A large playlist consists of songs with times which have mean 2 minutes and standard deviation 10 seconds.

What is the probability that more than 64 randomly chosen songs are required to fill a program which is 132 minutes long? Assume that the length of a randomly selected song is normally distributed.

My mean for the total length of 64 randomly chosen songs is 128 minutes.

The standard deviation for the total length of 64 songs is 4/3 minutes.

I have tried using z-scores to calculate this. (132-128)/(4/3) = 3. Then I did 1 - F(3) = 0.0013, which was wrong.

I'm confused as to where I went wrong.

Thank you for any help!

  • $\begingroup$ What is the distribution of the length of a song? $\endgroup$ – Did Nov 5 '17 at 21:14
  • $\begingroup$ Apologies, it is normally distributed. $\endgroup$ – Blare Nov 5 '17 at 21:28
  • $\begingroup$ Mean and variance? $\endgroup$ – Did Nov 5 '17 at 21:30
  • $\begingroup$ Here is more background information: A large playlist consists of songs with times which have mean 2 minutes and standard deviation 10 seconds. $\endgroup$ – Blare Nov 5 '17 at 21:31

So assuming the length of the songs $X_i \ i\in\{1,...,64\}$ are iid $N(120,10^2)$ distributed, then you want to know the probability, that the sum of those lengths is smaller than 132 minutes.

$$P\left(\sum_{i}X_i<132\times 60\right)$$

But even if they are not normal distributed, as you only want to know the distribution of the sum of a bunch of them you can use the central limit theorem to get an approximation of the distribution. And then you can use the cumulative distribution to get your probability.


$$ \sum_i\frac{ X_i -\mathbb{E}[X_i]}{\sqrt{V(X_i)}} \sim N(0,1)$$ thus $$ \sum_{i=0}^{64} \frac{ X_i -120}{10}\sim N(0,1)$$ which leads to $$ \sum_{i=0}^{64} X_i\sim 10\times N(0,1)+64\times120=N(7680,10^2)$$

If I am not mistaken. Now you just apply the appropriate cumulative distribution function to your problem.

Edit: copied the wrong numbers now everything should be in seconds

  • $\begingroup$ I'm still a bit confused, I think the question is asking for the probability that more than 64 songs are needed for a length of 132 minutes, not the probability that the length of the 64 songs is greater than 132 minutes. $\endgroup$ – Blare Nov 5 '17 at 23:05
  • 1
    $\begingroup$ When do you need more than 64 songs to cover 132 minutes? When those 64 songs are shorter than 132 minutes all together. So the sum of the lenghts of those 64 randomly selected songs has to be shorter than 132 minutes for you to need more. P(need more)=P(sum<132) $\endgroup$ – Felix B. Nov 5 '17 at 23:09
  • $\begingroup$ Thank you very much, the wording was what got me. $\endgroup$ – Blare Nov 6 '17 at 18:14

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