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question 1 of 7 david’s phone has about 10,000 songs. the distribution …

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question 1 of 7
david’s phone has about 10,000 songs. the distribution of play time 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 \\(\bar{x}\\) of these songs.
(a) calculate the mean and standard deviation of the sampling distribution of \\(\bar{x}\\).
\\(\mu_{\bar{x}} = \\) seconds (do not round.)
\\(\sigma_{\bar{x}} = \\) seconds (round to 2 decimal places.)
(b) interpret the standard deviation from part (a).
in \\(\text{srss}\\) of size \\(n = 10\\) from david’s ipod, the sample mean play time of songs typically varies from \\(\text{the true mean}\\) by about \\(18.97\\) seconds.

Explanation:

Response
Part (a)

Step 1: Mean of Sampling Distribution

The mean of the sampling distribution of the sample mean \(\bar{x}\) (denoted as \(\mu_{\bar{x}}\)) is equal to the population mean \(\mu\). Given the population mean \(\mu = 225\) seconds, we have:
\(\mu_{\bar{x}}=\mu = 225\)

Step 2: Standard Deviation of Sampling Distribution

The standard deviation of the sampling distribution of the sample mean (also called the standard error) is given by the formula \(\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size. Here, \(\sigma = 60\) seconds and \(n = 10\). Plugging in the values:
\(\sigma_{\bar{x}}=\frac{60}{\sqrt{10}}\)
Calculating \(\sqrt{10}\approx3.1623\), so \(\frac{60}{3.1623}\approx18.97\) (rounded to 2 decimal places).

Brief Explanations

The standard deviation of the sampling distribution (standard error) tells us how much the sample mean \(\bar{x}\) varies from the true population mean, on average, when taking samples of size \(n = 10\). So, in SRSs (simple random samples) of size \(n = 10\) from David’s phone, the sample mean play time of songs typically varies from the true mean (which is 225 seconds) by about 18.97 seconds.

Answer:

For \(\mu_{\bar{x}}\): \(\boldsymbol{225}\) seconds
For \(\sigma_{\bar{x}}\): \(\boldsymbol{18.97}\) seconds

Part (b)