Songs on an iPod David's iPod has about 10,000 songs. The distribution of the play times for these songs is heavily skewed to the right with a mean of 225 seconds and a standard deviation of 60 seconds. Suppose we choose an SRS of 10 songs from this population and calculate the mean play time $x^{-}\overline{x}$of these songs.

a. Identify the mean of the sampling distribution of ${\mathrm{x}}^{-}\overline{x}$.

b. Calculate and interpret the standard deviation of the sampling distribution of $x^{-}·\overline{x}$. Verify that the $10\%$condition is met.

Given values are,

The population mean is equal to the mean of the sampling distribution of the sample mean:

${\mu }_{\overline{x}}=\mu =225$

The sample as the units of the population mean is the mean of the sampling distribution of the sample mean, and so the mean is $225$seconds.

Given values are,

Let use the below formulae

${\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}$

The population mean is equal to the mean of the sampling distribution of the sample mean:

${\mu }_{\overline{x}}=\mu =225$

The sample distribution's standard deviation is

${\sigma }_x=\frac{\sigma }{\sqrt{n}}=\frac{60}{\sqrt{10}}=18.9737$

The units of the sampling distribution of the sample mean's standard deviation are the same as the units of the population standard deviation, hence the standard deviation is $18.9737$seconds.

The average sample mean play time of ten randomly selected iPod tracks ranges by 18.9737 second based on the average play time of 225 seconds