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MathSofiya Group Title

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

  • 2 years ago
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  1. MathSofiya Group Title
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    p=.45 q=.55

    • 2 years ago
  2. MathSofiya Group Title
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    I guess I have to find the minimum n

    • 2 years ago
  3. MathSofiya Group Title
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    Oh I see. there is a table with a formula \[P \textrm{ (at least three successes) }=P(r \ge3)=1-P(0)-P(1)-P(2)\]

    • 2 years ago
  4. MathSofiya Group Title
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    I think I got this...one moment plz

    • 2 years ago
  5. MathSofiya Group Title
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    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

    • 2 years ago
  6. MathSofiya Group Title
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    @phi

    • 2 years ago
  7. MathSofiya Group Title
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    I meant P[0]+P[1]+P[3]

    • 2 years ago
  8. phi Group Title
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    shouldn't it be p(0),p(1) and p(2)

    • 2 years ago
  9. hartnn Group Title
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    90% sure of filling the quota, so isn't p=0.9?

    • 2 years ago
  10. wio Group Title
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    What does P(r) represent here?

    • 2 years ago
  11. MathSofiya Group Title
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    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

    • 2 years ago
  12. phi Group Title
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    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)

    • 2 years ago
  13. MathSofiya Group Title
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    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?

    • 2 years ago
  14. hartnn Group Title
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    so n=6 and u need to find r ?

    • 2 years ago
  15. MathSofiya Group Title
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    no that was for part (a) of the problem... this is part (e)

    • 2 years ago
  16. MathSofiya Group Title
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    (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

    • 2 years ago
  17. MathSofiya Group Title
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    So you wouldn't use the whole formula? \[P(r)=\frac{n!}{r!(n-r)!}p^rq^{n-r}=C_{n,r}p^rq^{n-r}\]?

    • 2 years ago
  18. hartnn Group Title
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    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

    • 2 years ago
  19. hartnn Group Title
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    sorry r>=3

    • 2 years ago
  20. MathSofiya Group Title
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    in other words solve for n?

    • 2 years ago
  21. hartnn Group Title
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    yup.

    • 2 years ago
  22. MathSofiya Group Title
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    Great! That makes sense. I'll do that calculation and get back to ya...just a min...

    • 2 years ago
  23. hartnn Group Title
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    ok.

    • 2 years ago
  24. MathSofiya Group Title
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    hmmm....doesn't look pretty.... \[3!\frac{0.9}{0.45}^3=n!\frac{(0.55)^{n-3}}{(n-3)!}\]

    • 2 years ago
  25. hartnn Group Title
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    u don't get 0.9^3

    • 2 years ago
  26. MathSofiya Group Title
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    oops typo \[3!\frac{0.9}{0.45^3}=n!\frac{(0.55)^{n-3}}{(n-3)!}\]

    • 2 years ago
  27. hartnn Group Title
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    yup, 0.55^n is not pretty at all n!=n(n-1)(n-2)(n-3)!

    • 2 years ago
  28. CliffSedge Group Title
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    This looks like a negative binomial situation. b*(x; r, P) = x-1Cr-1 * Pr * (1 - P)x - r

    • 2 years ago
  29. CliffSedge Group Title
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    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?

    • 2 years ago
  30. MathSofiya Group Title
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    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 \[P(r\ge3)=1-P(r<3)\] \[=1-(P(0)+P(1)+P(2))\] \[=1-(0.003+0.021+0.076)\] \[=1-0.10\] \[=0.90\]

    • 2 years ago
  31. MathSofiya Group Title
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    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)?

    • 2 years ago
  32. phi Group Title
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    yes, they used a table, then verified the result

    • 2 years ago
  33. phi Group Title
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    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

    • 2 years ago
  34. phi Group Title
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    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.

    • 2 years ago
  35. MathSofiya Group Title
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    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

    • 2 years ago
  36. phi Group Title
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    Doesn't your text give a hint as to how to solve this type of problem?

    • 2 years ago
  37. MathSofiya Group Title
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    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.

    • 2 years ago
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