The average duration of a final exam is 78 minutes. With a variance of 36. Assume that the execution time is set to a Normal If a sample of 32 students is taken, what is the probability that the average time is less than 80 min?



Answer to a math question The average duration of a final exam is 78 minutes. With a variance of 36. Assume that the execution time is set to a Normal If a sample of 32 students is taken, what is the probability that the average time is less than 80 min?

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Para resolver este problema, necesitamos utilizar la distribución normal estándar y la fórmula para el intervalo de confianza para la media poblacional.

Dado que sabemos la media y la varianza de la población, podemos calcular la desviación estándar de la población utilizando la fórmula:

\sigma = \sqrt{\text{varianza}}

\sigma = \sqrt{36}

\sigma = 6

Luego, podemos calcular la desviación estándar de la media muestral utilizando la fórmula:

\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}

donde n es el tama√Īo de la muestra. En este caso, n = 32 , por lo que

\sigma_{\bar{x}} = \frac{6}{\sqrt{32}}

\sigma_{\bar{x}} \approx 1.061

Ahora, necesitamos convertir el tiempo de duración de 80 min a una puntuación Z utilizando la fórmula:

Z = \frac{X - \mu}{\sigma_{\bar{x}}}

donde X es el valor que queremos encontrar y \mu es la media de la población.

Z = \frac{80 - 78}{1.061}

Z \approx 1.885

Finalmente, podemos utilizar una tabla de la distribución normal estándar (o un software estadístico) para encontrar la probabilidad de que Z sea menor que 1.885. Esta probabilidad se conoce como el área a la izquierda de Z .

Responderemos a la pregunta en el próximo mensaje.

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