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/