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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-x¯of these songs.

a. Identify the mean of the sampling distribution of x-x¯.

b. Calculate and interpret the standard deviation of the sampling distribution of x-·x¯. Verify that the 10%condition is met.

Short Answer

Expert verified

(a). The mean of the sampling distribution of x¯is 225 seconds

(b). The sample mean play time of 10 randomly selected songs of the iPod varies on average by 18.9737second from the mean play time 225 seconds.

Step by step solution

01

part(a) step 1: Given information

Given values are,

μ=225σ=60n=10
02

Part(a) step 2: Calculation

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

μx¯=μ=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 225seconds.

03

Part(b) step 1: Given information

Given values are,

μ=225σ=60n=10

Let use the below formulae

σx¯=σn

04

Part(b) step 2: Calculation

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

μx¯=μ=225

The sample distribution's standard deviation is

σx=σn=6010=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 18.9737seconds.

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

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