Question

The delivery times for all food orders at a fast-food restaurant during the lunch hour are normally distributed with a mean of minutes and a standard deviation of minutes. Let be the mean delivery time for a random sample of orders at this restaurant. Calculate the mean and standard deviation of . Round your answers to two decimal places.

Answer 1

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:

$X \sim N(14.7,3.7)$  

Where $\mu=14.7$ and $\sigma=3.7$

Since the distribution of X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we have;

$\mu_{\bar X}= 14.70$

$\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59$

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 $\bar X$ 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:

$X \sim N(14.7,3.7)$  

Where $\mu=14.7$ and $\sigma=3.7$

Since the distribution of X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we have;

$\mu_{\bar X}= 14.70$

$\sigma_{\bar X} =\frac{3.7}{\sqrt{40}}= 0.59$