## sleung one year ago 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.

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1. douglaswinslowcooper

You are using a normal approximation to the binomial distribution presumably. What is the continuity correction?

2. sleung

This link explains the continuity correction pretty well. http://www.mathsrevision.net/advanced-level-maths-revision/statistics/normal-approximations

3. douglaswinslowcooper

Thanks. Must run. Sorry.

4. SithsAndGiggles

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.

5. sleung

How'd you get .8461-.0735?

6. SithsAndGiggles

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.

7. sleung

Shouldn't it be calculated as $P(-1.457\le Z \le0)+P(0 \le Z \le1.174)=P(0 \le Z \le 1.457)+P(0 \le Z \le 1.174)$?

8. SithsAndGiggles

That equation isn't true, though. Consider the graph of the curve: |dw:1392424666942:dw| This is the desired area.

9. SithsAndGiggles

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|

10. douglaswinslowcooper

"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?

11. SithsAndGiggles

Yup, that's what it's used for

12. sleung

The 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].