# For large value of n, and moderate values of the probabiliy of success p (roughly, 0.05 Leftarrow p Leftarrow 0.95), the binomial distribution can be

For large value of n, and moderate values of the probabiliy of success p (roughly, ), the binomial distribution can be approximated as a normal distribution with expectation mu = np and standard deviation
. Explain this approximation making use the Central Limit Theorem.
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Step 1
Alternatively prove that:
Theorem:
If x is a random variable with distribution B(n, p), then for sufficiently large n, the distribution of the
variablez
where
Proof:
It can be prove using Moment generating function for binomial distribution. It's given as,

where
Step 2
By the linear transformation properties of the moment generating function.

Taking the natural log of both sides, and then expanding the power series of ${e}^{\theta /\sigma }$
Then,
Since

If n is made sufficiently large $\sigma =\sqrt{npq}$ can be made large enough that for any fixed theta the absolute value of the sum above will be less than 1.
Let $t=p\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m!}{\left(\frac{\theta }{\sigma }\right)}^{m}$
Thus for sufficiently large
The ln term in the previous expression is and so we may expand this term as follows:

Step 3
This means that