Answer
${\bf 50\%}$
Work Step by Step
The given wave function is:
$$ \psi(x) = \sqrt{\frac{b}{\pi(x^2 + b^2)}} $$
where $ b $ is a positive constant.
We are asked to find the probability that the particle is located in the interval $ -b \leq x \leq b $. The probability $ P $ that the particle is located in this interval is given by the integral of $ |\psi(x)|^2 $ over the range from $ -b $ to $ b $.
Now we need to find if the wave function is normalized or not,
$$\int_{-\infty}^{\infty} |\psi(x)|^2 dx = \int_{-\infty}^{\infty} \frac{b}{\pi(x^2 + b^2)} dx$$
$$\int_{-\infty}^{\infty} |\psi(x)|^2 dx = \frac{b}{\pi} \int_{-\infty}^{\infty} \frac{1}{x^2 + b^2} dx = \frac{b}{\pi} \left[ \frac{1}{b} \tan^{-1} \frac{x}{b} \right]_{-\infty}^{\infty}$$
$$\int_{-\infty}^{\infty} |\psi(x)|^2 dx = \frac{1}{\pi} \left[ \frac{\pi}{2} - \left( -\frac{\pi}{2} \right) \right] = 1 $$
Now, we can proceed with integrating $ |\psi(x)|^2 $ over the interval $ -b \leq x \leq b $.
$$ \text{Prob}( -b \leq x\leq b) = \int_{-b}^{b} |\psi(x)|^2 \; dx\tag 1 $$
where $
|\psi(x)|^2 = \left( \sqrt{\dfrac{b}{\pi(x^2 + b^2)}} \right)^2 = \dfrac{b}{\pi(x^2 + b^2)}$
Plug into (1);
$$ \text{Prob}( -b \leq x\leq b) = \int_{-b}^{b} \dfrac{b}{\pi(x^2 + b^2)}\; dx $$
$$ \text{Prob}( -b \leq x\leq b) = \frac{1}{\pi} \tan^{-1}\left(\frac{x}{b}\right)\bigg|_{-b}^{b} $$
$$ \text{Prob}( -b \leq x\leq b) = \frac{1}{\pi} \left[ \tan^{-1}\left( \frac{b}{b} \right) - \tan^{-1}\left( \frac{-b}{b} \right) \right]
$$
$$ \text{Prob}( -b \leq x\leq b) = \frac{1}{\pi} \left[ \tan^{-1}(1) - \tan^{-1}(-1) \right] $$
Recalling the fact that $ \tan^{-1}(1) = \dfrac{\pi}{4} $ and $ \tan^{-1}(-1) = -\dfrac{\pi}{4} $.
$$ \text{Prob}( -b \leq x\leq b) = \frac{1}{\pi} \left( \frac{\pi}{4} - \left( -\frac{\pi}{4} \right) \right) = \frac{1}{\pi} \times \frac{\pi}{2}$$
$$ \text{Prob}( -b \leq x\leq b)= \frac{1}{2} =\color{red}{\bf 50\%}$$
