Levey Jennings just uses the long term sigma to compute control limits, so you can just take the standard deviation of the entire column of the response you are plotting. Here's an example using the Vial Fill Weights example from the Sample Data Library.

```
dt = Data Table("Vial Fill Weights");
y = :Fill Weight << Get Values;
y_bar = mean(y);
sigma = std dev(y);
UCL = y_bar + 3*sigma;
LCL = y_bar - 3*sigma;
```

This returns 5.90482338181099 and 6.32439884041124 for LCL and UCL respectively. This matches the Levey Jennings limits:

-- Cameron Willden