Help finding value of N that minimizes a sum Suppose we have the following inequality: 
$\sum\limits_{k=N+1}^{1000}\binom{1000}{k}(\frac{1}{2})^{k}(\frac{1}{2})^{1000-k} = \frac{1}{2^{1000}}\sum\limits_{k=N+1}^{1000}\binom{1000}{k} < \frac{1}{100}$
Is there a program that can calculate the smallest value of N that makes the above true? This is related to probability and the binomial distribution with $X \sim B({1000},{0.5}$).
 A: The central limit theorem says that $X$ binomial $(1000,.5)$ is $X=500+5\sqrt{10}Z$ with $Z$ approximately standard normal. Using the erffunction, this suggests that $N$ is around $536.7$. 
WA indicates $0.01046$ for $N=536$ and $0.00833$ for $N=537$ hence the solution is $N=\color{red}{537}$.

sum of binom(1000,k)/2^(1000) from k=537 to k=1000
Decimal approximation:
0.0104635553030424713317175500708088450031910719508853

sum of binom(1000,k)/2^(1000) from k=538 to k=1000
Decimal approximation: 
0.0088311156677493199532392344343609404372379787296116
A: The sum approximates the Normal distribution with mean of $500$ and variance of $250$. Using the cumulative normal distribution, $1\%$ is higher than $2.326$ standard deviations above the mean. That would be about $500+2.326\sqrt{250}=536.7$. Since this is greater than $536.5$, we should guess that $N=537$ would satisfy the given condition.
Plugging in $N=536$ into the binomial sum, we get $0.01046$, and for $N=537$, we get $0.00883$. So our guess was correct; we need $N=537$.

To approximate
$$
\frac1{2^{1000}}\sum_{k=537}^{1000}\binom{1000}{k}
$$
we compute the amount of the cumulative normal distribution higher than $\frac{536.5-500}{\sqrt{250}}=2.30846$ standard deviations above the mean, which is $0.01049$.
To approximate
$$
\frac1{2^{1000}}\sum_{k=538}^{1000}\binom{1000}{k}
$$
we compute the amount of the cumulative normal distribution higher than $\frac{537.5-500}{\sqrt{250}}=2.37171$ standard deviations above the mean, which is $0.00885$.
