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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 μ \$ p σ \$ p can show you the details. Notice that the number of trials (n) appears in the denominator of the formula for the standard deviation.

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