Let’s formalize our answer with some notation: We see here that for Z = 1.35, the probability is 0.9115 or 91.15%. 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.
Below is a typical cumulative Z-value lookup. 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. A standard normal variable has zero mean and variance of one (consequently its standard deviation is also one). The first step is to standardize the given value of 5.0 into a Z value (aka, Z score):Īll we’ve done here is translate a normal variable into a standard normal variable. Parsimony here refers to the normal conveniently has only two parameters, mean and variance. The normal distribution is rarely realistic, but it is popular for learning purposes due to its special properties and what is called parsimony. I hope you noticed the phrase “normally distributed?” It comes up often in exams. 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%? The same question could be re-phrased into the language of asset returns.
