SOLUTION: Show that for 1 million flips of a balanced coin, the probability is at least 0.99 that the proportion of heads will fall between 0.495 and 0.505.
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-> SOLUTION: Show that for 1 million flips of a balanced coin, the probability is at least 0.99 that the proportion of heads will fall between 0.495 and 0.505.
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Question 897778: Show that for 1 million flips of a balanced coin, the probability is at least 0.99 that the proportion of heads will fall between 0.495 and 0.505. Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! n = 1 million = number of tosses.
p = .5 = probability of heads.
q = 1 - .5 = .5 = probability of tails.
with a proportion, the mean is equal to p and the standard deviation is equal to sqrt(p*q/n)
this makes the mean equal to .5 and the standard deviation equal to .0005
you have a low score of .495 and a high score of .505.
the z score is equal to (raw score minus mean) divided by standard deviation.
the low score will therefore have a z score of (.495 - .5) / .0005 which is equal to -10
the high score will therefore have a z score of (.505 - .5) / .0005 which is equal to +10.
for a two sided confidence interval of 99%, the critical alpha would be equal to (100% - 99%) / 200 = .005
this leads to a critical low z score of -2.57582... and a critical high z score of +2.57582...
since the z score of plus or minus 10 is greater than plus or minus 2.57582...., then the probability is at least .99 that the proportion of heads will fall between .495 and .505.
in fact, the confidence interval for a z score between plus or minus 10 is so high that it shows up as 1.
a probability of 1 means there is a 100 probability that the mean score of the sample of 1000000 tosses will be between .495 and .505.
the following graphs confirm the results.
the first graph shows you that the probability that the score will be be between .495 and .505 is equal to 1.
the second graph shows you that the score required to be within a 99% confidence interval is from .499 to .501.
since an interval between .495 and .505 is larger than an interval between .499 and .501, then the probability that a sample mean will be between that wider interval has to be greater.
the probability therefore has to be at least .99 and and is actually 1 or as close to 1 as can be measured.
