Question

Songs on an iPod David’s iPod has about $10000$songs. The distribution of the playtimes 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 playtime x of these songs. What are the mean and the standard deviation of the sampling distribution of x? Explain.

Short answer

The mean of the sampling distribution is ${\mu }_{\overline{x}}=225seconds$

The standard deviation of the sampling distribution is${\sigma }_{\overline{x}}=18.9737seconds$.

Step-by-step solution

Given information

Total songs $=10000$

Mean of total songs$=225seconds$

The standard deviation of total songs$=60seconds$

the mean and the standard deviation of the sampling distribution of x

Explanation

The given data is written as

$\mu =225seconds$

$\sigma =60seconds$

$n=10$

The sample mean $\overline{x}$is equivalent to the population mean, which is the mean of the sampling distribution.

localid="1650109681905" $$\begin{gathered} {\mu }_{\overline{x}}=\mu \\ =225seconds \end{gathered}$$

The standard deviation of the sampling distribution of the sample mean $\overline{x}$is calculated by dividing the population standard deviation by the square root of the sample size.

localid="1650109697363" $$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{60}{\sqrt{10}} \\ \approx 18.9737 \end{gathered}$$.