ t-8=0\\\\t=2\ or\ t=8\\\\t=2^1\ or\ t=2^3\\\\therefore\ 2^x=2^1\ or\ 2^x=2^3\\\\Answer:\boxed{x=1\ or\ x=3}$

6 0

4 years ago

Read 2 more answers

Sales of 10 000 increase by 65 each year

lorasvet [3.4K]

Answer:

Whats the question

Step-by-step explanation:

also if you are looking for the unit rate its 153.846...

or if you are looking for how many in 2 years then 650,000

Hope this helps

Have a nice day :)