Question

which of the following probabilities is the greatest for a standard normal distribution?
○ $p(-1.0 \leq z \leq -0.5)$
○ $p(-0.5 \leq z \leq 0.5)$
○ $p(0.5 \leq z \leq 1.5)$
○ $p(1.5 \leq z \leq 2.5)$

Explanation:

Step1: Recall Normal Distribution Properties

In a standard normal distribution (\(Z \sim N(0,1)\)), the probability \(P(a \leq Z \leq b)\) is the area under the curve between \(z = a\) and \(z = b\). The curve is symmetric about \(z = 0\), and the total area under the curve is 1. The density is highest near \(z = 0\) and decreases as \(|z|\) increases.

Step2: Analyze Each Interval

• For \(P(-1.0 \leq Z \leq -0.5)\): By symmetry, this is equal to \(P(0.5 \leq Z \leq 1.0)\). The area between \(z = 0.5\) and \(z = 1.0\).
• For \(P(-0.5 \leq Z \leq 0.5)\): This is the area between \(z=-0.5\) and \(z = 0.5\), which is centered at \(z = 0\), where the density is highest.
• For \(P(0.5 \leq Z \leq 1.5)\): Area between \(z = 0.5\) and \(z = 1.5\), which is less than the area around \(z = 0\) because the density is lower as \(z\) moves away from 0.
• For \(P(1.5 \leq Z \leq 2.5)\): Area between \(z = 1.5\) and \(z = 2.5\), where the density is much lower, so the area is smaller.

Step3: Compare Areas

The interval \([-0.5, 0.5]\) is closest to \(z = 0\), where the probability density function (pdf) is the highest. So the area (probability) in this interval will be the largest among the given options.

Answer:

\(P(-0.5 \leq Z \leq 0.5)\) (the second option)