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:
\[P(r)=\frac{n!}{r!(n-r)!}p^rq^{n-r}=C_{n,r}p^rq^{n-r}\]

- anonymous

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- anonymous

p=.45
q=.55

- anonymous

I guess I have to find the minimum n

- anonymous

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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## More answers

- anonymous

I think I got this...one moment plz

- anonymous

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

- anonymous

@phi

- anonymous

I meant P[0]+P[1]+P[3]

- phi

shouldn't it be p(0),p(1) and p(2)

- hartnn

90% sure of filling the quota, so isn't p=0.9?

- anonymous

What does P(r) represent here?

- anonymous

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)

- anonymous

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?

- hartnn

so n=6 and u need to find r ?

- anonymous

no that was for part (a) of the problem... this is part (e)

- anonymous

(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

- anonymous

So you wouldn't use the whole formula?
\[P(r)=\frac{n!}{r!(n-r)!}p^rq^{n-r}=C_{n,r}p^rq^{n-r}\]?

- hartnn

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

- hartnn

sorry r>=3

- anonymous

in other words solve for n?

- hartnn

yup.

- anonymous

Great! That makes sense. I'll do that calculation and get back to ya...just a min...

- hartnn

ok.

- anonymous

hmmm....doesn't look pretty....
\[3!\frac{0.9}{0.45}^3=n!\frac{(0.55)^{n-3}}{(n-3)!}\]

- hartnn

u don't get 0.9^3

- anonymous

oops typo
\[3!\frac{0.9}{0.45^3}=n!\frac{(0.55)^{n-3}}{(n-3)!}\]

- hartnn

yup, 0.55^n is not pretty at all
n!=n(n-1)(n-2)(n-3)!

- anonymous

This looks like a negative binomial situation.
b*(x; r, P) = x-1Cr-1 * Pr * (1 - P)x - r

- anonymous

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?

- anonymous

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
\[P(r\ge3)=1-P(r<3)\]
\[=1-(P(0)+P(1)+P(2))\]
\[=1-(0.003+0.021+0.076)\]
\[=1-0.10\]
\[=0.90\]

##### 1 Attachment

- anonymous

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.

- anonymous

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?

- anonymous

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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