Question

g If a random sample of 18 deliveries was selected instead and the standard deviation of these delivery times was found to be 9.31 minutes, find the probability that their average delivery time is between 21 and 25 minutes

Answer 1

Answer:

We can use the z score formula given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

If we use this formula we got:

$z=\frac{21-20}{\frac{9.31}{\sqrt{18}}}= 0.456$

$z=\frac{25-20}{\frac{9.31}{\sqrt{18}}}= 2.279$

And using a calculator, excel or the normal standard table and we have that:

$P(0.456$

Step-by-step explanation:

We assume this previous info:  It is known that the amounts of time required for room-service delivery at a certain Marriott Hotel are Normally distributed with the average delivery time of 20 minutes.

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:

$X \sim N(20,9.31)$  

Where $\mu=20$ and $\sigma=9.31$

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

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

And we can use the z score formula given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

If we use this formula we got:

$z=\frac{21-20}{\frac{9.31}{\sqrt{18}}}= 0.456$

$z=\frac{25-20}{\frac{9.31}{\sqrt{18}}}= 2.279$

And using a calculator, excel or the normal standard table and we have that:

$P(0.456$