# Thread: Normal approximation of a die.

1. ## Normal approximation of a die.

Final the exact probability of rolling 2 or 3 sixes when rolling a normal six-sided die 12 times. What is the normal approximation for this situation?
Find the exact probability of rolling 14 - 17 sixes when rolling a normal six-sided die 100 times. What is the normal approximation for this situation? Compare the two cases.

2. These are binomial probabilities.
The p=1/6 and in the first case n=12 and has a lousy approximation via the central limit theorem since n is small.

The answer to the first one is

${12\choose 2} \biggl({1\over 6}\biggr)^2 \biggl({5\over 6}\biggr)^{10} +{12\choose 3} \biggl({1\over 6}\biggr)^3 \biggl({5\over 6}\biggr)^9$.

The second is

$\sum_{k=14}^{17}{100\choose k} \biggl({1\over 6}\biggr)^k \biggl({5\over 6}\biggr)^{100-k}$

where I am assuming that you're including the 14 and 17.
And instead of a normal die you mean a fair die.
You can approximate this via the CLT, the mean is 100/6 and the variance is 100(1/6)(5/6).
The normal superimposed over this binomial is $P(13.5.
Now subtract the mean and divide by the st deviation and use your st normal table.