We want to find $P(-1 < X < 1)$. We can use the CDF, $F(x) = \frac{1}{\pi} \arctan(x) + 0.5$.

$P(-1 < X < 1) = F(1) - F(-1)$

Evaluate $F(1)$:
$F(1) = \frac{1}{\pi} \arctan(1) + 0.5$
We know $\arctan(1) = \pi / 4$.
$F(1) = \frac{1}{\pi} (\frac{\pi}{4}) + 0.5 = 0.25 + 0.5 = 0.75$

Evaluate $F(-1)$:
$F(-1) = \frac{1}{\pi} \arctan(-1) + 0.5$
We know $\arctan(-1) = -\pi / 4$.
$F(-1) = \frac{1}{\pi} (-\frac{\pi}{4}) + 0.5 = -0.25 + 0.5 = 0.25$

Subtract them:
$P(-1 < X < 1) = 0.75 - 0.25 = 0.50$

Exactly 50% of the data falls between -1 and 1. (Notice how this compares to the Standard Normal, where 68% falls between -1 and 1. The Cauchy is much more spread out!)

For the Standard Cauchy distribution, we need to find $P(X > 10)$.
We use the complement of the CDF: $1 - F(10)$.

$F(10) = \frac{1}{\pi} \arctan(10) + 0.5$

First, we evaluate $\arctan(10)$. Using a calculator, $\arctan(10) \approx 1.471$ radians.

Now, calculate the CDF:
$F(10) = \frac{1}{\pi} (1.471) + 0.5 \approx 0.468 + 0.5 = 0.968$

So, $P(X > 10) = 1 - 0.968 = 0.032$, which is about 3.2%.

For the Standard Normal distribution:
We are given that $P(Z > 10) \approx 7.6 \times 10^{-24}$. This is incredibly, unimaginably small.

Comparison:
The Cauchy probability (3.2%) is roughly $4 \times 10^{21}$ times larger than the Normal probability! This stark contrast perfectly illustrates what 'heavy tails' mean: extreme outliers are dramatically more common in a Cauchy distribution than in a Normal one.