There's a special form of the normal probability distribution which was crucial when calculators and computers were not available yet. The standard normal distribution, also called the z-distribution, but even today, it's still frequently use for quick calculations and to present analysis results. In this video, I'll explain its properties and application. While the values of 1, 2 and 3 sigma away from the mean are often useful values to be analyzed, there are many cases where you would need to calculate probabilities for, say, a value of 1.3 standard deviations away from the mean. To symbolize the fact that you'd like to list the probability values associated with any number of standard deviations away from the mean, the letter z has been chosen. The probability distribution of these z values is a normal distribution with a 0 mean, and a standard deviation of 1, also called the standard normal distribution, or z-distribution. The cumulative z-distribution is often represented by a table. This table gives the probability that the outcome of a normally distributed random variable will be lower than or equal to the point, mu plus z times sigma. This is an example of such a table showing z values together with the associated cumulative probabilities. As you see, it starts at the value -2.00, and increments with smalls steps. So this is going to be a long list before we get to cumulative probabilities near 1 as would be inefficient to print it in this way. That's why it's usually represented in a more compact form. With the first decimal of z along one margin and the second decimal along another margin. Using such a table you can, for a given value of z, quickly find the associated cumulative probability. For example, if you'd have to find the cumulative probability for a z value of 1.41, you would select a value of 1.4 in this margin and 0.01 in this, and find the corresponding probability of 0.92. But how do you get a z value if you're starting off with a normally distributed random variable? For that you need to consider this relation, a certain value x for the random variable X is a z standard deviations away from the mean. So if you would like to know the value of z based on the value of x, mu and sigma, it's a matter of rearranging the equation to the following form. This equation tells that the z-value equals the difference between the value of the random variable and the mean of the probability distribution divided by the standard deviation. Let's apply it to an example. A population of green-legged geese migrates every autumn from the Baltic Region to the Atlantic Coast in Europe. The migration duration is normally distributed with a mean of four and standard deviation of 1.3 days. Now, what would be the probability that the geese would complete their migration within six days to here. This is the formal way to state the problem, try to answer it. First, you need to z transform the value of siix days. You do that by subtracting the mean and dividing by the standard deviation. This gives a value of 1.54. Next, you should look up the z value in the table, and find the cumulative probability that matches with this value. As you see, this z value matches with the probability of 0.9382. And this is the answer to the question, the probability that the geese would complete their migration within six days in a given year. Let's make another calculation. Now you 'd like to know the probability that the migration duration would lay between two and five days. Could you calculate that probability as well? The formal problem statement is given here. This question needs to be answered in three parts. First the probability of values smaller than five needs to be calculated. Subsequently the probability of values smaller than two, and last these need to be subtracted to get the probability for the desired range between two and five days. This first probability for values smaller than five is 0.78. The second probability, for a value smaller than two is 0.06. So, the probability for the range from two til five is the difference, which is 0.72. And that's the answer to the question. Okay, let's take a breath. We've seen that you can turn any normally distributed variable into a standard normal, or z-distributed variable by subtracting its mean and dividing with its standard deviation. And with the help of a tabulated set distribution, you can find probabilities to encounter a value below, above, or between specific values of the random variable. So what, if you would have a probability in mind and would like to find a corresponding critical value of the random variable, no worries. It's almost the same procedure just backwards. Let's take the migration example once again to illustrate this inverse procedure. It takes the geese on average four days to migrate with a standard deviation of 1.3 days. You can then, for instance, find the 10th percentile of the migration duration. The 10th percentile of duration means that it takes the geese this amount of time or less to migrate in 10% of the cases, and it takes them more time in 90% of the cases. First you'd look up the probability value of 0.1 or the value closest to that. Next, you read off the corresponding set value. In this case, it is -1.28. Subsequently, it's a matter of applying this formula. You multiply the z value with the standard deviation of 1.3 and then you add the mean of four days. You end up with the value of 2.34 days as an answer to the question. As a last point, I would like to note that the z-transformation, which is subtracting a mean and dividing with the standard deviation, can be applied to any kind of numerical data. It results in a data set with mean 0 and standard deviation 1, and involves no assumptions about the underlying distribution of the data. It can be a good method to standardize data to make, for example, comparisons among different case studies. However, the z-transformation does not automatically create data which follows a z-distribution and allow you to make probability statements about your data. You will have z-distributed data only where you know or can assume that the random variable which generates your data is normally distributed. And you can estimate the mean and standard deviation for that normal variable well. Let me summarize what I explained in this video. A z-transformation can be applied to standardize any data to get a dataset with 0 mean and standard deviation of 1, regardless the underlying distribution of the original data. When that data is known to follow a normal distribution, probability statements can be made on the basis of the resulting z scores by using the table that lists cumulative probabilities with the corresponding z values. For a given value x you can find the corresponding z value by subtracting the mean and dividing by standard deviation. The z-table provides the cumulative probability matching with the z value. This is the probability of encountering a value for the random variable that is lower than or equal to x. Inversely, for a given probability, you can also find a z value in the table and calculate a data value x that matches with that cumulative probability.