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 11 months 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.
 11 months 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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douglaswinslowcooper
 11 months 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?

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

douglaswinslowcooper
 11 months ago
Best ResponseYou've already chosen the best response.0Thanks. Must run. Sorry.

SithsAndGiggles
 11 months ago
Best ResponseYou've already chosen the best response.1Probability 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.

sleung
 11 months ago
Best ResponseYou've already chosen the best response.0How'd you get .8461.0735?

SithsAndGiggles
 11 months ago
Best ResponseYou've already chosen the best response.1From 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.

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

SithsAndGiggles
 11 months ago
Best ResponseYou've already chosen the best response.1That equation isn't true, though. Consider the graph of the curve: dw:1392424666942:dw This is the desired area.

SithsAndGiggles
 11 months ago
Best ResponseYou've already chosen the best response.1You'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

douglaswinslowcooper
 11 months 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?

SithsAndGiggles
 11 months ago
Best ResponseYou've already chosen the best response.1Yup, that's what it's used for

sleung
 11 months 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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