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Degger [83]
2 years ago
12

New legislation passed in 2017 by the U.S. Congress changed tax laws that affect how many people file their taxes in 2018 and be

yond. 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.
Mathematics
1 answer:
stiks02 [169]2 years ago
3 0

Answer:

a. Yes

b. 0.35

c. 0.5

d.

E(x)=34.25

V(x)=73.1875

SD(x)=8.555

Step-by-step 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

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iren [92.7K]

Answer:

X=5 y=-2

Step-by-step explanation:

Set them equal to each other so...

-2x+8=3x-7

Simplify by balancing out the equation so

-5x=-1

So x=5

Now plug it into either y equation

-2(5)+8

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Hope this helps!

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3 years ago
Answer number 5 click image.
natita [175]
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3 years ago
Read 2 more answers
A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute
nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

P(k\geq8)=1-P(k

P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\

P(k

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002

5 0
3 years ago
You received partial credit in th
Blababa [14]

Answer:

Expected value would be $ 0.896

Step-by-step explanation:

Given,

The price of the lottery ticket = $44800000,

Also, the probability of winning the grand prize = .000000020,

Thus, the expected value of the lottery ticket = value of the lottery ticket × probability of getting the lottery ticket

= 44800000 × .000000020

= $0.896

Note : value of lottery ticket = prize amount - cost of each ticket,

Here the cost price of a ticket is not given,

That's why we did not consider it.

5 0
2 years ago
Please help!!! Brainliest?
Elena L [17]
I think the answer is b i hope this helps
7 0
2 years ago
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