Question

If \(Z\) is a standard normal variable, \[ P(Z>z_1)=\alpha, \] \[ P(Z<z_2)=\beta \] and \(z_2<z_1\), then a possible value of \[ P(z_2<Z<z_1) \] is

• A. \(\alpha-\beta\)
• B. \(0.5+(\alpha+\beta)\)
• C. \(1-(\alpha+\beta)\)
• D. \(|\beta-\alpha|\)

Hint:

For normal distribution questions, always remember: \[ \text{Middle Area} = 1-(\text{Left Tail}+\text{Right Tail}). \] This is often the fastest way to solve such problems.

Answer 1

The Correct Option is C

Concept: The total probability under a probability density curve is always equal to 1. The interval \((z_2,z_1)\) is obtained by removing the two tail probabilities from the total area.

Step 1: Write the given probabilities.
\[ P(Z>z_1)=\alpha. \] \[ P(Z<z_2)=\beta. \] Since \(z_2<z_1\), these two regions do not overlap.

Step 2: Use total probability.
The entire area under the curve equals \[ 1. \] Hence, \[ P(Zz_1) = 1. \] Substituting, \[ \beta + P(z_2<Z<z_1) + \alpha = 1. \]

Step 3: Solve for the required probability.
\[ P(z_2<Z<z_1) = 1-(\alpha+\beta). \] \[ \boxed{1-(\alpha+\beta)} \] Hence option (C) is correct.