Question

New legislation passed in 2017 by the U.S. Congress changed tax laws that affect how many people file their taxes in 2018 and beyond. These tax law changes will likely lead many people to seek tax advice from their accountants (The New York Times). Backen and Hayes LLC is an accounting firm in New York state. The accounting firms believe that it may have to hire additional accountants to assist with the increased demand in tax advice for the upcoming tax season. Backen and Hayes LLC has developed the following probability distribution for x= number of new clients seeking tax advice.x f(x)20 .0525 .2030 .2535 .1540 .1545 .1050 .10a. Is this a valid probability distribution? Explain.b. What is the probability that Backens and Hayes LLC will obtain 40 or more new clients?c. What is the probability that Backens and Hayes LLC will obtain fewer than 35 new clients?d. Compute the expected value, variance, and standard deviation of x.

Answer 1

a. Yes

b. 0.35

c. 0.5

d.

E(x)=34.25

V(x)=73.1875

SD(x)=8.555

Explanation:

a.

x 20 25 30 35 40 45 50

f(x) 0.05 0.2 0.25 0.15 0.15 0.1 0.1

This will be valid probability distribution if

1. All probabilities are between 0 and 1 inclusive.

2. Sum of probabilities must be 1.

Condition 1 is satisfied in this distribution. Now, we will check second condition

sum(f(x))=0.05+0.2+0.25+0.15+0.15+0.1+0.1=1.

As the both conditions are satisfied, thus the given distribution is a valid probability distribution.

b.

P(x≥40)=P(X=40)+P(X=45)+P(X=50)

P(x≥40)=0.15+0.1+0.1

P(x≥40)=0.35

So, the probability that Backens and Hayes LLC will obtain 40 or more new clients is 0.35.

c.

P(x<35)=P(X=20)+P(X=25)+P(X=30)

P(x<35)=0.05+0.2+0.25

P(x<35)=0.5

So, the probability that Backens and Hayes LLC will obtain fewer than 35 new clients is 0.5.

d.

Expected value of x=E(x)= sum[x*f(x)]

E(x)= 20*0.05+25*0.2+30*0.25+35*0.15+40*0.15+45*0.1+50*0.1

E(x)=1+5+7.5+5.25+6+4.5+5

E(x)=34.25

Variance of x=V(x)=sum[x²*f(x)]-(sum[x*f(x)])²

sum[x²*f(x)]= 20²*0.05 +25²*0.2 +30²*0.25 +35²*0.15 +40²*0.15 +45²*0.1 +50²*0.1

sum[x²*f(x)]=20+1255+225+183.75+240+202.5+50

sum[x²*f(x)]=1246.25

V(x)=1246.25-(34.25)²

V(x)=73.1875

Standard deviation of x=SD(x)=√V(x)

SD(x)=8.555