Mathematics, 15.04.2020 00:07 greg777

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.

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Answer from: iamnee

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.

Answer from: Quest

answer: a. 13/44

step-by-step explanation: judging by the information given to us, we are looking for experimental probability. we can see that 13 males have chosen chocolate ice cream. there are 44 ice creams in total (remember that the total is always the denominator). so our answer is 13/44.

Answer from: Quest

it is not that hard all u have to do is plot the graph

step-by-step explanation:

Need with this if u can answer the question comment it

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