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anonymous
 2 years ago
A fair 6sided die is rolled 1,000 times. Using a normal approximation with a continuity correction, what is the probability the number of 3's rolled is greater than 150 and less than 180? I'm supposed to get .78, but instead I'm getting .81.
anonymous
 2 years ago
A fair 6sided die is rolled 1,000 times. Using a normal approximation with a continuity correction, what is the probability the number of 3's rolled is greater than 150 and less than 180? I'm supposed to get .78, but instead I'm getting .81.

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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0You are using a normal approximation to the binomial distribution presumably. What is the continuity correction?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0This link explains the continuity correction pretty well. http://www.mathsrevision.net/advancedlevelmathsrevision/statistics/normalapproximations

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks. Must run. Sorry.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Probability of rolling a 3 \(p=\dfrac{1}{6}=1.6667\). If \(X\) denotes the number of 3's that you roll, you have \[P(150\le X\le180)\] which is approximately equal to, using the continuity correction, \[P(149.5\le X\le180.5)\] Now transform to a standard normal random variable, \(Z\): \[P\left(\frac{149.5\mu}{\sigma}\le \frac{X\mu}{\sigma}\le\frac{180.5\mu}{\sigma}\right)\] where \(\mu=np=1000\times\dfrac{1}{6}=166.67\) and \(\sigma=\sqrt{np(1p)}=11.785\). \[P\left(\frac{149.5166.67}{11.785}\le Z\le\frac{180.5166.67}{11.785}\right)\\ P\left(1.457\le Z\le1.174\right)\] and I get \(0.84610.0735=0.7726\) by my calculations, which is close enough to the real answer.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0How'd you get .8461.0735?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0From a table of \(z\)values. \[P(1.457\le Z\le1.174)=P(Z\le1.174)P(Z\le1.457)\] Referring to this table: http://facstaff.unca.edu/dohse/Stat225/Images/Tablez.JPG For this table, the desired probability is given as the area to the left of a certain \(z\)value. If you're unsure how to use this table, what you do is find the probability/area associated with 1.174 by looking for the area for \(z=1.17\), which gives you 0.879 \(\color{red}{\textbf{(my mistake, sorry!)}}\). And then I'm also getting 0.81 as an answer... It's interesting that I mistakenly picked the "right" value for \(P(Z\le 1.174)\)... In any case, I think the answer you should be getting might be a typo. Then again, I'm comparing my steps to scribbled notes, so it's possible I've made a mistake somewhere along the line.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Shouldn't it be calculated as \[P(1.457\le Z \le0)+P(0 \le Z \le1.174)=P(0 \le Z \le 1.457)+P(0 \le Z \le 1.174)\]?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0That equation isn't true, though. Consider the graph of the curve: dw:1392424666942:dw This is the desired area.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0You're right in saying that this area: dw:1392424758459:dw is the same as this area: dw:1392424820652:dw This is true due to the symmetry of the curve about the yaxis. So yes, the probability, calculated that way, is the same as the probability I found. What I did was find the difference between these two areas: dw:1392424941634:dw dw:1392424971190:dw

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0"Continuity correction" is going a half unit lower in x at the low end and half unit higher at the high end? I guess this is to compensate for the inherently integer argument of the binomial and the continuous argument of the normal?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Yup, that's what it's used for

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0The cdf of a random variable X satisfies \[F(x)=1\frac{ 200^2 }{ (x+200)^2 }\] for x>0. Find P[50<X<300  X>100].
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