Thread: Probability question

A contractor is required by a county planning department
to submit anywhere from one to five forms (depending on the nature of the project) in applying for a building permit. Let r.v. X = the number of forms required of the next applicant. The probability that x forms are required is known to be proportional to x; that is, p(x) = cx for x = 1,...,5.

What is the value of c?

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I got c = (100 / (1+2+3+4+5)) / 100 = 0.066666667, so that:

p(1) = 0.07
p(2) = 0.13
p(3) = 0.2
p(4) = 0.27
p(5) = 0.33

Is this correct?

Originally Posted by BrownianMan

A contractor is required by a county planning department
to submit anywhere from one to five forms (depending on the nature of the project) in applying for a building permit. Let r.v. X = the number of forms required of the next applicant. The probability that x forms are required is known to be proportional to x; that is, p(x) = cx for x = 1,...,5.

What is the value of c?

---------------------------------------

I got c = (100 / (1+2+3+4+5)) / 100 = 0.066666667, so that:

p(1) = 0.07
p(2) = 0.13
p(3) = 0.2
p(4) = 0.27
p(5) = 0.33

Is this correct?

Why have you put in those two "100"'s? Otherwise yes.

CB

Thanks!

I have another question:

You are to take a multiple-choice exam consisting of 100 questions with five possible responses to each. Suppose that you have not studied and so must guess (select one of the five answers in a completely random fashion) on each question. Let r.v.

X represent the number of correct responses on the exam.

Specify the probability distribution of X.

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I'm not sure what the distribution would look like. Would it be 1/500 for answering 1 question correctly, 2/500 for 2 correct, etc.? In other words x/500 for x = 1, 2, ... , 100???

Originally Posted by BrownianMan

Thanks!

I have another question:

You are to take a multiple-choice exam consisting of 100 questions with five possible responses to each. Suppose that you have not studied and so must guess (select one of the five answers in a completely random fashion) on each question. Let r.v.

X represent the number of correct responses on the exam.

Specify the probability distribution of X.

-----------------------------------------------------------------

I'm not sure what the distribution would look like. Would it be 1/500 for answering 1 question correctly, 2/500 for 2 correct, etc.? In other words x/500 for x = 1, 2, ... , 100???

Each question is a Bernouli trial with probability of sucess 0.2. You should know the distribution of the number of successes in 100 such Bernouli trials with this probability of succes in a single trial.

CB

Originally Posted by CaptainBlack

Each question is a Bernouli trial with probability of sucess 0.2. You should know the distribution of the number of successes in 100 such Bernouli trials with this probability of succes in a single trial.

CB

Ok, thanks. I got it now. I also found the expected number of correct guesses (20), and the standard deviation (4).

Then it asks me to find the approximate probability that the number of correct responses will be between 15 and 30 (inclusive)? More than 25? Less than 35?

Do I use the Normal approximation for the Binomial Distribution to answer those questions?

Originally Posted by BrownianMan

Ok, thanks. I got it now. I also found the expected number of correct guesses (20), and the standard deviation (4).

Then it asks me to find the approximate probability that the number of correct responses will be between 15 and 30 (inclusive)? More than 25? Less than 35?

Do I use the Normal approximation for the Binomial Distribution to answer those questions?

Yes, that is an acceptable way of doing the calculations (you may have a binomial distribution calculator, if you do you could use that).

CB

I used the Normal approximation and a Binomial calculator online, and I get slightly different answers for each method. Does that mean my answers using the Normal approximation are incorrect? Using the Normal approximation I get:

P(between 15 and 30) = 0.9119
P(more than 25) = 0.1292

and for P(less than 35) I get a z-score of 3.88 which isn't on the Normal Distribution table. Not sure what to do for that one. I would say that it's very close to 1.