$$ P ( x \geq \mu - \sigma ) $$

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To find $$ P(x \geq \mu - \sigma) $$, we need to understand the context in which this probability is being calculated. Typically, $$ \mu $$ represents the mean and $$ \sigma $$ represents the standard deviation of a probability distribution, often a normal distribution.

Assuming $$ x $$ follows a normal distribution $$ N(\mu, \sigma^2) $$, we can standardize the variable to use the standard normal distribution $$ Z $$:

$$
Z = \frac{x - \mu}{\sigma}
$$

We want to find:

$$
P(x \geq \mu - \sigma)
$$

First, we standardize $$ \mu - \sigma $$:

$$
P(x \geq \mu - \sigma) = P\left(Z \geq \frac{(\mu - \sigma) - \mu}{\sigma}\right) = P\left(Z \geq -1\right)
$$

Using the properties of the standard normal distribution, we know:

$$
P(Z \geq -1) = 1 - P(Z < -1)
$$

From standard normal distribution tables or using a calculator, we find:

$$
P(Z < -1) \approx 0.1587
$$

Thus,

$$
P(Z \geq -1) = 1 - 0.1587 = 0.8413
$$

Therefore, if $$ x $$ follows a normal distribution, we have:

$$
P(x \geq \mu - \sigma) \approx 0.8413
$$

This means that approximately 84.13% of the values in a normal distribution are greater than or equal to one standard deviation below the mean.
$$ P(x \geq \mu - \sigma) \approx 0.8413 $$

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