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%0.465=46.5% (93/200 heads) and 0.535=53.5%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).

Explanation:

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=0.5p=0.5 and that the total number XX of heads in n=200n=200 tosses will follow a binomial distribution.

The mean of this binomial distribution is n*p=200*0.5=100np=2000.5=100 and its standard deviation is sqrt(n*p*(1-p))=sqrt(50) approx 7.07107np(1p)=507.07107. 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 77.071077.

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=931007=93 and 100+7=107100+7=107. Since 93/200=0.465=46.5%93200=0.465=46.5% and 107/200=0.535=53.5%107200=0.535=53.5%, we reach the final conclusion.