Question

1. - / 7 points\

west virginia has one of the highest divorce rates in the nation, with an annual rate of approximately 5 divorces per 1,000 people. 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.\

x	f(x)	\
----	------	\
10	0.05	\
20	0.15	\
30	0.15	\
40	0.20	\
50	0.30	\
60	0.15	\

(a) is this probability distribution valid? explain.\
this probability distribution \\u2014select\\u2014 valid because \\u2014select\\u2014 for x = 10, 20, 30, 40, 50, 60. also, $\sum f(x) = $ \
(b) what is the probability mcc will obtain more than 30 new clients?\
\
(c) what is the probability mcc will obtain fewer than 20 new clients?\
\
(d) compute the expected value and variance of x.\
expected value of x = \
variance of x = \

Explanation:

Part (a)

Step1: Check non - negativity of probabilities

For a probability distribution, each probability \( f(x) \) must be non - negative (i.e., \( f(x)\geq0 \)) and the sum of all probabilities must be equal to 1.
We check the values of \( f(x) \) for \( x = 10,20,30,40,50,60 \). The values are \( 0.05,0.15,0.15,0.20,0.30,0.15 \) respectively, and all of them are non - negative (greater than or equal to 0).

Step2: Calculate the sum of probabilities

We calculate \( \sum f(x)=0.05 + 0.15+0.15 + 0.20+0.30 + 0.15\)
\(0.05+0.15 = 0.20\); \(0.20+0.15=0.35\); \(0.35 + 0.20=0.55\); \(0.55+0.30 = 0.85\); \(0.85+0.15=1.00\)

So this probability distribution is valid because \( f(x)\geq0 \) for \( x = 10,20,30,40,50,60 \). Also, \( \sum f(x)=1 \).

Part (b)

Step1: Identify the relevant values

We want to find \( P(X>30) \). The values of \( x \) that satisfy \( x > 30 \) are \( x = 40,50,60 \) with corresponding probabilities \( f(40)=0.20\), \( f(50)=0.30\), \( f(60)=0.15 \)

Step2: Sum the probabilities

\( P(X > 30)=f(40)+f(50)+f(60)=0.20 + 0.30+0.15\)
\(0.20+0.30 = 0.50\); \(0.50+0.15 = 0.65\)

Part (c)

Step1: Identify the relevant values

We want to find \( P(X<20) \). The only value of \( x \) that satisfies \( x < 20 \) is \( x = 10 \) with \( f(10)=0.05 \)

Step2: Get the probability

\( P(X < 20)=f(10)=0.05 \)

Part (d)

Expected Value (\( E(X) \))

Answer:

s:
(a) This probability distribution \(\boldsymbol{\text{is}}\) valid because \( f(x)\geq0 \) for \( x = 10,20,30,40,50,60 \). Also, \( \sum f(x)=\boldsymbol{1} \).

(b) The probability that MCC will obtain more than 30 new clients is \(\boldsymbol{0.65}\)

(c) The probability that MCC will obtain fewer than 20 new clients is \(\boldsymbol{0.05}\)

(d) The expected value of \( x \) is \(\boldsymbol{40}\), the variance of \( x \) is \(\boldsymbol{210}\)