Suppose you must cut a triangular piece off the comer of a square plate by measuring the distances x and y from the comer (see Figure 7.3-2). The desired value of x is 10 in., and the desired value of () is 20°. This requires that y = 3.64 in. We are told that measurements of x and y are normally distributed with means of 10 and 3.64, respectively, with a standard deviation equal to 0.05 in. Determine the standard deviation of () and plot the relative frequency histogram for ().

Solution

From Figure 7.3-2, we see that the angle ()is determined by Ø = tan-I (y [x). We can find the statistical distribution of () by creating random variables x and y that have means of 10 and 3.64, respectively, with a standard deviation of 0.05. The random variable () is then found by calculating () = tan-1 (y Ix) for each random pair (x, y). The following script file shows this procedure.

Figure 7.3-2 Dimensions of a triangular cut.

Figure 7.3-3 Scaled histogram of the angle ∅.

The choice of 8000 simulations was a compromise between accuracy and the amount of time required to do the calculations. You should try different values of n and compare the results. The results gave a mean of 19.9993° for () with a standard deviation of 0.2730°. The histogram is shown in Figure 7.3-3. Although the plot resembles the normal distribution, the values of ∅ are not distributed normally. From the histogram we can calculate that approximately 65 percent of the values of ∅ lie between 19.8 and 20.2. This range corresponds to a standard deviation of 0.2°, not 0.273° as calculated from the simulation data. Thus the curve is not a normal distribution.

This example shows that the interaction of two of more normally distributed variables does not produce a result that is normally distributed. In general, the result is normally distributed if and only if the result is a linear combination of the variables.