# 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. Jun 20, 2015 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. Jun 20, 2015 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... Jun 20, 2015 at 19:14