Approximation to Binomial Probabilities. When you toss a coin 2n times, the expected number of heads is n. Given the probability density function for a standard normal random variable phi (x)=1/(sqrt{2pi}) e^{(-x^2)/2}, approximate the probability that you observe n+j heads when you toss the coin 2n times using the density function of a standard random variable phi and the standard deviation sigma.

cochetezgh 2022-09-11 Answered
When you toss a coin 2n times, the expected number of heads is n. Given the probability density function for a standard normal random variable ϕ ( x ) = 1 2 π e x 2 2 , approximate the probability that you observe n + j heads when you toss the coin 2n times using the density function of a standard random variable ϕ and the standard deviation σ. Hint: use Stirling's formula: n ! 2 π n n + 1 2 e n .
I know that the exact probability that I get n + j heads is ( 2 n n + j ) 1 2 n and I tried to use Stirling's formula on the binomial but it just made it more complicated.
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Answers (1)

Savanah Morton
Answered 2022-09-12 Author has 15 answers
Step 1
Let X be the number of heads in 2n tosses of a fair coin. Then X has binomial(2n,1/2) distribution, so E ( X ) = n and V ( X ) = n / 2. To calculate P ( X = n + j ), use the continuity correction to express P ( X = n + j ) as
(1) P ( X = n + j ) = P ( n + j 1 2 X n + j + 1 2 ) .
Step 2
The RHS of (1) is algebraically equal to
(2) P ( j 1 2 n / 2 X n n / 2 j + 1 2 n / 2 ) .
Step 2
We recognize that the middle quantity in (2) has approximately standard normal distribution (by the normal approx to the binomial, i.e., the de Moivre-Laplace theorem). The desired prob is therefore approximately the area under the standard normal density between the two endpoints L := j 1 2 n / 2 and R := j + 1 2 n / 2 . Approximate this area by a rectangle. The rectangle will have height ϕ ( z ), where z = j n / 2 is the midpoint of L and R. The width of the rectangle will be the difference between the two endpoints L and R.

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