Standard Deviation and Conditional Probability
Date: 07/14/98 at 22:45:22 From: Angela C. Williams Subject: Quantitive stats I have a couple of questions: Find the standard deviation of the following sample: 5. My answer is the square root of five. Am I suppose to use some formula? Also, here are data on 246 stocks: Price Increase No Increase Total Dividends paid 34 78 112 No dividends paid 85 49 134 Total 119 127 246 Given that a stock increased in price, what is the probability that it also paid dividends? Thanks for your help.
Date: 10/22/98 at 21:52:24 From: Doctor Andrewg Subject: Re: Quantitive stats Hi Angela! Okay, first of all let's look at the standard deviation. Standard deviation is a measure of variability in the observations - a higher standard deviation means more variability, and a lower standard deviation means less variability. Here is a sample with high variability and one with low variability: High variability Low variability 2 15 8 16 14 14 13 12 26 15 13 14 3 11 Can you see what I mean? The first column's values are spread out much more than the second's. We need a way of quantifying (giving a number that measures) the variability of a sample of observations. This way we can talk about a sample with a standard deviation of 100, say, not just a large standard deviation, since in some cases 100 may be a small standard deviation as well, depending on the units we use to measure the observations. Imagine heights measured in meters and then measured in centimeters. Which one looks more variable? For a sample we define the standard deviation as the sum of the squares around the mean, divided by the number of observations. If n is the number of observations, x-bar (the x with a bar over the top) is the mean of the observations, and xi is the ith observation, then the standard deviation is: n ---\ 2 \ (x - x) / i ---/ i=1 ------------------------ n If you put the single number 5 into that formula, n = 1, x-bar = 5, and x1 = 5. The answer should be zero. This isn't very useful, but it is the only answer I can see here. So, if 5 is the only number in the sample then the standard deviation for that sample would be zero - there is no variability at all. If you were using the sample standard deviation to estimate the population standard deviation, then you wouldn't be able to do this with only one observation. The division by n in the formula above would be replaced with a division by (n-1) which is 1-1 = 0 in this case. The standard deviation would then be equal to 0/0 (which is what we call an indeterminate form). The second question is a question of conditional probability. What you are asking is the following: What is the probability that a stock paid dividends if it increased in price? We can write this as P(A|B) where A is "paid dividend," the "|" symbol can be read as "given that," and B is "stock increased in price." So in total that reads "Probability(paid dividends given that stock increased in price)." So what you are saying is: how many (as a proportion) of the stocks that increased in price also paid dividends? Can you see what I mean? We are not interested in companies whose stocks did not increase in price, only those whose prices did increase. So we can ignore the "no increase" and "total" columns and just look at the "price increase" column. There are 119 companies that had a stock price increase, and 34 of them paid dividends, so the probability that a company whose price increased also paid dividends is 34/119. Does this make sense to you? I'll finish this message now and let you read through my suggestions. If you have any more questions or if I'm not clear enough in my explanation, please ask again. Good luck with your work! - Doctor AndrewG, The Math Forum http://mathforum.org/dr.math/
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