Question

Probability, approximate the normal probability with an error of less than \(0.0001,\) where the probability is given by \(P(a< x < b)=\frac{1}{\sqrt{2 \pi}} \int_{a}^{b} e^{-x^{2} / 2} d x\). $$ P(1

Short answer

The probability \(P(1<x<2)\) for a standard normal distribution is approximately 0.1359.

Step-by-step solution

Understand the problem

The formula mentioned in the problem is the probability density function (pdf) for a normal distribution with mean 0 and standard deviation 1. To solve this problem, there's a need to evaluate the definite integral provided in the problem.

Application of normal distribution

The normal distribution integral can be translated to the error function (erf(x)), defined as \(\frac{2}{\sqrt{\pi}} \int_{0}^{x} e^{-t^{2}} dt.\) Thus, \(P(a<x<b)=0.5*[erf(b/\sqrt{2}) - erf(a/\sqrt{2})]\), where the 0.5 and the \(\sqrt{2}\) come from the necessary normalization to fit the error function to the normal distribution.

Evaluate the probability

Substitute the limits of 1 and 2 into the formula. This will give \(P(1<x<2)=0.5*[erf(2/\sqrt{2}) - erf(1/\sqrt{2})].\) Note that the error function values are usually looked up from statistical tables or calculated using statistical software. For this problem, use the approximation \(erf(2/\sqrt{2}) \approx 0.9545\) and \(erf(1/\sqrt{2}) \approx 0.6827\).

Calculate the result

The result will be obtained by subtracting these two terms, i.e., \(P(1<x<2) = 0.5*(0.9545 - 0.6827) = 0.1359.\)