Question

A tablet PC contains 3217 music files. The distribution of file size is highly skewed with many small file sizes. The mean file size on the tablet is 2.35 MB with a standard deviation of 3.25 MB.

a What is the probability that the average size of 30 randomly selected music files taken from the tablet is more than 2.75 MB?

b What is the probability that the average size of 50 randomly selected music files taken from the tablet is less than 2.21 MB?

c How many files would you need to sample if you wanted the standard deviation of the sample mean file size to be no larger than 0.25 MB?

Answer 1

a.

By Central limit theorem, the distribution of average file size will follow normal distribution with mean as the population mean and standard deviation as where is the  population standard deviation and n is the sample size.

For n = 30, mean = 2.35 MB, standard deviation = 3.25 / = 0.5933661

Probability that the average size of 30 randomly selected music files taken from the tablet is more than 2.75 MB = P(X > 2.75)

= P[Z > (2.75 - 2.35)/0.5933661]

= P[Z > 0.6741201]

= 0.2501 (Using Z table)

b.

For n = 50, mean = 2.35 MB, standard deviation = 3.25 / = 0.4596194

Probability that the average size of 50 randomly selected music files taken from the tablet is less than 2.21 MB = P(X < 2.21)

= P[Z < (2.21 - 2.35)/0.4596194]

= P[Z < -0.3046]

= 0.3803 (Using Z table)

c

Given ,

sample standard deviation < 0.25 MB

< 0.25 MB

=> 3.25 / < 0.25

=> > 3.25 / 0.25

=> > 13

=> n > 13² = 169

So, the minimum files required in the sample is 169.