anonymous
  • anonymous
A fair 6-sided 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.
The Normal Distribution
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
You are using a normal approximation to the binomial distribution presumably. What is the continuity correction?
anonymous
  • anonymous
This link explains the continuity correction pretty well. http://www.mathsrevision.net/advanced-level-maths-revision/statistics/normal-approximations
anonymous
  • anonymous
Thanks. Must run. Sorry.

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anonymous
  • anonymous
Probability 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 \(\sigma=\sqrt{np(1-p)}=11.785\). \[P\left(\frac{149.5-166.67}{11.785}\le Z\le\frac{180.5-166.67}{11.785}\right)\\ P\left(-1.457\le Z\le1.174\right)\] and I get \(0.8461-0.0735=0.7726\) by my calculations, which is close enough to the real answer.
anonymous
  • anonymous
How'd you get .8461-.0735?
anonymous
  • anonymous
From a table of \(z\)-values. \[P(-1.457\le Z\le1.174)=P(Z\le1.174)-P(Z\le-1.457)\] Referring to this table: http://facstaff.unca.edu/dohse/Stat225/Images/Table-z.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
  • anonymous
Shouldn'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
  • anonymous
That equation isn't true, though. Consider the graph of the curve: |dw:1392424666942:dw| This is the desired area.
anonymous
  • anonymous
You'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 y-axis. 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
  • anonymous
"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
  • anonymous
Yup, that's what it's used for
anonymous
  • anonymous
The cdf of a random variable X satisfies \[F(x)=1-\frac{ 200^2 }{ (x+200)^2 }\] for x>0. Find P[50100].

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