A convenience store has two separate locations where customers can be checked out as they leave. These locations each have two cash registers and two employees who check out customers. Let X be the number of each registers being used at a particular time for location 1 and Y the number being used at the- same time for location 2. The joint probability function is given by Give the marginal density of both X and Y as well as the probability distribution of X given Y = 2. Give E(X) and Var(X). Give E(X|Y = 2) and Var(X|Y = 2).

The joint probability function is missing and I've attached it.

A) - For g(x);

g(0)=0.2 ; g(1)=0.32 ; g(2) =0.48

- For h(y);

h(0)=0.26 ; h(1)=0.35 ; h(2) =0.39

-The probability distribution of X given Y = 2 is;

f(0|2) = 4/39 ; f(1|2) = 5/39 ; f(2|2) = 30/39

B) Var(X) = 0.602

C) E(X|Y=2) = 5/3

Var(X|Y= 2) = 150/351

Step-by-step explanation:

A) Let f(x, y) be the joint probability distribution of the function with the table attached and let g(x) be the marginal density function of the random variable X. Thus, we compute the following;

g(x) = Σ_y f(x, y) = Σ(2)_(y=0) f(x, y)

Now, solving from the table;

g(0) = f(0,0) + f(0,1) + f(0,2) = 0.12 + 0.04 + 0.04 = 0.20

g(1) = f(1,0) + f(1,1) + f(1,2) = 0.08 + 0.19 + 0.05 = 0.32

g(2) = f(2,0) + f(2,1) + f(2,2) = 0.06 + 0.12 + 0.30 = 0.48

In a similar manner, let h(y) be the marginal density function of the random variable Y. Thus, we compute the following;

h(y) = Σ_x f(x, y) = Σ(2)_(x=0) f(x, y)

Now, solving from the table;

h(0) = f(0,0) + f(1,0) + f(2,0) = 0.12 + 0.08 + 0.06 = 0.26

h(1) = f(0,1) + f(1,1) + f(2,1) = 0.04 + 0.19 + 0.12 = 0.35

h(2) = f(0,2) + f(2,2) + f(2,1) = 0.04 + 0.05 + 0.30 = 0.39

Now, let's find the values f(x|2) for all x which the variable X assumes.

From earlier we have ;

f(x|2) = f(x,2)/h(2)

Solving from the joint probability table i attached, we obtain the following ;

f(0|2) = f(0,2)/h(2) = 0.04/0.39 = 4/39

f(1|2) = f(1,2)/h(2) = 0.05/0.39 = 5/39

f(2,2) = f(2,2)/h(2) = 0.30/0.39 = 30/39

B) The expected value of the random variable X is given as;

E(X) = Σ_x g(x) = Σ(2)_(x=0) x•g(x) =

(0 x 0.2) + (1 x 0.32) + (2 x 0.48) = 1.28

Now, the variance is given as;

Var(X) = Σ_x (x - E(X)²•g(X) =

Σ(2)_(x=0) (x - 1.28)²•g(x)

Var(X) = [(0 - 1.28)² x 0.2] + [(1 - 1.28)² x 0.32] + [(2 - 1.28)² x 0.42]

= 0.328 + 0.025 + 0.249 = 0.602

C) To find the mean of X, let's use the equivalent formula ;

E(X|Y = 2) = Σ_x (x•f(x|2)) = Σ(2)_(x=0) (x•f(x|2)

Thus,

E(X|Y = 2) = (0 x 4/39) + (1 x 5/39) + (2 x 30/39) = (5/39 + 20/13) = 5/3

Now, for the variance; it's given as;

Var(X|Y= 2) = Σ_x (x - E(X|Y= 2))²• f(x|2)

Var(X|Y= 2) = Σ(2)_(x=0) (x - 5/3)²•f(x|2)

Var(X|Y= 2) = [(0 - 5/3)²•4/39] + [1 - 5/3)²•5/39] + [(2 - 5/3)²•30/39]

= 100/351 + 20/351 + 30/351 = 150/351

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.