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

1 Answer
Bill K.
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).

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=0.5# 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.

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