Question

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.

Answer 1

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

Answer 2

This link explains the continuity correction pretty well.

http://www.mathsrevision.net/advanced-level-maths-revision/statistics/normal-approximations

Answer 3

Thanks. Must run. Sorry.

Answer 4

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.

Answer 5

How'd you get .8461-.0735?

Answer 6

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.

Answer 7

Shouldn't it be calculated as

\[P(-1.457\le Z \le0)+P(0 \le Z \le1.174)=P(0 \le Z \le 1.457)+P(0 \le Z \le 1.174)\]?

Answer 8

That equation isn't true, though. Consider the graph of the curve:

This is the desired area.