Answer:
The average number of customers in the system is 3.2
Step-by-step explanation:
The average number of customes in the system is given by:

In which
is the number of arirvals per time period
is the average number of people being served per period.
The number of arrivals is modeled by the Poisson distribution, while the service time is modeled by the exponential distribution.
Customers arrive at the stand at the rate of 28 per hour
This means that 
Service times are exponentially distributed with a service rate of 35 customers per hour.
This means that
. So
The average number of customers in the system (i.e., waiting and being served) is


The average number of customers in the system is 3.2
Answer:
Step-by-step explanation:
1. There will be 9 roses in total.
Probability Orange: 2/9
Probability Yellow: 3/9 = ⅓
Probability Pink: 4/9
2. 4/7 as there are 7 bowls in total and 4 are chocolate.
3. 4/10 = ⅖ (simplified) There are 10 fruits in total and 4 are apples.
4. 5/15 = ⅓ (simplified) There are 15 cars in total and 5 or them are cooper minis.
5. 2/6 = ⅓ (simplified) There are 6 textbooks and 2 of them are math.
6. 13/20 There are 20 drinks and 13 of them are lemonade.
7. There are 10 toys in total.
Probability toy cars: 3/10
Probability dolls: 2/10 = ⅕ (simplified)
Probability balls: 5/10 = ½ (simplified)
8. In total the number of red roses and lilies are 7. There are 10 flowers in total so the number of jasmines are 3. Therefore the probability of getting a jasmine is 3/10
9. 20/50 = ⅖ (simplified) There are 50 papers in total and 20 of them are yellow.
10. 8/18 = 4/9 (simplified) There are 18 buses and 8 of them are air conditioned.
Hello again lol. I would think it would be two rooms don’t count me on this one but I think it is two
I'm going to assume the joint density function is

a. In order for
to be a proper probability density function, the integral over its support must be 1.

b. You get the marginal density
by integrating the joint density over all possible values of
:

c. We have

d. We have

and by definition of conditional probability,


e. We can find the expectation of
using the marginal distribution found earlier.
![E[X]=\displaystyle\int_0^1xf_X(x)\,\mathrm dx=\frac67\int_0^1(2x^2+x)\,\mathrm dx=\boxed{\frac57}](https://tex.z-dn.net/?f=E%5BX%5D%3D%5Cdisplaystyle%5Cint_0%5E1xf_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac67%5Cint_0%5E1%282x%5E2%2Bx%29%5C%2C%5Cmathrm%20dx%3D%5Cboxed%7B%5Cfrac57%7D)
f. This part is cut off, but if you're supposed to find the expectation of
, there are several ways to do so.
- Compute the marginal density of
, then directly compute the expected value.

![\implies E[Y]=\displaystyle\int_0^2yf_Y(y)\,\mathrm dy=\frac87](https://tex.z-dn.net/?f=%5Cimplies%20E%5BY%5D%3D%5Cdisplaystyle%5Cint_0%5E2yf_Y%28y%29%5C%2C%5Cmathrm%20dy%3D%5Cfrac87)
- Compute the conditional density of
given
, then use the law of total expectation.

The law of total expectation says
![E[Y]=E[E[Y\mid X]]](https://tex.z-dn.net/?f=E%5BY%5D%3DE%5BE%5BY%5Cmid%20X%5D%5D)
We have
![E[Y\mid X=x]=\displaystyle\int_0^2yf_{Y\mid X}(y\mid x)\,\mathrm dy=\frac{6x+4}{6x+3}=1+\frac1{6x+3}](https://tex.z-dn.net/?f=E%5BY%5Cmid%20X%3Dx%5D%3D%5Cdisplaystyle%5Cint_0%5E2yf_%7BY%5Cmid%20X%7D%28y%5Cmid%20x%29%5C%2C%5Cmathrm%20dy%3D%5Cfrac%7B6x%2B4%7D%7B6x%2B3%7D%3D1%2B%5Cfrac1%7B6x%2B3%7D)
![\implies E[Y\mid X]=1+\dfrac1{6X+3}](https://tex.z-dn.net/?f=%5Cimplies%20E%5BY%5Cmid%20X%5D%3D1%2B%5Cdfrac1%7B6X%2B3%7D)
This random variable is undefined only when
which is outside the support of
, so we have
![E[Y]=E\left[1+\dfrac1{6X+3}\right]=\displaystyle\int_0^1\left(1+\frac1{6x+3}\right)f_X(x)\,\mathrm dx=\frac87](https://tex.z-dn.net/?f=E%5BY%5D%3DE%5Cleft%5B1%2B%5Cdfrac1%7B6X%2B3%7D%5Cright%5D%3D%5Cdisplaystyle%5Cint_0%5E1%5Cleft%281%2B%5Cfrac1%7B6x%2B3%7D%5Cright%29f_X%28x%29%5C%2C%5Cmathrm%20dx%3D%5Cfrac87)