## Quiz 6: Second C Level Question |
AZobi |
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You toss a coin 80,000 times. The coin lands on heads exactly 20,000 times. Estimate the probability that the probability of heads for this coin is in the interval [.23,.27]. |

mtariveran: March 7, 2015, 10:49 p.m. |
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I got 1/[(4)(80000)(0.02)^2] |

Alan_Mendoza: March 7, 2015, 11:10 p.m. |
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That answer seems kind of small? I'm not really sure at all but 1 - 1/[(4)(80000)(0.02)^2] seems like a better answer. |

mtariveran: March 7, 2015, 11:24 p.m. |
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yea thats what I was confused |

AZobi: March 8, 2015, 12:57 a.m. |
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Alan Mendoza, I agree with you. That is the same answer I got. |

AZobi: March 8, 2015, 1 a.m. |
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Actually, nevermind. I accidently overlooked some things. My answer does not match. :( |

ShivaniGillon: March 8, 2015, 10:23 a.m. |
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I got the same answer as Alan, too. |

weisbart: March 8, 2015, 12:50 p.m. |
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Why does Alan take the one minus in front to obtain the correct answer? Also, is this an exact answer or an estimate. How does one estimate the probability? |

PeterH: March 8, 2015, 4:51 p.m. |
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I'm a little confused about where the 4 in the denominator coming from? |

Alan_Mendoza: March 8, 2015, 5:44 p.m. |
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Peter H, it's part of the inequality for the Weak Law Estimate. |

martha2A: March 9, 2015, 1:21 a.m. |
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I subtracted one from 1/4nc^2 because I originally set up the equation as P(|Sn-p| <_0.02), but in order for me to relate it to 1/4nc^2, I needed to take the complement of P(...) to get it to be in the form P(|Sn-p| >_0.02) to match Chebyshev's inequality. 1- P(|Sn-p <_0.02) = P(|Sn-p| >_0.02) and P(|Sn-p| >_0.02) <_ 1- 1/4nc^2. |

martha2A: March 9, 2015, 1:32 a.m. |
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In a nutshell, the problem was set up to make it so that |Sn-p| would have to be less than or equal to (within) .02 (c), but chebyshev's only gives us an estimate when |Sn-p| is greater than or equal to .02 (c). |

SheydaMesgarzadeh: March 9, 2015, 12:14 p.m. |
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I got (.55)(.45)/ (100000)(.04)^2 because I used the equation p(1-p)/nc^2 due to the fact that our probability isn't 1/2, am I wrong in doing this? |

SheydaMesgarzadeh: March 9, 2015, 12:15 p.m. |
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Sorry ignore that,that was for the A level question |

SheydaMesgarzadeh: March 9, 2015, 12:19 p.m. |
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But, for this question i got 1- (1/4)(3/4)/(80,000)(.02)^2 < c because I used the equation p(1-p)/nc^2 < 1-c, did I do something wrong or are both ways of solving the problem acceptable? |