Probability of 100 coin tosses I was wondering if I could get some help on this problem. 
Suppose that a fair coin is tossed 100 times. Find the probability of observing at least 60 heads. 
Thanks!
Note: on the study guide I did "1-binomcdf(100,0.50,59)" and still got it wrong
Also, this table is included. 
This is what it looks like. It seems I got it right, but my professor does not approve of my methods. Note the crossed out "-1/2" and "OK"

EDIT: Okay, so the answer is 0.0287. 
I got that by... 1-P(X* < 9.5/5)
=1-P(X* < 1.9)
Then look at the table to find 1.9. 
=1-(0.9713)
=0.0287
 A: For a solution that you can do off the top of your head, consider that for a sampling process with N draws and probability of success P, the standard deviation of the approximating normal distribution is:
$$\sigma = \sqrt{( N * P * (1-P) )}$$
for $N = 100$ and $P = 0.5, \sigma = 5$
Fortuitously, your question of 60 or more heads corresponds precisely to a z value of $2 * \sigma$ or more, which every stats student should know is about $2.5%$.  For greater precision, look it up on the table.
I can't speak for your professor, but if you were my student I would award 110% credit and the offer of a paid assistantship if you presented a quick, accurate answer that did not rely on the use of electronic computing.
A: At least $60$ is the counterpart of less than $60$. As we're dealing with an integer-valued random variable, less than $60$ is the same as less than or equal to $59$.
So $P(X \geq 60) = 1 - P(X \leq 59)$.
So we use the binomcdf(n,p,k) command, where $n$ is the number of trials, $p$ is the probability of success on a trial, and $k$ is the number of events.
So $P(X) \geq 60 = 1-($ binomcdf(100,0.5,59$)$.
Edit: so it looks like your work is correct. Ask your teacher if they're sure the textbook answer is right.
If you'd like more clarification on the matter, this looks like a good resource.
A: It looks like they wanted you to use the normal approximation to the binomial distribution.
Let $X \sim \operatorname{Binomial}_{n,p}$.  We can form the "standardized" random variable $X^*$ by taking 
$$X^* = \frac{X - \mu}{\sigma} = \frac{X - np}{\sqrt{npq}}$$
As it happens, if $n$ is large and $p$ is "small", then we can approximate $X^*$ by the standard normal random variable $Z$.  In particular,
$$ P(X^* < k) \approx P(Z < k)$$
Actually, since $X$ and $X^*$ are discrete, we would probably apply the continuity correction.  The point of the continuity correction is to account for the fact that $Z$ is continuous.  For example, suppose that $x = 1.3$ is a data point.  Presumably, that would be rounded down to $1$ when making a discrete model.  The continuity correction captures all the extras that should get rounded down.  We include them by doing $P(X^* < 1.5)$.
Okay, so how does the whole problem look?
$$ 1 - P(X < 59) = 1 - P\left(X^* < \frac{59 - (100)*0.5}{\sqrt{100*0.5*0.5}}\right)\\
 = 1 - P\left(X^* < \frac{9.5}{25}\right)\\
 \approx 1 - P\left(Z < 0.38\right)
 \approx 1 - P\left(Z < 0.5\right)$$
The last step is the continuity correction.
You can look up the probability that $Z < 0.5$ in that table. Apparently, I made some kind of arithmetic mistake, but that's the gist of it.  Try to "debug" it!
Actually, I think the problem is that I made the continuity correction at the wrong time.  You want to do it when standardizing, which would look like:
$$ 1 - P(X < 59) = 1 - P\left(X^* < \frac{59.5 - (100)*0.5}{\sqrt{100*0.5*0.5}}\right)$$
A: There are a lot of confusing answers here! As far as I can tell, your only mistake was to get the wrong value for $binomcdf(100,.50,59)$. It's not $0.284439664$. The correct value is $~0.971556$, which would have given you an answer of $0.028444$ to the question.
