If students toss a coin 200 times each, about 68% should have proportions between what two numbers?

1 Answer
Nov 23, 2017

Using a Normal approximation to a binomial distribution as well as the "68-95-99.7 rule " results in the conclusion that, about 68% of the time that someone tosses a coin 200 times, the result will be a proportion of heads between about #0.465=46.5%# (93/200 heads) and #0.535=53.5%# (107/200 heads).

Keep in mind that this is just an approximation and we're not even specifying whether "between" means "inclusive" (including the endpoints) or "exclusive" (excluding the endpoints).


First, we are assuming that the coin is "fair " and tossed vigorously so that the results of the tosses are independent . Under these assumptions, we can say the probability of heads on any given toss is #p=0.5# and that the total number #X# of heads in #n=200# tosses will follow a binomial distribution.

The mean of this binomial distribution is #n*p=200*0.5=100# and its standard deviation is #sqrt(n*p*(1-p))=sqrt(50) approx 7.07107#. If we use a Normal curve to approximate this is distribution we would then use one with a mean of 100 and a standard deviation of about #7.07107 approx 7#.

The "68-95-99.7 rule " says that about 68% of the samples from this distribution will result in values within 1 standard deviation of the mean. In this case, about 68% will result in values between #100-7=93# and #100+7=107#. Since #93/200=0.465=46.5%# and #107/200=0.535=53.5%#, we reach the final conclusion.