Jump to content
[[Template core/front/custom/_customHeader is throwing an error. This theme may be out of date. Run the support tool in the AdminCP to restore the default theme.]]

Calling all Nerds


Controller
 Share

Recommended Posts

Here is the extra credit problem:

 

Suppose that you are a car salesperson. Whenever you approach a potential new customer, you have a 10% chance of selling him or her a new car. Let the random variable N be the number of customers you talk to until you first sell a car (for example, if you do not sell a car to your first two prospects and then you do sell one to the third, N = 3).

 

What is the expected value E[N]?

 

What is the variance V[N]?

 

Note: I thought the probability P(N=K) = 1/10^k, but the prof. tells me that is wrong.

 

Any :D:D:tup: out there who can help???

Link to comment
Share on other sites

It's been 15 years since I had prob/stats in college, but isn't the expected value the average of the possible outcomes?

 

Chances of selling a car to the first person is .1(1) = .1. Since it's a 10% chance, numbers say you'll sell one by the time N=10. (100% chance you'll sell a car if you see 10 people)

 

Add up E(N) = .1(N) where N=1 to 10

 

.1+.2+.3+.4+.5+.6+.7+.8+.9+1 = 5.5

 

Expected value is 5.5 people.

 

I'm probably wrong.

 

I don't remember variance, but I think it's the square of the standard deviation?

Link to comment
Share on other sites

It's been 15 years since I had prob/stats in college, but isn't the expected value the average of the possible outcomes?

 

Chances of selling a car to the first person is .1(1) = .1. Since it's a 10% chance, numbers say you'll sell one by the time N=10. (100% chance you'll sell a car if you see 10 people)

This is not correct.

 

If you see 10 people, there is a 34.867844% chance that you won't have sold any of them a car.

 

I can't remember off the top of my head how to answer the original questions, but perhaps it will come to me later on this afternoon (no promises though).

Link to comment
Share on other sites

This is not correct.

 

If you see 10 people, there is a 34.867844% chance that you won't have sold any of them a car.

 

I can't remember off the top of my head how to answer the original questions, but perhaps it will come to me later on this afternoon (no promises though).

 

 

And here I was about to come in here and say Wiegie will be along shortly to answer this :D

Link to comment
Share on other sites

Expected value is defined at the summation of k from 1 to infinity of:

 

k times (the probability of N = k)

 

Variance is defined as the summation of k from 1 to infinity of:

 

[k minus (expected value of n)] squared times the probability of N = k)

 

If that helps.

Link to comment
Share on other sites

Last time, I promise, I guess my first post was right.

 

The probability of selling a car is P(success)=1-0.9^K

 

The probability of N=K is P(N=K) = 1 - 0.9^K - P(success at K-1) since the sample stops when a success is reached

 

thus the probability of success at N=1 is 0.1 and the probability of success at N=2 is 0.19, then P(N=2)=0.19-0.1=0.09

 

This is easy to accomplish with an Excel spreadsheet.

Multiply K by the probability of N=K and sum your answers. You should get 9.96.

Link to comment
Share on other sites

Not sure if this helps...

 

The probabilty of NOT selling a car to any given person is 90% ... so, with any random group of 10 individuals, you WON'T sell a single car 34.868% of the time (i.e., 0.9^10) ... or, said another way ... you WILL sell at least one car 65.132% (1 - 34.868%) of the time.

Link to comment
Share on other sites

Here is the extra credit problem:

 

Suppose that you are a car salesperson. Whenever you approach a potential new customer, you have a 10% chance of selling him or her a new car. Let the random variable N be the number of customers you talk to until you first sell a car (for example, if you do not sell a car to your first two prospects and then you do sell one to the third, N = 3).

 

What is the expected value E[N]?

 

What is the variance V[N]?

 

Note: I thought the probability P(N=K) = 1/10^k, but the prof. tells me that is wrong.

 

Any :D:D:tup: out there who can help???

 

 

With your personality, I'd say the probability of you selling anything is 0. :doh:

Edited by cre8tiff
Link to comment
Share on other sites

Not sure if this helps...

 

The probabilty of NOT selling a car to any given person is 90% ... so, with any random group of 10 individuals, you WON'T sell a single car 34.868% of the time (i.e., 0.9^10) ... or, said another way ... you WILL sell at least one car 65.132% (1 - 34.868%) of the time.

 

muck cheated off of me

 

This is not correct.

 

If you see 10 people, there is a 34.867844% chance that you won't have sold any of them a car.

 

I can't remember off the top of my head how to answer the original questions, but perhaps it will come to me later on this afternoon (no promises though).

 

:oldrazz:

Link to comment
Share on other sites

I'm not 100% certain about this, but I think what we are dealing with here is a geometric distribution.

 

The expected value of such distribution = (1-p) / p = (1-.1)/.1 = .9/.1 = 9

 

The variance of the distribution = (1-p)/p^2 = .9/(.1)^2 = .9/.01 = 90

 

(where p = the probability of selling the car)

 

I could be wrong, so don't bet the house on it.

Link to comment
Share on other sites

I'm not 100% certain about this, but I think what we are dealing with here is a geometric distribution.

 

The expected value of such distribution = (1-p) / p = (1-.1)/.1 = .9/.1 = 9

 

The variance of the distribution = (1-p)/p^2 = .9/(.1)^2 = .9/.01 = 90

 

(where p = the probability of selling the car)

 

I could be wrong, so don't bet the house on it.

 

 

If you assume a geometric distribution, the expected value is not (1-p)/p but it is in fact E(N) = 1/p leaving you with an expected value of 10, very close to my crude methodology answer of 9.96.

 

The variance is the same however.

 

(1-p)/p is the expected value of consecutive failures, not the success expected value

 

http://en.wikipedia.org/wiki/Geometric_distribution

Edited by Tford
Link to comment
Share on other sites

If you assume a geometric distribution, the expected value is not (1-p)/p but it is in fact E(N) = 1/p leaving you with an expected value of 10, very close to my crude methodology answer of 9.96.

 

The variance is the same however.

 

(1-p)/p is the expected value of consecutive failures, not the success expected value

 

http://en.wikipedia.org/wiki/Geometric_distribution

 

I have seen different formulas given for the expected value (the one I gave and the one you gave). I'm not sure exactly which one is correct, but I used the version of the formula given in my old graduate Probability & Statistics textbook (DeGroot).

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

 Share

  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...

Important Information