## What Is a Z Table?

Functions based on the normal distribution are easy to retrieve in code or excel, so we do not really need z tables anymore, in practice. But we still want to understand the z table. Why? Because the popular exam calculators (TI BA II+ and HP 12c) do not include z table functionality, so we do need to use them to lookup values on the exam (yes, the z table has been provided in recent FRM^{®} exams). But understanding the z table also helps reinforce a basic grasp of random variables. Let’s start with a simple example question which is just my variation on an old FRM exam question:

Assume a random normal variable follows a normal distribution with a mean of 2.30 and a standard deviation of 2.00. What is the probability that this random variable is greater than 5.0?

The same question could be re-phrased into the language of asset returns. Here is same question re-phrased: If an asset’s daily return is normally distributed with mean of 2.30% and daily volatility of 2.00%, what is the probability the asset’s return will be at least 5.0%?

I hope you noticed the phrase “normally distributed?” It comes up often in exams. The normal distribution is rarely realistic, but it is popular for learning purposes due to its special properties and what is called parsimony. Parsimony here refers to the normal conveniently has only two parameters, mean and variance. The first step is to standardize the given value of 5.0 into a Z value (aka, Z score):

Z = (5.0 – 2.3)/2.0 = 1.350.

All we’ve done here is translate a normal variable into a standard normal variable. A standard normal variable has zero mean and variance of one (consequently its standard deviation is also one). The Z value of 1.350 means “The value of 5.0 is 1.350 standard deviations above the mean of 2.30.” Now we can use the common Z table to retrieve the associated probability. Below is a typical cumulative Z-value lookup. Because our Z-value is 1.35, we want to go down the rows until we arrive at 1.3, then we want to go across the columns until we arrive at 0.05. That’s because 1.35 = 1.30 + 0.05. We see here that for Z = 1.35, the probability is 0.9115 or 91.15%. Let’s formalize our answer with some notation:

Pr[X ≤ 5.0 | µ(X) = 2.3 and σ(X) = 2.0] = Pr(Z ≤ 1.350) = 91.15%.

Please notice the shift from X to Z. The first function says “The probability that X is less than or equal to 5.0 conditional on a mean of X equal to 2.3 and standard deviation of X equal to 2.0.” The second function, Pr(Z ≤ 1.350), reflects the normalization (translation) from the normal X to the standard normal Z, and we don’t need to specify the mean or standard deviation of the Z.

The same question could be re-phrased into the language of asset returns. Here is same question re-phrased: If an asset’s daily return is normally distributed with mean of 2.30% and daily volatility of 2.00%, what is the probability the asset’s return will be at least 5.0%?

I hope you noticed the phrase “normally distributed?” It comes up often in exams. The normal distribution is rarely realistic, but it is popular for learning purposes due to its special properties and what is called parsimony. Parsimony here refers to the normal conveniently has only two parameters, mean and variance. The first step is to standardize the given value of 5.0 into a Z value (aka, Z score):

Z = (5.0 – 2.3)/2.0 = 1.350

All we’ve done here is translate a normal variable into a standard normal variable. A standard normal variable has zero mean and variance of one (consequently its standard deviation is also one). The Z value of 1.350 means “The value of 5.0 is 1.350 standard deviations above the mean of 2.30.” Now we can use the common Z table to retrieve the associated probability. Below is a typical cumulative Z-value lookup. Because our Z-value is 1.35, we want to go down the rows until we arrive at 1.3, then we want to go across the columns until we arrive at 0.05. That’s because 1.35 = 1.30 + 0.05. We see here that for Z = 1.35, the probability is 0.9115 or 91.15%. Let’s formalize our answer with some notation:

Pr[X ≤ 5.0 | µ(X) = 2.3 and σ(X) = 2.0] = Pr(Z ≤ 1.350) = 91.15%.

Please notice the shift from X to Z. The first function says “The probability that X is less than or equal to 5.0 conditional on a mean of X equal to 2.3 and standard deviation of X equal to 2.0.” The second function, Pr(Z ≤ 1.350), reflects the normalization (translation) from the normal X to the standard normal Z, and we don’t need to specify the mean or standard deviation of the Z.