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West Virginia has one of the highest divorce rates in the nation, with an annual rate of approximately 5 divorces per 1000 people (Centers for Disease Control and Prevention website, January 12, 2012). The Marital Counseling Center, Inc. (MCC) thinks that the high divorce rate in the state may require them to hire additional staff. Working witha consultant, the management of MCC has developed the following probability distribution for the number of new clients for marriage counseling for the next year Excel File: data05-19.

col1 xlsx 10 20 30 40 50 60
col2 f(x) .05 10 10 .20 .35 .20

a. Is this probability distribution valid? Yes Explain f(r) greater than or equal to 0 f(equal to 1
b. What is the probability MCC will obtain more than 30 new clients (to 2 decimals)?
c. What is the probability MCC will obtain fewer than 20 new clients(to 2 decimals)?
d. Compute the expected value and variance of x Expected value clients per year 43 Variance 490 squared clients per year

Respuesta :

Answer:

a) In order to see if that represent a probability distribution we need to satisfy two conditions:

1) [tex]P(X_i) \geq 0 , i =1,2,...,6 [/tex] we have this condition satisfied

2) [tex]\sum_{i=1}^n X_i P(X_i) , i=1,2,....,6[/tex]

And we verify that 0.05+0.1+0.1+0.2+0.35+0.2=1

So then we can conclude that we have a probability distribution since both conditions are satisifed.

b) [tex] P(X>30) = P(X=40)+P(X=50) +P(X=60) = 0.2+0.35+0.2=0.75[/tex]

c) [tex] P(X<20) = P(X=10)=0.05[/tex]

d) [tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]

[tex]E(X^2) =10^2*0.05 +20^2*0.1 +30^2*0.1 +40^2*0.2 +50^2*0.35+ 60^2*0.2=2050[/tex]

[tex] Var(X) =E(X^2) -[E(X)]^2 = 2050 -(43^2) =201[/tex]

Step-by-step explanation:

For this case we have the following distribution given:

x       10      20     30      40       50       60

f(x)  0.05    0.1     0.1      0.2      0.35    0.2

Part a

In order to see if that represent a probability distribution we need to satisfy two conditions:

1) [tex]P(X_i) \geq 0 , i =1,2,...,6 [/tex] we have this condition satisfied

2) [tex]\sum_{i=1}^n X_i P(X_i) , i=1,2,....,6[/tex]

And we verify that 0.05+0.1+0.1+0.2+0.35+0.2=1

So then we can conclude that we have a probability distribution since both conditions are satisifed.

Part b

For this case we want this probability:[tex] P(X>30) = P(X=40)+P(X=50) +P(X=60) = 0.2+0.35+0.2=0.75[/tex]Part c

For this case we want this probability:

[tex] P(X<20) = P(X=10)=0.05[/tex]

Part d

The expected value can be calculated with this formula:

[tex] E(X) = \sum_{i=1}^n X_i P(X_i)[/tex]

And if we replace we got:[tex] E(X) =10*0.05 +20*0.1 +30*0.1 +40*0.2 +50*0.35+ 60*0.2=43[/tex]And in order to calculate the variance we need to calculate the second moment like this:

[tex]E(X^2) = \sum_{i=1}^n X^2_i P(X_i)[/tex]

And we got:[tex]E(X^2) =10^2*0.05 +20^2*0.1 +30^2*0.1 +40^2*0.2 +50^2*0.35+ 60^2*0.2=2050[/tex]And then the variance is given by:

[tex] Var(X) =E(X^2) -[E(X)]^2 = 2050 -(43^2) =201[/tex]