Question

which of the following probabilities is the greatest for a standard normal distribution? p(-1.5≤z≤-0.5) p(-0.5≤z≤0.5) p(0.5≤z≤1.5) p(1.5≤z≤2.5)

Explanation:

Step1: Recall standard - normal table property

The standard - normal distribution is symmetric about \(z = 0\), and the probability density function is highest at \(z = 0\) and decreases as \(|z|\) increases. The area under the standard - normal curve between two \(z\) - values \(z_1\) and \(z_2\) is given by \(P(z_1\leq z\leq z_2)=\Phi(z_2)-\Phi(z_1)\), where \(\Phi(z)\) is the cumulative distribution function of the standard - normal distribution.

Step2: Analyze each option

• For \(P(-1.5\leq z\leq - 0.5)=\Phi(-0.5)-\Phi(-1.5)\). Using the property \(\Phi(-z)=1 - \Phi(z)\), we have \(P(-1.5\leq z\leq - 0.5)=(1 - \Phi(0.5))-(1 - \Phi(1.5))=\Phi(1.5)-\Phi(0.5)\).
• For \(P(-0.5\leq z\leq0.5)=\Phi(0.5)-\Phi(-0.5)=\Phi(0.5)-(1 - \Phi(0.5)) = 2\Phi(0.5)-1\).
• For \(P(0.5\leq z\leq1.5)=\Phi(1.5)-\Phi(0.5)\).
• For \(P(1.5\leq z\leq2.5)=\Phi(2.5)-\Phi(1.5)\).

Step3: Compare the areas

The standard - normal curve has more area concentrated around \(z = 0\). The interval \((-0.5,0.5)\) is centered at \(z = 0\), while the other intervals are farther from \(z = 0\). Since the standard - normal distribution is symmetric about \(z = 0\) and the density function is highest at \(z = 0\), the probability \(P(-0.5\leq z\leq0.5)\) is the greatest among the given options.

Answer:

P(-0.5 ≤ z ≤ 0.5)