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!
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.
CLT:
$$ \sum_i\frac{ X_i -\mathbb{E}[X_i]}{\sqrt{V(X_i)}} \sim N(0,1)$$ 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