My stats has been self taught, but a lot of material I read point to a dataset having mean 0 and standard deviation of 1.

If that is the case then:

  1. Why is mean 0 and SD 1 a nice property to have?

  2. Why does a random variable drawn from this sample equal 0.5? The chance of drawing 0.001 is the same as 0.5 so this should be flat distribution...

  3. When people talk about Z Scores what do they actually mean here?

  1. At the beginning the most useful answer is probably that mean of 0 and sd of 1 are mathematically convenient. If you can work out the probabilities for a distribution with a mean of 0 and standard deviation of 1 you can work them out for any similar distribution of scores with a very simple equation.

  2. I'm not following this question. The mean of 0 and standard deviation of 1 usually applies to the standard normal distribution, often called the bell curve. The most likely value is the mean and it falls off as you get farther away. If you have a truly flat distribution then there is no value more likely than another. Your question here is poorly formed. Were you looking at questions about coin flips perhaps? Look up binomial distribution and central limit theorem.

  3. "mean here"? Where? The simple answer for z-scores is that they are your scores scaled as if your mean were 0 and standard deviation were 1. Another way of thinking about it is that it takes an individual score as the number of standard deviations that score is from the mean. The equation is calculating the (score - mean) / standard deviation. The reasons you'd do that are quite varied but one is that in intro statistics courses you have tables of probabilities for different z-scores (see answer 1).

If you looked up z-score first, even in wikipedia, you would have gotten pretty good answers.

  • $\begingroup$ On 2) I believe the confusion is what p(X=.01) means when X is a continuous random variable. Intuitively, the probability seem to be zero everywhere because there is no chance X is exactly .01. The questioner should review the definition of a density function in the continuous case, which is defined as the derivative of the cumulative density function. $\endgroup$
    – Tristan
    Jul 31 '10 at 17:15

To start with what we're talking about here is the standard normal distribution, a normal distribution with a mean of 0 and a standard deviation of 1. The short-hand for a variable which is distributed as a standard normal distribution is Z.

Here are my answers to your questions.

(1) I think there are two key reasons why standard normal distributions are attractive. Firstly, any normally distributed variable can be converted or transformed to a standard normal by subtracting its mean from each observation before dividing each observation by the standard deviation. This is called the Z-transformation or the creation of Z-scores. This is very handy especially in the days before computers.

If you wanted to find out the probability of some event from your variable which is normally distributed with mean 65.6 with a standard deviation of 10.2 wouldn't that be a right pain in the backside without a computer? Let's say that this variable is the heights in inches of American women. And let's say that we're interested in finding out the probability that a woman randomly drawn from the population will be very tall - say over 75 inches tall. Well this is a bit of a pain to find out with a computer as I would have to carry around a table for every possible normal distribution with me. However, if I transform this to a Z-score I can use the one table to find out the probability, thus: $$ \begin{aligned} \frac{(x_i - \bar x)}{\sigma_x} &= Z \\ \frac{(75 - 65.6)}{10.2} &= 0.9215 \end{aligned} $$ Using the Z table I find that the cumulative probability P(z < Z) - 0.8212 and therefore the probability of finding a woman as tall or taller than 75 inches is 17.88%. We can do this with any normally distributed variable and so this standard normal distribution is very handy.

The second reason why the standard normal distribution is used frequently is due to the interpretation is provides in terms of Z-scores. Each "observation" in a Z-transformed variable is how many standard deviations the original untransformed observation was from the mean. This is particularly handy for standardized tests where the raw or absolute performance is less important than the relative performance.

(2) I don't follow you here. I think you may be confused as to what we mean by a cumulative distribution function. Note that the expected value of a standard normal distribution is 0, and this value corresponds to the value of .5 on the associated cumulative distribution function.

(3) Z-scores are the individual "observations" or datum in a variable which has been Z-transformed. Return to my example of the variable - height of American women in inches. One particular observation of which may be a tall woman of height 75 inches. The Z-score for this is the result of Z-transforming the variable as we did earlier: $$ \begin{aligned} \frac{(x_i - \bar x)}{\sigma_x} &= Z \\ \frac{(75 - 65.6)}{10.2} &= 0.9215 \end{aligned} $$ The Z-score in this case is 0.9215. The interpretation of the Z-score is that this particular woman is 0.9215 standard deviations taller than the mean height. A person who was 55.4 inches tall have a Z-score of 1 and would be 1 standard deviation below mean height.


Since you received excellent explanations from Graham and John, I'm just going to answer your last question:

When people talk about Z Scores what do they actually mean here?

Best way to answer this is to think about this question: The grades in class CS 101 is normally distributed with $\mu$ = 80 and $\sigma$ = 5. What is the z-score for the grade 65?

So: (65-80)/5=-3

You can say the z-score for the grade 65 is -3; or in other words 3 standard deviation to the left.


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