Question

A tablet PC contains 3217 music files. The distribution of file size is highly skewed with many small files. Suppose the true mean file size of music and video files on the tab, μ = 2.30 MB, and also assume that the standard deviation for this population (σ )is 3.25 megabytes (MB). If you select a random sample of 50 files, explain why it may be reasonable to assume that the sample mean (x bar) is approximately normal even though the population distribution is highly skewed.

Answer 1

Answer:

Let X the random variable who represents the file sizeof music. We know the following info:

$\mu =2.3,\sigma =3.25$

We select a sample of n=50 nails. That represent the sample size.  

Since the sample size is large enough n >30, we can use the central limit theorem. From this theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we can approximate the distribution of the sample mean as a normal distribution and no matter if the distribution for X is right skewed or no.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable who represents the file sizeof music. We know the following info:

$\mu =2.3,\sigma =3.25$

We select a sample of n=50 nails. That represent the sample size.  

Since the sample size is large enough n >30, we can use the central limit theorem. From this theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we can approximate the distribution of the sample mean as a normal distribution and no matter if the distribution for X is right skewed or no.