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anonymous
 3 years ago
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:
\[P(r)=\frac{n!}{r!(nr)!}p^rq^{nr}=C_{n,r}p^rq^{nr}\]
anonymous
 3 years ago
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: \[P(r)=\frac{n!}{r!(nr)!}p^rq^{nr}=C_{n,r}p^rq^{nr}\]

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I guess I have to find the minimum n

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Oh I see. there is a table with a formula \[P \textrm{ (at least three successes) }=P(r \ge3)=1P(0)P(1)P(2)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think I got this...one moment plz

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok now I need help =) I tried doing: \[P \textrm{ (at least three successes) }=P(r \ge3)=1P[0]P[1]P[3]\] but that gives a solution greater than 1 (for P[0]−P[1]−P[3]) which isn't right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I meant P[0]+P[1]+P[3]

phi
 3 years ago
Best ResponseYou've already chosen the best response.2shouldn't it be p(0),p(1) and p(2)

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.090% sure of filling the quota, so isn't p=0.9?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0What does P(r) represent here?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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
 3 years ago
Best ResponseYou've already chosen the best response.2your 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
 3 years ago
Best ResponseYou've already chosen the best response.0Oh 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
 3 years ago
Best ResponseYou've already chosen the best response.0so n=6 and u need to find r ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0no that was for part (a) of the problem... this is part (e)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(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 ad successfully...but (e) is the problem child

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So you wouldn't use the whole formula? \[P(r)=\frac{n!}{r!(nr)!}p^rq^{nr}=C_{n,r}p^rq^{nr}\]?

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0ok, 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in other words solve for n?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Great! That makes sense. I'll do that calculation and get back to ya...just a min...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmmm....doesn't look pretty.... \[3!\frac{0.9}{0.45}^3=n!\frac{(0.55)^{n3}}{(n3)!}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oops typo \[3!\frac{0.9}{0.45^3}=n!\frac{(0.55)^{n3}}{(n3)!}\]

hartnn
 3 years ago
Best ResponseYou've already chosen the best response.0yup, 0.55^n is not pretty at all n!=n(n1)(n2)(n3)!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This looks like a negative binomial situation. b*(x; r, P) = x1Cr1 * Pr * (1  P)x  r

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Eh, 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
 3 years ago
Best ResponseYou've already chosen the best response.0uhm. 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 \[P(r\ge3)=1P(r<3)\] \[=1(P(0)+P(1)+P(2))\] \[=1(0.003+0.021+0.076)\] \[=10.10\] \[=0.90\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.2yes, they used a table, then verified the result

phi
 3 years ago
Best ResponseYou've already chosen the best response.2I 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 10.09955 = 0.9 prob of 3 or more sucesses

phi
 3 years ago
Best ResponseYou've already chosen the best response.2I found the upper bound with this % cumulative distribution % F(k; n,p) ≤ 0.5*exp( (2/n)* (n*pk)^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^24*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
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.2Doesn't your text give a hint as to how to solve this type of problem?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Not really, They've provided me with that table. I guess trial and error is all they're asking me to do, since it is algebrabased course.
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