for a standard normal distribution, which of the following expressions must always be equal to 1?\np(z≤…

for a standard normal distribution, which of the following expressions must always be equal to 1?\np(z≤ - a)-p(-a≤z≤a)-p(z≥a)\np(z≤ - a)-p(-a≤z≤a)+p(z≥a)\np(z≤ - a)+p(-a≤z≤a)-p(z≥a)\np(z≤ - a)+p(-a≤z≤a)+p(z≥a)

for a standard normal distribution, which of the following expressions must always be equal to 1?\np(z≤ - a)-p(-a≤z≤a)-p(z≥a)\np(z≤ - a)-p(-a≤z≤a)+p(z≥a)\np(z≤ - a)+p(-a≤z≤a)-p(z≥a)\np(z≤ - a)+p(-a≤z≤a)+p(z≥a)

Answer

Explanation:

Step1: Recall properties of normal - distribution

The total area under the standard - normal curve is 1. The entire distribution can be partitioned into three non - overlapping regions: $z\leq - a$, $-a\leq z\leq a$, and $z\geq a$.

Step2: Use the property of total probability

The sum of the probabilities of these three non - overlapping regions must equal the total probability of the distribution. That is, $P(z\leq - a)+P(-a\leq z\leq a)+P(z\geq a)=1$.

Answer:

$P(z\leq - a)+P(-a\leq z\leq a)+P(z\geq a)$