Pregunta
Let \( Z \sim N(0,1) \) Without using the normalcdf(...) function, evaluate \( P(0 \leq Z \leq 1) \) 0.66 0.34 0.32 0.68 0.16
Ask by Franklin Gray. in the United States
Mar 09,2025
Solución de inteligencia artificial de Upstudy
Respuesta verificada por el tutor
Responder
\( P(0 \leq Z \leq 1) \approx 0.34 \)
Solución
We start with the definition of the cumulative distribution function (CDF) for the standard normal distribution:
\[
P(Z \leq z) = \Phi(z)
\]
For the interval \( 0 \leq Z \leq 1 \):
\[
P(0 \leq Z \leq 1) = \Phi(1) - \Phi(0)
\]
Since for the standard normal distribution:
\[
\Phi(0) = 0.5
\]
and
\[
\Phi(1) \approx 0.8413
\]
we have:
\[
P(0 \leq Z \leq 1) \approx 0.8413 - 0.5 = 0.3413
\]
Thus, the closest answer is:
\[
\boxed{0.34}
\]
Respondido por UpStudy AI y revisado por un tutor profesional
Como
error msg
Explicar 
Simplifique esta solución The Deep Dive
To evaluate \( P(0 \leq Z \leq 1) \) where \( Z \sim N(0,1) \), we first look for the area under the standard normal curve between 0 and 1. The total area under the curve is 1, and we know from statistical tables or the properties of the standard normal distribution that \( P(Z \leq 1) \approx 0.8413 \) and \( P(Z < 0) = 0.5 \). Thus, we can calculate: \[ P(0 \leq Z \leq 1) = P(Z \leq 1) - P(Z < 0) = 0.8413 - 0.5 = 0.3413 \] Therefore, upon rounding, the probability \( P(0 \leq Z \leq 1) \approx 0.34 \), which is an option provided in your list. So, the answer is **0.34**.
¡Prueba Premium ahora!
¡Prueba Premium y hazle a Thoth AI preguntas de matemáticas ilimitadas ahora!
Quizas mas tarde
Hazte Premium
