Question

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 with a consultant, the management of MCC has developed the following probability distribution for x = the number of new clients for marriage counseling for the next year.
West Virginia has one of the highest divorce rates

a. Is this probability distribution valid?
SelectYesNoItem 1

Explain.

f(x) Selectgreater than or equal to 0less than or equal to 0greater than or equal to 1less than or equal to 1Item 2
?f(x) Selectequal to 1not equal to 1greater than 1less than 1Item 3
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
Variance squared clients per year

Answer 1

Answer:

a. Yes. It is a probability density function because \sum f(x) =1

. b. probability MCC will obtain more than 30 new clients=P(40)+P(50)+P(60)= 0.20+0.35+0.20=0.75

c. probability MCC will obtain fewer than 20 new clients= P(10)= 0.05

d.

x f(x) x*f(x) x*x*f(x)

10 0.05 0.5 5

20 0.1 2 40

30 0.1 3 90

40 0.2 8 320

50 0.35 17.5 875

60 0.2 12 720

1 43 2050

expected value = \sum xf(x) = 43

Variance = 2050-43^2= 201

Explanation: