# Dice probability normal distribution

You roll a dice 1000 times. Calculate the probability you roll a six between 150 and 200 times. I understand how you calculate this with the binomial distribution: $$= Binomialcdf(1000, 1/6, 200) - Binomialcdf(1000, 1/6, 149) = 0.9975-0.0710=0.9265$$

How do you do this with the normal distribution? The answer should be 92.53%. Problem is when I use Normalcdf(200) - Normalcdf(150) I get 0.9190 What am I doing wrong?

My second question: If you roll exactly 200 times a six, what is the probability there will be less than 150 times a five? Again I understand how I could calculate it with the binomial distribution, but not with the normal distribution... Answer should be 20.04%

Any suggestions?

• The answer based on the normal distribution is an approximation justified by central limit theorem (CLT). You need to properly normalize (i.e., standardize) you sum to use the CLT. – passerby51 Jun 20 '15 at 18:58
• I would cross my fingers and use the continuity correction, so find the probability a suitable normal is $\le 200.5$ minus the probability the normal is $\le 149.5$. That may give a better approximation. – André Nicolas Jun 20 '15 at 19:04
• Excellent suggestion! I didn't use the continuity correction for my first question. It seems that gives me the correct value. Thanks! On to the second one... – Seneca Jun 20 '15 at 19:14