Business, 22.01.2020 23:31, kayonapretty14p45995

The customer care manager at a cell phone company keeps track of how long each -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 five minutes

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answered: charati1317

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

Because the sample mean is defined as:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

So then the mean for the sample distibution is $\mu_{\bar X} =\mu=12 min$

b) $\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{5}{\sqrt{10}}= 1.581$

c) For this case since the distribution for the random variable $\bar x$ 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, $\bar x$ .

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:

$X \sim N(12,5)$

Where $\mu=12$ and $\sigma=5$

(a) What is the mean of the sampling distribution of $\bar x$ ?

Since the distribution for the random variable X is normal then the distribution for the random variable $\bar X$ is also normal and given by:

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

Because the sample mean is defined as:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

So then the mean for the sample distibution is $\mu_{\bar X} =\mu=12 min$

(b) Is it possible to calculate the standard deviation of $\bar x$ ? 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:

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

Then we know that the deviation for the sample mean is given by:

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{5}{\sqrt{10}}= 1.581$

(c) Do you know the approximate shape of the sampling distribution of $\bar x$ ? If so, describe the shape and justify your answer. If not, explain why not.

For this case since the distribution for the random variable $\bar x$ 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

answered: Guest

answer is i don't know

answered: Guest

explanation:

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