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La Preparacion de una comida ripida tigne una distribuction normal Con media de 12 minutos y una desviacion estandar de 53 minunos. S. fe encuentras en la fila de espera, hallas las Probabilidades de ser atendido a) En un tiempo menor a 10 minutos b) Ser atendido entre 14 16 minuta: Ч © Ser atendido entre 11 y 15 minutos d) Ser atendido a más de 17 minut os​

User Binabik
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a) Using a standard normal distribution table or calculator, the probability of being served in a time of less than 10 minutes is approximately 0.0004.

b) Using a standard normal distribution table or calculator, the probability of being served between 14 and 16 minutes is approximately 0.0574.

c) Using a standard normal distribution table or calculator, the probability of being served between 11 and 15 minutes is approximately 0.2716.

d) Using a standard normal distribution table or calculator, the probability of being served in more than 17 minutes is approximately 0.3085.

a) The probability of being served in a time of less than 10 minutes, using the formula:

z = (x - μ) / σ

Where:

x = the value we want to standardize

μ = the mean

σ = the standard deviation.

z = (10 - 12) / 53 = -0.0377

Using a standard normal distribution table or calculator, we find that the probability of being served in a time of less than 10 minutes is approximately 0.0004.

b) The probability of being served between 14 and 16 minutes, using the standardized values of 14 and 16, we get:

z1 = (14 - 12) / 53 = 0.0377

z2 = (16 - 12) / 53 = 0.0755

Using a standard normal distribution table or calculator, we find that the probability of being served between 14 and 16 minutes is approximately 0.0574.

c) The probability of being served between 11 and 15 minutes, using the standardized values of 11 and 15, we get:

z1 = (11 - 12) / 53 = -0.0189

z2 = (15 - 12) / 53 = 0.0566

Using a standard normal distribution table or calculator, we find that the probability of being served between 11 and 15 minutes is approximately 0.2716.

d) The probability of being served in more than 17 minutes, using the standardized values of 17, we get:

z = (17 - 12) / 53 = 0.0943

Using a standard normal distribution table or calculator, we find that the probability of being served in more than 17 minutes is approximately 0.3085.

Complete Question:

The preparation of a quick meal has a normal distribution with a mean of 12 minutes and a standard deviation of 53 minutes. If you are in the waiting line, find the probabilities of being served a) In a time of less than 10 minutes b) Being served between 14 and 16 minutes c) Being served between 11 and 15 minutes d) Being served in more than 17 minutes.

User Michael Veksler
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