Respuesta :
Answer:
a) [tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
Because the sample mean is defined as:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
So then the mean for the sample distibution is [tex] \mu_{\bar X} =\mu=12 min[/tex]
b) [tex] \sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{5}{\sqrt{10}}= 1.581[/tex]
c) For this case since the distribution for the random variable [tex]\bar x[/tex] is normal and for the sample mean is also normal so then we can conclude that the distribution for the sample mean is symmetrical and bell shaped with a deviation lower than the deviation for X
Explanation:
Assuming this complete problem:
The customer care manager at a cell phone company keeps track of how long each help-line caller spends on hold before speaking to a customer service representative. He finds that the distribution of wait times for all callers has a mean of 12 minutes with a standard deviation of 5 minutes. The distribution is moderately skewed to the right. Suppose the manager takes a random sample of 10 callers and calculates their mean wait time, [tex]\bar x[/tex] .
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".
Let X the random variable that represent the wait times for all callers of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(12,5)[/tex]
Where [tex]\mu=12[/tex] and [tex]\sigma=5[/tex]
(a) What is the mean of the sampling distribution of [tex]\bar x[/tex] ?
Since the distribution for the random variable X is normal then the distribution for the random variable [tex]\bar X[/tex] is also normal and given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
Because the sample mean is defined as:
[tex] \bar X = \frac{\sum_{i=1}^n X_i}{n}[/tex]
So then the mean for the sample distibution is [tex] \mu_{\bar X} =\mu=12 min[/tex]
(b) Is it possible to calculate the standard deviation of [tex]\bar x[/tex] ? If it is, do the calculation. If it isn’t, explain why.
For this case since we knwo that the distribution for the sampel mena is given by:
[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]
Then we know that the deviation for the sample mean is given by:
[tex] \sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{5}{\sqrt{10}}= 1.581[/tex]
(c) Do you know the approximate shape of the sampling distribution of [tex]\bar x[/tex]? If so, describe the shape and justify your answer. If not, explain why not.
For this case since the distribution for the random variable [tex]\bar x[/tex] is normal and for the sample mean is also normal so then we can conclude that the distribution for the sample mean is symmetrical and bell shaped with a deviation lower than the deviation for X
