which of the following probabilities is the greatest for a standard normal distribution?\np(-1.5≤z≤…

Explanation:

Step1: Utilizar la tabla de la normal estándar

Para una distribución normal estándar $Z \sim N(0,1)$, la probabilidad $P(a\leq Z\leq b)=\Phi(b)-\Phi(a)$, donde $\Phi(z)$ es la función de distribución acumulada de la normal estándar.

Step2: Calcular $P(-1.5\leq z\leq - 0.5)$

$P(-1.5\leq z\leq - 0.5)=\Phi(-0.5)-\Phi(-1.5)$. Usando la simetría $\Phi(-z)=1 - \Phi(z)$, tenemos $\Phi(-0.5)=1-\Phi(0.5)$ y $\Phi(-1.5)=1 - \Phi(1.5)$. Entonces $P(-1.5\leq z\leq - 0.5)=\Phi(1.5)-\Phi(0.5)$.

Step3: Calcular $P(-0.5\leq z\leq 0.5)$

$P(-0.5\leq z\leq 0.5)=\Phi(0.5)-\Phi(-0.5)=\Phi(0.5)-(1 - \Phi(0.5))=2\Phi(0.5)-1$.

Step4: Calcular $P(0.5\leq z\leq 1.5)$

$P(0.5\leq z\leq 1.5)=\Phi(1.5)-\Phi(0.5)$.

Step5: Calcular $P(1.5\leq z\leq 2.5)$

$P(1.5\leq z\leq 2.5)=\Phi(2.5)-\Phi(1.5)$.

Step6: Comparar valores

Sabemos que la función $\Phi(z)$ es creciente. $\Phi(0.5)\approx0.6915$, $\Phi(1.5)\approx0.9332$, $\Phi(2.5)\approx0.9938$. $P(-1.5\leq z\leq - 0.5)=\Phi(1.5)-\Phi(0.5)=0.9332 - 0.6915=0.2417$. $P(-0.5\leq z\leq 0.5)=2\Phi(0.5)-1=2\times0.6915 - 1=0.383$. $P(0.5\leq z\leq 1.5)=\Phi(1.5)-\Phi(0.5)=0.2417$. $P(1.5\leq z\leq 2.5)=\Phi(2.5)-\Phi(1.5)=0.9938 - 0.9332=0.0606$.

Answer:

$P(-0.5\leq z\leq 0.5)$