A tablet PC contains 8152 music and video files. The distribution of file size is highly skewed.

Question

Assume that the standard deviation for this population is 0.84 megabytes (MB). Suppose that the true mean file size of the music and video files on the tablet is 7.6 MB and you plan to take an SRS of n=50 files. Use technology to find the probability (±0.001) that your sample mean will differ from the population mean by more than 0.12 MB.

Answer 1

Jenna,

 

I'm assuming that this is a normal distribution which means you want to the probability that the difference between sample mean and population mean is between .-12 and .12. To do this, let's find z for each value. You'll need to factor in the sample size to adjust the population standard deviation by calculating standard error:

SE=σ/√50=.84/√50=.84/7.07=.1188

now z=+/-.12/.1188=+/-1.01

 

Looking up a z table:

The Probability at z=1.01 (up to 1.01) is .843752

The Probability at z=-1.01 (up to -1.01) is .156248

So the region in between for a max difference .12 is .843752-.156248=.6875

 

Hope that helps!