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

a. In words,Χ=_________

b.Χ~_____________

c. In words,X=_____________

d.X~_____(_____,_____)

e. Find the first quartile for the average song length, X.

f. The IQR (interquartile range) for the average song length, X, is from ___ - ___.

Short Answer

Expert verified

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. 2.71minutes

f. 0.09minutes

Step by step solution

01

Given information

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

Suppose we randomly pick five albums from the collection.

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

02

Explanation (part a)

The random variable Xis given by,

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

03

Explanation (part b)

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

X~U(2,3.5)

04

Explanation (part c)

The mean random variable is defined as,

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

05

Explanation (part d)

The mean random variable for normal distribution is given by,

X=N(μ,σn)where,n=43

the mean is found using, μ=2+3.52=2.75

The standard deviation is found using, σ=(3.5-2)212=0.433

Therefore, σn=0.43343=0.066

Therefore, the mean random variable for normal distribution isX=N(2.75,0.066)

06

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 X=N(2.75,0.066)

Then, for the first quartile,x0.25=μ+z0.25σ=2.75+normsinv(.25) *0.433=2.71mins.

07

Explanation (part f)

From the symmetry of the normal, The IQR (interquartile range) equals2z0.75σ=2×normsinv(.75)×0.433=0.09mins.

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