Question

(from previous problem) "An Ipod has about 10,000 songs. the distribution of the play time for these songs is highly skewed".....

3.) Averages of several measurements are less variable than individual measurements. Suppose the true mean duration of the play time for the songs in the IPOD is 350 seconds.

a.) Assuming the play times to be normally distributed, sketch the Normal curve when a single song is sampled. show work.

b.) Assuming the play times to be normally distributed, sketch the Normal curve for the mean of 10 songs. show work.

c.) What is the probability that the sample mean differs from the population mean by more than 19 seconds when only one song is sampled. show work.

d.) How does the probability that you calculated in part (c) change for the mean of an SRS of 10 songs? show work.

Answer 1

Assume that the population standard deviation of play time for the songs is = 280 seconds. (Since population variance is needed).

a)

For a single song,

Mean = = 350 seconds

Standard deviation =

= 280 seconds.

So, the graph will be as below:

The points on the graph are

b)

For 10 songs,

Mean = = 350 seconds

Standard deviation =

= 88.5438 seconds.

So, the graph will be as below:

The points on the graph are

c)

Probability that sample mean differs from population mean by more than 19 seconds when only one song is sampled

= P(Z > 19/280) + P(Z < -19/280)

= P(Z > 0.0679) + P(Z < -0.0679)

= 0.4729 + 0.4729

= 0.9458

d)

Probability that sample mean differs from population mean by more than 19 seconds when 10 songs are sampled

= P(Z > 19/88.5438) + P(Z < -19/88.5438)

= P(Z > 0.2146) + P(Z < -0.2146)

= 0.415 + 0.415

= 0.830

Thus, when 10 songs are sampled the Probability that the sample mean differs largely from the population mean decreases.