Question

The length of songs in a collector’s iTunes album collection is uniformly distributed from two to $3.5$minutes. Suppose we randomly pick five albums from the collection. There are a total of $43$songs on the five albums.

a. In words,${\rm X}=\_\_\_\_\_\_\_\_\_$

b.${\rm X}~\_\_\_\_\_\_\_\_\_\_\_\_\_$

c. In words,$\overline{X}=\_\_\_\_\_\_\_\_\_\_\_\_\_$

d.$\overline{X}~\_\_\_\_\_(\_\_\_\_\_,\_\_\_\_\_)$

e. Find the first quartile for the average song length, $\overline{X}$.

f. The IQR (interquartile range) for the average song length, $\overline{X}$, is from ___ - ___.

Short answer

a. the length of a song, in minutes, in the collection

b.$U(2,3.5)$

c. the average length, in minutes, of the songs from a sample of five albums from the collection

d.$N(2.75,0.0660)$

e. $2.71$minutes

f. $0.09$minutes

Step-by-step solution

Given information

The length of songs in a collector’s iTunes album collection is uniformly distributed from two to $3.5$minutes.

Suppose we randomly pick five albums from the collection.

There are a total of $43$songs on the five albums.

Explanation (part a)

The random variable $X$is given by,

$X=$the length of a song, in minutes, in the collection

Explanation (part b)

According to the given information, the length of the songs is uniformly distributed from two to $3.5$minutes, the random variable is expressed as,

$X~U(2,3.5)$

Explanation (part c)

The mean random variable is defined as,

$\overline{X}=$the average length, in minutes, of the songs from a sample of five albums from the collection.

Explanation (part d)

The mean random variable for normal distribution is given by,

$\overline{X}=N(\mu ,\frac{\sigma }{\sqrt{n}})$where,$n=43$

the mean is found using, $\mu =\frac{2+3.5}{2}=2.75$

The standard deviation is found using, $\sigma =\sqrt{\frac{{(3.5-2)}^2}{12}}=0.433$

Therefore, $\frac{\sigma }{\sqrt{n}}=\frac{0.433}{\sqrt{43}}=0.066$

Therefore, the mean random variable for normal distribution is$\overline{X}=N(2.75,0.066)$

Explanation (part e)

The Central Limit Theorem tells us that the distribution approaches the normal distribution. That the original distribution is the uniform and n is relatively large assures us that it will be "close", and we are looking at quartiles, so all should be well.

Then, the distribution of the sample is approximately $\overline{X}=N(2.75,0.066)$

Then, for the first quartile,$x_{0.25}=\mu +z_{0.25}\sigma =2.75+\text{normsinv(.25) *}0.433=2.71$mins.

Explanation (part f)

From the symmetry of the normal, The IQR (interquartile range) equals$2z_{0.75}\sigma =2\times norm\sin v(.75)\times 0.433=0.09$mins.