Normal Approximation to the Binomial Distribution Assignment Help

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Normal Approximation to the Binomial Distribution:

The Binomial Probability

P(x1 ≤ X ≤ x2 ) =  647_Normal Approximation to the Binomial Distribution.pngpx q n-x

where xl, x2 are any two non-negative integers, represents the probability in that the binomial variable X suppose an integral value inside the interval [ x1, x2 ], where X is based on n Bernoulli's trials along with probability of success p. For given n and p, the probability could, no doubt, be computed using the summation in Equation. Therefore, this computation becomes numerically too complicated if n is large. The random variable X could be written as'

X=X1+X2...+Xn

where XI, X2, ..., Xn, are iid Bernoulli's variables every supposing the value 1 along with probability p or the value 0 with probability q ( - 1 -p ).

Additionally, E(X) = np and V(X) = npq and so

E(x¯) - p and v(x¯) = pq/n where x¯  = X/n

Thus, the central limit theorem gives for large n,

731_Normal Approximation to the Binomial Distribution1.png

In other words, X/n approximately follows N (p, pq/n) since the inequality X ≤ x is equivalent to

709_Normal Approximation to the Binomial Distribution2.png

we have the cdf of X

F(x) = P(X ≤ x) ∼Φ (x-pq/√(npq))

Thus the binomial probability in equation, can be written as

P(x1, ≤ X ≤ x2) = F(x2)-(x1)

∼ Φ (x2 -np/√npq) - Φ (x1-np/√npq)

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