# Math Help - Binomial Distributions

1. ## Binomial Distributions

Supposing that an interviewer needs to interview five people and that the probability that each person agrees to be be interviewed is 0.6, what is the expected value and variance of the number of people she must ask to interview five people? The answers are E = 8.33 and V = 5.56.

My textbook isn't very clear on how to calculate expected values and variance - can someone push me in the right direction?

2. My textbook isn't very clear on how to calculate expected values and variance - can someone push me in the right direction?
I'd be pretty close to stunned if this actually were the case. I can believe the text maybe isn't talking to you very well. No all books speak to everyone. No worries.

Originally Posted by mintsharpie
Supposing that an interviewer needs to interview five people and that the probability that each person agrees to be be interviewed is 0.6, what is the expected value and variance of the number of people she must ask to interview five people? The answers are E = 8.33 and V = 5.56.
For a binomial distribution with p = 0.6 and n trials, we have ALWAYS,

E[X] = n*p

Var(x) = n*p*q

Where q = 1 - p

That's it. Is that in your book? It had better be, otherwise, why is the author published?

So.

n*(3/5) = 5 ==> n = 5 / (3/5) = 25/3

You tell me about the variance. I hardly can wait to see how the book's answer was created. Maybe I'm not understanding the question.

3. Originally Posted by TKHunny
I'd be pretty close to stunned if this actually were the case. I can believe the text maybe isn't talking to you very well. No all books speak to everyone. No worries.

For a binomial distribution with p = 0.6 and n trials, we have ALWAYS,

E[X] = n*p

Var(x) = n*p*q

Where q = 1 - p

That's it. Is that in your book? It had better be, otherwise, why is the author published?

So.

n*(3/5) = 5 ==> n = 5 / (3/5) = 25/3

You tell me about the variance. I hardly can wait to see how the book's answer was created. Maybe I'm not understanding the question.
I know your answer agrees with the book and I know it appears that the question has come from the section of the book dealing with binomial distribution, but I disagree that it's a binomial distribution question.

From the wording of the question, my understanding is that

Originally Posted by Wikipedia
... there is a sequence of independent Bernoulli trials, each trial having two potential outcomes called “success” and “failure”. In each trial the probability of success is p and of failure is 1 − p. We are observing this sequence until a predefined number r of failures has occurred.
For my money, technically this question requires the negative binomial distribution: Negative binomial distribution - Wikipedia, the free encyclopedia

where X is the random variable "number of successes (number of people who do NOT want to be interviewd)", p = 0.4 (the 'failures' are the people who want to be interviewed and so success is a person who does not want to be interviewed) and r = 5 (stop after five 'failures' are observed).

Then Mean = E(X + 5) = E(X) + 5 = (20/6) + 5.

4. Agreed. I was uncomfortable working the Binomial Distribution backwards, but failed to challenge the original premise. Perhaps this explains the variance?

5. Originally Posted by TKHunny
For a binomial distribution with p = 0.6 and n trials, we have ALWAYS,

E[X] = n*p

Var(x) = n*p*q

Where q = 1 - p

That's it. Is that in your book? It had better be, otherwise, why is the author published?

So.

n*(3/5) = 5 ==> n = 5 / (3/5) = 25/3

You tell me about the variance. I hardly can wait to see how the book's answer was created. Maybe I'm not understanding the question.
If E(X) = n*p, wouldn't that make the answer 5 x 0.6, which is 3? I don't understand where the 8.33 is coming from - where did you get the 25 from in order to divide it by 3?

All my book says is that $E(X) = \sum p(x)$ (there's supposed to be a y under the sum symbol, but I don't know how to do that). Then there's a lot of steps to follow, and at the end, it ends up being $E(X) = \lambda$. It's not very clear, I have no idea what it's talking about. I took an advanced probability class in grade 12, and the textbook was really clear and I actually understood most of it, but this year I'm completely lost. I know it's university so it's supposed to be harder, but it just seems like the textbook moves too quickly and includes limited examples, so I'm confused most of the time.

Originally Posted by mr fantastic
For my money, technically this question requires the negative binomial distribution: Negative binomial distribution - Wikipedia, the free encyclopedia

where X is the random variable "number of successes (number of people who do NOT want to be interviewd)", p = 0.4 (the 'failures' are the people who want to be interviewed and so success is a person who does not want to be interviewed) and r = 5 (stop after five 'failures' are observed).

Then Mean = E(X + 5) = E(X) + 5 = (20/6) + 5.
I'm sorry, I don't understand. Where did the 20 and the 6 come from?

6. Originally Posted by mintsharpie
If E(X) = n*p, wouldn't that make the answer 5 x 0.6, which is 3? Mr F says: It's not Binomial, forget that idea. It's negative binomial, as I said in my earlier post.
I don't understand where the 8.33 is coming from - where did you get the 25 from in order to divide it by 3?

All my book says is that $E(X) = \sum p(x)$ (there's supposed to be a y under the sum symbol, but I don't know how to do that). Then there's a lot of steps to follow, and at the end, it ends up being $E(X) = \lambda$. It's not very clear, I have no idea what it's talking about. I took an advanced probability class in grade 12, and the textbook was really clear and I actually understood most of it, but this year I'm completely lost. I know it's university so it's supposed to be harder, but it just seems like the textbook moves too quickly and includes limited examples, so I'm confused most of the time.

I'm sorry, I don't understand. Where did the 20 and the 6 come from?
Go to the link I gave, read the formula for mean, substiute the given values (as defined by me in my earlier post).