Question

for a standard normal distribution, which of the following expressions must always be equal to 1?
○ $p(z \leq -a) - p(-a \leq z \leq a) - p(z \geq a)$
○ $p(z \leq -a) - p(-a \leq z \leq a) + p(z \geq a)$
○ $p(z \leq -a) + p(-a \leq z \leq a) - p(z \geq a)$
○ $p(z \leq -a) + p(-a \leq z \leq a) + p(z \geq a)$

Explanation:

Step1: Recall standard normal distribution properties

In a standard normal distribution, the total probability over the entire range of \( z \) is 1. The regions \( z \leq -a \), \( -a \leq z \leq a \), and \( z \geq a \) are mutually exclusive and cover the entire real line (since any \( z \)-value must be in one of these three regions: less than or equal to \( -a \), between \( -a \) and \( a \), or greater than or equal to \( a \)).

Step2: Apply the law of total probability for mutually exclusive events

For mutually exclusive events \( A \), \( B \), and \( C \) that cover the entire sample space, \( P(A) + P(B) + P(C) = 1 \). Here, let \( A \) be the event \( z \leq -a \), \( B \) be the event \( -a \leq z \leq a \), and \( C \) be the event \( z \geq a \). These three events are mutually exclusive (no overlap) and their union is the entire set of real numbers (so they cover all possible \( z \)-values in the standard normal distribution). Therefore, \( P(z \leq -a) + P(-a \leq z \leq a) + P(z \geq a) = 1 \).

Answer:

\( \boldsymbol{P(z \leq -a) + P(-a \leq z \leq a) + P(z \geq a)} \) (the fourth option)