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Statistics: Suppose you were assigned to write an article for the student newspaper and you were given a quota (by the editor) of interviewing at least three extroverted professors. How many professors selected at random would you need to interview to be at least 90% sure of filling the quota? I was able to do the parts of the problems (this is part (e)) by using the equation: \[P(r)=\frac{n!}{r!(n-r)!}p^rq^{n-r}=C_{n,r}p^rq^{n-r}\]

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p=.45 q=.55
I guess I have to find the minimum n
Oh I see. there is a table with a formula \[P \textrm{ (at least three successes) }=P(r \ge3)=1-P(0)-P(1)-P(2)\]

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Other answers:

I think I got this...one moment plz
ok now I need help =) I tried doing: \[P \textrm{ (at least three successes) }=P(r \ge3)=1-P[0]-P[1]-P[3]\] but that gives a solution greater than 1 (for P[0]−P[1]−P[3]) which isn't right
I meant P[0]+P[1]+P[3]
  • phi
shouldn't it be p(0),p(1) and p(2)
90% sure of filling the quota, so isn't p=0.9?
What does P(r) represent here?
The first part of the problem is We now have the tools to solve the Chapter Focus Problem. In the book A Guide to the Development and Use of the Myers–Briggs Type Indicators by Myers and McCaully, it was reported that approximately 45% of all university professors are extroverted. Suppose you have classes with six different professors. That's where I got my p=.45 and q=.55 from
  • phi
your idea is correct P (at least three successes) =P(r≥3)=1−P[0]−P[1]−P[3] but you have p(3) and it should be p(2)
Oh I see. It looks like I wrote P(3) on OS, but in my notebook I did calculate P(2).... Another mistake I made....I plugged in n=6 which is wrong, what n should I use?
so n=6 and u need to find r ?
no that was for part (a) of the problem... this is part (e)
(a) What is the probability that all six are extroverts? (b) What is the probability that none of your professors is an extrovert? (c) What is the probability that at least two of your professors are extroverts? (d) In a group of six professors selected at random, what is the expected number of extroverts? What is the standard deviation of the distribution? (e) Quota Problem Suppose you were assigned to write an article for the student newspaper and you were given a quota (by the editor) of interviewing at least three extroverted professors. How many professors selected at random would you need to interview to be at least 90% sure of filling the quota? I did a-d successfully...but (e) is the problem child
So you wouldn't use the whole formula? \[P(r)=\frac{n!}{r!(n-r)!}p^rq^{n-r}=C_{n,r}p^rq^{n-r}\]?
ok, i got it. Atleast 90% sure means P(r) = 0.9 u need to select atleast 3 extroverted from 'n', n=n, r=3 p=0.45,q=0.55
sorry r>=3
in other words solve for n?
yup.
Great! That makes sense. I'll do that calculation and get back to ya...just a min...
ok.
hmmm....doesn't look pretty.... \[3!\frac{0.9}{0.45}^3=n!\frac{(0.55)^{n-3}}{(n-3)!}\]
u don't get 0.9^3
oops typo \[3!\frac{0.9}{0.45^3}=n!\frac{(0.55)^{n-3}}{(n-3)!}\]
yup, 0.55^n is not pretty at all n!=n(n-1)(n-2)(n-3)!
This looks like a negative binomial situation. b*(x; r, P) = x-1Cr-1 * Pr * (1 - P)x - r
Eh, sorry, that formula came out a bit sloppy... From what I understand, though, you're solving for x number of trials to have r number of successes?
uhm. here is what I found online.... "Since the probability of success is p=.45, we need to look in the binomial table under p=0.45 and different values of n that will satisfy this preceding relation. Binomial Probability Distribution table shows that if n=10 when p=0.45, then \[P(r\ge3)=1-P(r<3)\] \[=1-(P(0)+P(1)+P(2))\] \[=1-(0.003+0.021+0.076)\] \[=1-0.10\] \[=0.90\]
i don't quite understand what that person did? So they looked in the table under p=.45, and then did trial and error to see which n would give a solution for P(3)?
  • phi
yes, they used a table, then verified the result
  • phi
I found this upper bound http://en.wikipedia.org/wiki/Binomial_distribution#Cumulative_distribution_function which gives 11 then testing it with the actual numbers give n=10 with prob 0.09955 6 for 0,1,or 2 successes so 1-0.09955 = 0.9 prob of 3 or more sucesses
  • phi
I found the upper bound with this % cumulative distribution % F(k; n,p) ≤ 0.5*exp( (-2/n)* (n*p-k)^2 ) % k successes in n trials, prob of success = p % solve for n % n^2 + (ln(0.2)/(2*p^2) - 2*k/p) n + (k/p)^2 ≥ 0 % use quadratic formula p= 0.45; prob= 0.1; k=2; b= log(2*prob)/(2*p^2) -2*k/p dis= b^2-4*k^2/p^2 sd= sqrt(dis) upper_bound= 0.5*(-b+sd) It found upper_bound = 11.0800 but I had to test 11 and 10 to find the actual value.
So there isn't a direct way to do it I guess...just trial and error till we get the solution we want. Makes sense. Thanks =D
  • phi
Doesn't your text give a hint as to how to solve this type of problem?
Not really, They've provided me with that table. I guess trial and error is all they're asking me to do, since it is algebra-based course.

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