# find the normal approximation for the binomial probability that x=4 wher n=12 p=0.7 compare this probability to the value of p(x=4) please show work

If X ~ Binomial(n,p) where X is the number of successes out of n trials

then a Normal approximation to the distribution of X is

`X` ~ Normal`(mu,sigma^2)`  where `mu = np` and `sigma^2 = np(1-p)`

Given that the pdf of the Normal distribution is

`f(x) = 1/sqrt(2pisigma^2)exp(-(x-mu)^2/(2sigma^2))`

The density...

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If X ~ Binomial(n,p) where X is the number of successes out of n trials

then a Normal approximation to the distribution of X is

`X` ~ Normal`(mu,sigma^2)`  where `mu = np` and `sigma^2 = np(1-p)`

Given that the pdf of the Normal distribution is

`f(x) = 1/sqrt(2pisigma^2)exp(-(x-mu)^2/(2sigma^2))`

The density at f(x = 4) under the normal distribution assumption is

`f(4) = 1/sqrt(2pisigma^2)exp(-(4-mu)^2/(2sigma^2)) = 1/sqrt(2pi*12*0.7*0.3)exp(-(4-12*0.7)^2/(2*12*0.7*0.3))`

` = 0.00539`

However, since the Normal distribution is continuous, to get the probability that X= 4 we integrate in a region around x = 4 since the probability that X equals precisely 4 is zero.

For example, `Pr( 3.4 < X < 4.6) = phi(4.6) - phi(3.4) = 0.00752`

where `phi(x)` is the cdf of the Normal distribution.

Under the binomial distribution X ~ Binomial(n,p) the pdf is

`Pr(X=x) = (n!)/(x!(n-x)!) p^x(1-p)^(n-x)`

so `Pr(X=4) = (12!)/(4!8!)0.7^12 0.3^8 = 0.00780`

So we have a Normal approximation to the probability X = 4 of 0.00752  and a Binomial probability that X = 4 of 0.00780

As n gets larger and p is closer to 1/2 the approximation is better.

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