# Guessing Coin Toss Probability (In deperate need of help!)

• Oct 10th 2012, 09:24 AM
jab1023
Guessing Coin Toss Probability (In desperate need of help!)
Hi all,
I have a quick question about a stats class assignment. We did the classic coin toss experiement. We tossed 10 coins 60 times. The first part asks us to guess how many heads will come up: 11 guesses for 0-10. How would I go about guessing? Also, if any instructions can be given on how to type the info I need in on the ti83 plus, that would be a huge help as well. My instructor has been difficult to reach through email for assistance.

I know the resulting distribution graphs will be relatively normal, but we are asked to make all 11 guesses add up to one by percent, and I'm not sure how to go about that.

Thanks!
• Oct 10th 2012, 08:00 PM
chiro
Re: Guessing Coin Toss Probability (In deperate need of help!)
Hey jab1023.

For this, recall that the coin toss outcomes have a distribution that is Binomial(10,0.5) if we assume a fair coin (it may not be, but that's usually the assumption we make).

Now the average number of heads we get is 5 since 10*0.5 = 5.

Now lets say you want to figure out the number of outcomes given some probability. Suppose we want to look at a 90% confidence interval of how many different numbers of heads we would expect given this 90%.

If you use a normal approximation with sample mean x_bar and standard error se = SQRT(x_bar*(1-x_bar))/SQRT(n) and your interval will [x_bar - Z(a/2)*se, x_bar + z(a/2)*se] where z(a/2) refer to the z-values for the interval (two tailed). For 95% z(a/2) is roughly 1.95.

Now you take that and look at the interval it generates and you will get a lower and an upper end which will contain certain whole numbers corresponding to the number of successes: those numbers correspond to the events inside that interval.

So if you get say (3.776,6.224) for some a then the outcomes will include 3,4,5, and 6 successes for that confidence level.

You may also want to check out correction methods for using the normal approximation.