This CDF calculator computes the cumulative probability P(x1 <= X <= x2) for a given normal distribution and also offers features including inverse CDF calculations, and solving for the mean (μ) and standard deviation (σ). It can also create a plot as a function of any of the paramaters.
A financial analyst is evaluating the risk of a new loan product with a default rate normally distributed, based on past data showing a 5% average rate and 2% standard deviation. They aim to calculate the chance that the default rate stays within the acceptable 3% to 7% range, to assess the financial stability and risk of introducing this product, ensuring it aligns with the bank's risk tolerance.
| Value | Keystrokes | Display | Description |
|---|---|---|---|
| 3 | x1 | 5.00 | Stores the x1 value. |
| 7 | x2 | 7.00 | Stores the x2 value. |
| 5 | μ | 5.00 | Stores the μ (mean) value. |
| 2 | σ | 2.00 | Stores the sd (standard deviation) value. |
| P(x₁₂) | 0.683 | Calculates the cdf between x1 and x2. |
Normal Distribution CDF for 2 Values
3.00 : The x1 value.
7.00 : The x2 value.
5.00 : The mean of the distribution.
2.00 : The standard deviation of the distribution.
0.683 : The normal cumulative distribution function, two values P(x1≤X≤x2).
Result: 0.683
Building on the prior analysis, the analyst plans to modify the loan's risk profile so 80% of loans have a 3% to 7% default rate. With a 5% mean default rate, the focus is on finding a standard deviation adjustment to meet this target, ensuring 80% of loans stay within the acceptable risk range.
These keystrokes assume the values from example 1.
| Value | Keystrokes | Display | Description |
|---|---|---|---|
| .8 | P(x₁₂) | 0.800 | Stores the cdf value. |
| σ | 1.56 | Calculates the σ value. |
Normal Distribution CDF for 2 Values
Solving for σ, standard deviation of the distribution, with initial value (guess) of 2.00
3.00 : The x1 value.
7.00 : The x2 value.
5.00 : The mean of the distribution.
1.56 : The standard deviation of the distribution.
0.800 : The normal cumulative distribution function, two values P(x1≤X≤x2).
Result: 1.56
Reference: