SOLUTION: Use Tchebycheff's inequality to find how many times a fair coin must be tossed in order that the probability that the ratio of the number of heads to the number of tosses will lie

Algebra ->  Probability-and-statistics -> SOLUTION: Use Tchebycheff's inequality to find how many times a fair coin must be tossed in order that the probability that the ratio of the number of heads to the number of tosses will lie       Log On


   

Question 892720: Use Tchebycheff's inequality to find how many times a fair coin must be tossed in order that the probability that the ratio of the number of heads to the number of tosses will lie between 0.45 and 0.55 will be atleast 0.95
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Tchebycheff's inequality states that P%28abs%28X+-+mu%29%3C=k%2Asigma%29%3E=1-1%2Fk%5E2, where mu is the mean of the random variable and sigma is its standard deviation.
Incidentally, if X = ratio of number of heads to n tosses of a fair coin, then the mean mu+=+1%2F2, and variance is sigma%5E2+=+1%2F%284n%29. (Remember, E(X) = p, while Var%28X%29+=+%28pq%29%2Fn.)
==> standard deviation is sigma+=+1%2F%282%2Asqrt%28n%29%29
Now from the given P%280.45%3C=X%3C=0.55%29%3E=0.95
<==>P%28abs%28X+-+0.50%29%3C=0.05%29%3E=0.95
Now let 0.05+=+k%2Asigma, and 1+-+1%2Fk%5E2+=+0.95
The first equation is equivalent to k%2F%282%2Asqrt%28n%29%29+=+0.05, or sqrt%28n%29+=+10k.
==> n+=+100k%5E2
But the second equation gives k%5E2+=+20.
==> n = 100*20 = 2,000 tosses.