Question

By using the table of the t distribution given in the back of this book, determine the value of the integral

\(\int\limits_{ - \infty }^{2.5} {\frac{{dx}}{{{{\left( {12 + {x^2}} \right)}^2}}}} \)

Short answer

The value of the integral is 0.03

Step-by-step solution

Given information

Using t-distribution at the back of the book to determine the integral of \(\int\limits_{ - \infty }^{2.5} {\frac{{dx}}{{{{\left( {12 + {x^2}} \right)}^2}}}} \)

This is needed to evaluate the integral using t distribution.

Calculation of the integral

The pdf of random variable X from t distribution with n degrees of freedom

\(f\left( x \right) = \frac{{\left( {\left( {\frac{{n + 1}}{2}} \right)} \right)}}{{{{\left( {n\pi } \right)}^{\frac{1}{2}}}\left( {\frac{n}{2}} \right)}}{\left( {1 + \frac{{{x^2}}}{n}} \right)^{ - \left( {\frac{{n + 1}}{2}} \right)}}\)

Now,

x\(\begin{align}\int\limits_{ - \infty }^{2.5} {\frac{{dx}}{{{{\left( {12 + {x^2}} \right)}^2}}}} &= \frac{1}{{144}}\int\limits_{ - \infty }^{2.5} {\frac{{dx}}{{{{\left( {12 + {x^2}} \right)}^2}}}} \\ &= \frac{1}{{72}}\int\limits_{ - \infty }^{1.25} {{{\left( {1 + \frac{{{y^2}}}{3}} \right)}^{ - 2}}} \end{align}\)

by Let,

\(Y = \frac{X}{2}\)

\(\begin{align}\int\limits_{ - \infty }^{2.5} {\frac{{dx}}{{{{\left( {12 + {x^2}} \right)}^2}}}} &= \frac{1}{{72}}\frac{{{{\left( {3\pi } \right)}^{\frac{1}{2}}}\left( {\frac{3}{2}} \right)}}{{\left( {\frac{{3 + 1}}{2}} \right)}}P\left( {X \le 1.25} \right)\\ &= \frac{1}{{72}}\sqrt {3\pi } \frac{{\sqrt \pi }}{2}0.85\\ &= 0.03\end{align}\)

Hence,

\(\int\limits_{ - \infty }^{2.5} {\frac{{dx}}{{{{\left( {12 + {x^2}} \right)}^2}}}} \)=0.03