ERF

Calculates the error function (Gauss error function).

Syntax:

ERF(number1, number2)

if number2 is omitted, returns the error function calculated between 0 and number1, otherwise returns the error function calculated between number1 and number2. The error function, also known as the Gauss error function, is defined for ERF(x) as:

$\frac{2}{π} \int_{0}^{x} e^{- t^{2}} d t$

$ERF (x_{1}; x_{2}) i s ERF (x_{2}) - ERF (x_{1})$

Example:

ERF(0.5)

returns 0.520499877813.

ERF(0.2; 0.5)

returns 0.297797288603.

Application:

An application of the error function (erf) is in probability and statistics, specifically when dealing with a normal distribution. The erf function is closely related to the cumulative distribution function (CDF) of the standard normal distribution.

Let's consider a scenario involving the manufacturing of a product, for example, the diameter of ball bearings.

Scenario: Quality Control of Ball Bearings

A company manufactures ball bearings with a target diameter of 10.0 mm. Due to slight variations in the manufacturing process, the actual diameters follow a normal distribution with a mean (μ) of 10.0 mm and a standard deviation (σ) of 0.05 mm. The company wants to determine the percentage of ball bearings that fall within a certain tolerance range, say between 9.95 mm and 10.05 mm.

The probability that a randomly selected ball bearing has a diameter between x1 and x2 is given by the integral of the probability density function (PDF) of the normal distribution:

$P (x_{1} \leq X \leq x_{2}) = \int_{x_{1}}^{x_{2}} \frac{1}{σ 2 π} e^{- \frac{( x - μ ) ^{2}}{2 σ ^{2}}} d x$

This integral is difficult to solve directly. However, it can be expressed in terms of the error function, erf(z), which is defined as:

$erf (z) = \frac{2}{π} \int_{0}^{z} e^{- t^{2}} d t$

The probability can be calculated using the following relationship:

$P (x_{1} \leq X \leq x_{2}) = \frac{1}{2} [erf (\frac{x _{2} - μ}{σ 2}) - erf (\frac{x _{1} - μ}{σ 2})]$

Applying the Formula

For our example, we have:

μ=10.0 mm
σ=0.05 mm
Tolerance range: x1=9.95 mm and x2=10.05 mm

First, let's calculate the values for the erf function arguments:

For x2=10.05:

$z_{2} = \frac{10.05 - 10.0}{0.05 2} = \frac{0.05}{0.05 2} = \frac{1}{2} \approx 0.7071$

For x1=9.95:

$z_{1} = \frac{9.95 - 10.0}{0.05 2} = \frac{- 0.05}{0.05 2} = - \frac{1}{2} \approx - 0.7071$

Now, we use a table or a calculator for the erf function to find the values for erf(0.7071) and erf(−0.7071).

	z	ERF(z)
	A	B
1	0	0
2	0.1	0.112462916
3	0.2	0.222702589
4	0.3	0.328626759
5	0.4	0.428392355
6	0.5	0.520499878
7	0.6	0.603856091
8	0.7	0.677801194
9	0.7071	0.682684851
10	0.8	0.742100965

Since the error function is an odd function, erf(−z)=−erf(z):

erf(0.7071)≈0.6827
erf(−0.7071)=−erf(0.7071)≈−0.6827

Now, we can calculate the probability:

$P (9.95 \leq X \leq 10.05) = \frac{1}{2} [erf (0.7071) - erf (- 0.7071)]$

$P (9.95 \leq X \leq 10.05) = \frac{1}{2} [0.6827 - (- 0.6827)]$

$P (9.95 \leq X \leq 10.05) = \frac{1}{2} [1.3654]$

$P (9.95 \leq X \leq 10.05) = 0.6827$

Conclusion

The probability that a randomly selected ball bearing has a diameter between 9.95 mm and 10.05 mm is approximately 0.6827, or 68.27%. This means that roughly 68.27% of the manufactured ball bearings will fall within the company's specified tolerance range. The erf function provided a direct way to calculate this percentage, which is a critical piece of information for quality control and process management.

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