The_Law_of_Large_Numbers[1]

The_Law_of_Large_Numbers[1] - The Law of Large Numbers The...

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The Law of Large Numbers The nut does not fall far from the tree - but in fact, how far away does it fall? Suppose I roll a die 100 times, and find the percent of times I rolled a 4. I expect about of the rolls to be 4s. 1 6 Of course, in any particular series of 100 rolls, I don’t expect the percent to be exactly . (In fact, it can’t be, because of 100 rolls is not a whole number.) But 1 6 1 6 how far away will my percent be? To answer this question we calculate the standard deviation of the percent. But first, some technical details: The jargon: Repeating some action such as rolling a die and counting the number of times you get a particular outcome is called “Bernouilli trials.” (The trials must be independent.) The outcome you are counting is called a “success” - even if you are counting failures! The number of successes is a random variable called the “Binomial random variable.” The notation: the number of trials (rolls) is n the chance of a “success” on any one trial (roll) is p the actual percent of “successes” in an actual series of trials (rolls) is $ p Then is a random variable (It will come out differently for different actual series.) $ p ( 29 μ σ $ $ p p p p p n = = - 1 The formulas for and have been found using algebra. If you are interested, I $ p $ p can show you the details. Notice that the number of trials (n) appears in the denominator of the formula for the standard deviation. That means that as the number of trials increases, the actual percent of successes is likely to get closer to the long-run percent, p. That is the Law
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The_Law_of_Large_Numbers[1] - The Law of Large Numbers The...

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