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Roll a die $n$ times and let $M$ be the number of times you roll $6$. Assume that $n$ is large.

(b) Write down an approximation, in terms on $n$ and $\phi$, of the probability that $M$ differs from its expectation by less than $10$ %
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Ans : 1/2 ( erf(n/60.√2piσ2) - erf(-n/60.√2piσ2) )

(Check out the solution below)

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# Probability- Gravner- 82.a

where EM came out to be n/6, and the Probability Mass Function of M was derived to be the Binomial Distribution.

Now, we know that Variance of a Binomial Distrbution is np(1-p).

Therefore Standard Deviation of Binomial Distribution = √(np(1-p))

The meaning of question asked is the following :

We have to find the probability that M is within 10% of the EM = n/6, which means 9n/60<= M <= 11n/60.

Now, lets take advantage of the assumption that n is very large, and hence Binomial Distribution can be approximated by Normal Distribution of mean np and Variance np(1-p).

Which means PDF of M = Normal{ n/6, 5n/36}

P{9n/60<= M <= 11n/60} = Definite Integral of Gaussian Distribution with mean = n/6 and variance = 5n/36 from limit a = 9n/60 to b = 11n/60.

The final answer can be written in terms of error-functions = erf(x) as follows :

Let σ2 = Variance = 5n/36 , and μ = Mean = n/6.

P{9n/60<= M <= 11n/60} = 1/2 ( erf(n/60.√2piσ2) - erf(-n/60.√2piσ2) )

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