West Virginia has one of the highest divorce rates in the nation, with an annual rate of approximately divorces per 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 the number of new clients for marriage counseling for the next year.
(x) f(x)
10 .05
20 .10
30 .10
40 .20
50 .35
60 .20
a. Is this probability distribution valid?
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.

[tex]\begin{tabular}{ x |f(x)|} x & f(x) & \\ 10 & .05& \\ 20 & .10 & \\ 30 & .10 & \\ 40 & .20 & \\ 50 & .35 & \\ 60 & ..20 & \\\end{tabular}\end{document}[/tex]

(a) For validity of probability distribution, sum of probability equal to be 1

[tex]\therefore \sum f(x) = 1[/tex]

[tex]\sum f(x) = .05+.10+.10+.20+.35+.20 = 1[/tex]

Yes, the probability distribution is valid

(b) The probability of MCC will obtain more than 30 new clients

[tex]= P(X> 30)[/tex]

[tex]= P(X=40) +P(X=50)+P(X=60)[/tex]

= .20+.35+.20 = .75

(C) The probability of MCC will obtain fewer than 20 new clients

[tex]= P(X< 20)= P(X=10) = .05[/tex]

(d ) - (1) Expected value

[tex]E(x)= \sum x f(x) = 10\times0.05+20\times0.1+30\times0.1+40\times0.2+50\times0.35+60\times0.2= 0.5+2+3+8+17.5+12=43[/tex]

(2) Variance of x

[tex]= E(X^2)-{E(x)}^2 E(X^2) = \sum x^2f(x) = 10^2\times.05 + 20^2\times.10 + 30^2\times.10+ 40^2\times.20+ 50^2\times.35 + 60^2\times.20 = 2050[/tex]

[tex]= E(X^2)-{E(x)}^2 = 2050 - 43^2 = 201[/tex]