I have already given the solution to
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 σ^{2} = Variance = 5n/36 , and μ = Mean = n/6.
P{9n/60<= M <= 11n/60} = 1/2 ( erf(n/60.√2piσ^{2}) - erf(-n/60.√2piσ^{2}) )