Question

What is the probability mcc will obtain fewer than 20 new clients(to 2 decimals)?

Answer 1

Given that 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.

$\begin{tabular} {|c|c|} x&f(x)\\[1ex] 10&0.05\\ 20&0.10\\ 30&0.10\\ 40&0.20\\ 50&0.35\\ 60&0.20 \end{tabular}$

Part A:

Is this probability distribution valid?

The sum of f(x) = 0.05 + 0.10 + 0.10 + 0.20 + 0.35 + 0.20 = 1

Since 0 ≤ f(x) ≤ 1, thus the distribution is a valid probability distribution.



Part B:

What is the probability MCC will obtain more than 30 new clients (to 2 decimals)?

The probability that the MCC will obtain more than 30 new clients is given by

$P(x\ \textgreater \ 30)=P(40)+P(50)+P(60)=0.20+0.35+0.20=0.75$



Part C:

What is the probability MCC will obtain fewer than 20 new clients(to 2 decimals)?

The probability that MCC will obtain fewer than 20 new clients is given by

$P(x\ \textless \ 20)=P(10)=0.05$



Part D

Compute the expected value

The expected value is given by

$\Sigma xf(x)=10(0.05)+20(0.1)+30(0.1)+40(0.2)+50(0.35)+60(0.2) \\ \\ =0.5+2+3+8+17.5+12=43$



Part 5:

The variance is given by

$Var(x)=\Sigma(x-E(x))^2f(x) \\ \\ =(10-43)^2(0.05)+(20-43)^2(0.1)+(30-43)^2(0.1)+(40-43)^2(0.2) \\ +(50-43)^2(0.35)+(60-43)^2(0.2) \\ \\ =(-33)^2(0.05)+(-23)^2(0.1)+(-13)^2(0.1)+(-3)^2(0.2)+(7)^2(0.35) \\ +(17)^2(0.2) \\ \\ =1089(0.05)+529(0.1)+169(0.1)+9(0.2)+49(0.35)+289(0.2) \\ \\ =54.45+52.9+16.9+1.8+17.15+57.8=201$