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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 values inside the Z table are probabilities, so they must lie between 0% and 100% inclusive. What does our 91.15% mean? It is the area under the curve to the left of Z(1.35), see graph on the left: Put another way, and still referring to the plot on the left above, the probability of a standard random normal variable (again, that’s a normal variable with mean of zero and unit standard deviation) resulting in 1.35 or less is about 91.15%, which is the area under the curve to the left of Z(1.35). The total area must be 100% because this is a probability distribution. In terms of our original question then, the probability of our random variable returning less than (or equal to) 5.0 is 91.15% and, consequently, the probability of returning greater than 5.0 is 8.85% = 100% – 91.15%. If we use excel, we get this answer with 8.85% = 1 – NORM.S.DIST(1.35, TRUE) or 8.85% – 1 = NORM.DIST(1.35, 0, 1, TRUE). The later function simply makes explicit the zero mean and unit standard deviation.

The plot on the right above gives the area under the curve that is between zero and 1.35; some Z tables employ this format instead. Instead of a cumulative probability, the table returns the probability that the standard random normal variable will lie between zero and the critical value. The difference must be 0.50 because half of the area is negative under the curve is negative; in our example, 0.4915 is exactly 0.50 less than 0.9915. How can we quickly identify the difference? We can look at the first entry in the table. For Z = 0, it must be either zero or 0.50.

Now let’s alter the question a bit in order to test our understanding. This is more like an exam question because we need to think a little:

Assume a random normal variable follows a normal distribution with a mean of 2.30 and a standard deviation of 2.0. What is the probability that this random variable falls between zero and 5.0?

This must be less that 91.15% because 91.15% includes all the outcomes below zero. We need to standardize the zero outcome: Z = (0 – 2.3)/2 = -1.150. So we are looking for Pr(-1.150 < Z ≤ 1.35). We already know the probability that Z is less than 1.35 is 91.15%. How do we retrieve Pr(Z ≤ -1.15) from the Z table above which only includes positive Z values? We rely on the natural symmetry of the (standard) normal distribution:

Pr(Z ≤ -1.15) = 1 – Pr(Z ≤ 1.15) = 1 – 0.8749 = 0.1251 or 12.51%; please make sure you understand this step!

We rely on the fact that due to the distribution’s symmetry, 12.51% is both Pr(Z < -1.15) and Pr(Z > 1.15). And we get Pr(Z > 1.15) by subtracting Pr(Z ≤ 1.15) from 100%; Z = +1.15 is a value we can lookup on the table. Now we can answer the full question:

Pr(0 ≤ X ≤ 5.0 | µ = 2.3 and σ = 2.0) = Pr(Z ≤ 1.350) – Pr(Z ≤ -1.15) = 0.9115 – 0.1251 = 78.6%.

Graphically, our probability region excludes the left tail of 12.51% and the right tail of 8.85% which is a total of 21.36% excluded and an included region of 78.64% = 100.0% – 21.36%. I hope that helps introduce you to the Z table!

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