Respuesta :
Answer:
a) 23.89% probability that the average size of 30 randomly selected music files taken from the tablet is less than 1.93 MB
b) 35.57% probability that the average size of 50 randomly selected music files taken from the tablet is more than 2.52 MB
c) At least 87 files
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this problem, we have that:
[tex]\mu = 2.35, \sigma = 3.25[/tex]
a. What is the probability that the average size of 30 randomly selected music files taken from the tablet is less than 1.93 MB?
[tex]n = 30, s = \frac{3.25}{\sqrt{30}} = 0.59[/tex]
This probability is the pvalue of Z when X = 1.93. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1.93 - 2.35}{0.59}[/tex]
[tex]Z = -0.71[/tex]
[tex]Z = -0.71[/tex] has a pvalue of 0.2389
23.89% probability that the average size of 30 randomly selected music files taken from the tablet is less than 1.93 MB
b. What is the probability that the average size of 50 randomly selected music files taken from the tablet is more than 2.52 MB?
[tex]n = 50, s = \frac{3.25}{\sqrt{50}} = 0.46[/tex]
This pobability is 1 subtracted by the pvalue of Z when X = 2.52.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{2.52 - 2.35}{0.46}[/tex]
[tex]Z = 0.37[/tex]
[tex]Z = 0.37[/tex] has a pvalue of 0.6443
1 - 0.6443 = 0.3557
35.57% probability that the average size of 50 randomly selected music files taken from the tablet is more than 2.52 MB
c. How many files would you need to sample if you wanted the standard deviation of the sample mean size to lie no larger than 0.35 MB?
At least n files, in which n is found when s = 0.35. So
[tex]s = \frac{3.25}{\sqrt{n}}[/tex]
[tex]0.35 = \frac{3.25}{\sqrt{n}}[/tex]
[tex]0.35\sqrt{n} = 3.25[/tex]
[tex]\sqrt{n} = \frac{3.25}{0.35}[/tex]
[tex](\sqrt{n})^{2} = (\frac{3.25}{0.35})^{2}[/tex]
[tex]n = 86.2[/tex]
Rounding up
We would need at least 87 files.
