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Songs on an iPod David鈥檚 iPod has about 10000songs. The distribution of the playtimes for these songs is heavily skewed to the right with a mean of 225seconds and a standard deviation of60 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

Expert verified

The mean of the sampling distribution is x=225seconds

The standard deviation of the sampling distribution isx=18.9737seconds.

Step by step solution

01

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

02

Explanation

The given data is written as

=225seconds

=60seconds

n=10

The sample mean xis equivalent to the population mean, which is the mean of the sampling distribution.

localid="1650109681905" x==225seconds

The standard deviation of the sampling distribution of the sample mean xis calculated by dividing the population standard deviation by the square root of the sample size.

localid="1650109697363" x=n=601018.9737.

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