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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.

The random variable $X$is given by,

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

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,

$X~U(2,3.5)$

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.

The mean random variable for normal distribution is given by,

$\overline{X}=N(\mathrm{μ},\frac{\mathrm{σ}}{\sqrt{n}})$where,$n=43$

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

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

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

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

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,$x_{0.25}=\mathrm{μ}+z_{0.25}\mathrm{σ}=2.75+\text{normsinv(.25) *}0.433=2.71$mins.

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