Q. 49 - The Practice of Statistics for AP

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. 

Step-by-Step Solution

Answer

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

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

Step 1: Given information

Total songs $=10000$

Mean of total songs $=225 seconds$

The standard deviation of total songs$=60 seconds$

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

Step 2: Explanation

The given data is written as 

$\mu =225 seconds$

$\sigma =60 seconds$

$n=10$

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

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

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.

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