Respuesta :
Answer:
Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(14.7,3.7)[/tex]
Where [tex]\mu=14.7[/tex] and [tex]\sigma=3.7[/tex]
Since the distribution of X is normal then we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And we have;
[tex]\mu_{\bar X}= 14.70[/tex]
[tex]\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59[/tex]
Step-by-step explanation:
Assuming this question: The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of 14.7 minutes and a standard deviation of 3.7 minutes. Let R be the mean delivery time for a random sample of 40 orders at this restaurant. Calculate the mean and standard deviation of [tex]\bar X[/tex] Round your answers to two decimal places.
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the delivery times of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(14.7,3.7)[/tex]
Where [tex]\mu=14.7[/tex] and [tex]\sigma=3.7[/tex]
Since the distribution of X is normal then we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
And we have;
[tex]\mu_{\bar X}= 14.70[/tex]
[tex]\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59[/tex]
