A convenience store has two separate locations where customers can be checked out as they leave. These locations each have two c
ash 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).
2 answers:
The joint probability function is missing and I've attached it.
Answer:
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
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