Evaluation of an Improper Integral The cumulative normal distribution is an important formula in statistics (see Fig. 22.10): N(x) = ∫^-2-∞ 1/√2π e^-x^2/2 dx where x = (y - y)/sy is called the normalized standard deviate. It represents a change of variable to scale the normal distribution so that

Question

Evaluation of an Improper Integral

The cumulative normal distribution is an important formula in statistics (see Fig. 22.10):

[latex]N(x)=\int_{-\infty}^{x} \frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2} d x[/latex] (E22.6.1)

where [latex]x=(y-\bar{y}) / s_{y}[/latex] is called the normalized standard deviate. It represents a change of variable to scale the normal distribution so that it is centered on zero and the distance along the abscissa is measured in multiples of the standard deviation (Fig. 22.10b).

Equation (E22.6.1) represents the probability that an event will be less than x. For example, if x = 1, Eq. (E22.6.1) can be used to determine that the probability that an event will occur that is less than one standard deviation above the mean is N(1) = 0.8413. In other words, if 100 events occur, approximately 84 will be less than the mean plus one standard deviation. Because Eq. (E22.6.1) cannot be evaluated in a simple functional form, it is solved numerically and listed in statistical tables. Use Eq. (22.34)

[latex]\int_{-\infty}^{b} f(x) d x=\int_{-\infty}^{-A} f(x) d x+\int_{-A}^{b} f(x) d x[/latex] (22.34)

in conjunction with Simpson’s 1/3 rule and the extended midpoint rule to determine N(1) numerically.

Accepted Answer

Equation (E22.6.1) can be reexpressed in terms of Eq. (22.34) as

[latex]N(x)=\frac{1}{\sqrt{2 \pi}}\left(\int_{-\infty}^{-2} e^{-x^{2} / 2} d x+\int_{-2}^{1} e^{-x^{2} / 2} d x\right)[/latex]

The fi rst integral can be evaluated by applying Eq. (22.33)

[latex]\int_{a}^{b} f(x) d x=\int_{1 / b}^{1 / a} \frac{1}{t^{2}} f\left(\frac{1}{t}\right) d t[/latex]

to give

[latex]\int_{-\infty}^{-2} e^{-x^{2} / 2} d x=\int_{-1 / 2}^{0} \frac{1}{t^{2}} e^{-1 /\left(2 t^{2}\right)} d t[/latex]

Then the extended midpoint rule with h = 1/8 can be employed to estimate

[latex]\int_{-1 / 2}^{0} \frac{1}{t^{2}} e^{-1 /\left(2 t^{2}\right)} d t \cong \frac{1}{8}\left[f\left(x_{-7 / 16}\right)+f\left(x_{-5 / 16}\right)+f\left(x_{-3 / 16}\right)+f\left(x_{-1 / 16}\right)\right][/latex]

[latex]=\frac{1}{8}[0.3833+0.0612+0+0]=0.0556[/latex]

Simpson’s 1/3 rule with h = 0.5 can be used to estimate the second integral as

[latex]\int_{-2}^{1} e^{-x^{2} / 2} d x =[1-(-2)] \frac{0.1353+4(0.3247+0.8825+0.8825)+2(0.6065+1)+0.6065}{3(6)}[/latex] = 2.0523
Therefore, the fi nal result can be computed as

[latex]N(1) \cong \frac{1}{\sqrt{2 \pi}}(0.0556+2.0523)=0.8409[/latex]

which represents an error of [latex]\varepsilon_{t}=0.046[/latex] percent.

Question 22.6: Evaluation of an Improper Integral The cumulative normal dis...

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