We have a playlist with $2000$ songs. The length of the songs on the playlist are on average $3.5$ minutes (i.e. $3$ minutes and $30$ seconds) with a standard deviation of $1.7$ minutes.
1) Can we find the probability that a randomly chosen song is longer than $4.5$ minutes?
2) Can we findthe probability that a random selection of $100$ songs lasts on average at least $4$ minutes?
3) Can we find the probability that a random selection of $200$ songs lasts in total at most $700$ minutes?
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I have done the following:
Let $X_i$ be the RV that describes the length of the $i$-th song.
$\overline{X}_n=\frac{1}{n}(X_1+X_2+\ldots +X_n)$ is the mean.
$\overline{X}_n$ approximates, according the central limit theorem, a normal distribution with parameters $E(\overline{X}_n)=3.5$ and $V(\overline{X}_n)=\frac{\sigma_X^2}{n}=\frac{1.7^2}{n}=\frac{2.89}{n}$. Is this correct?
How could we find the probability at 1) ?
At 2) we want to calculate the probability $P(\overline{X}\geq 4)$ with $n=100$ right?
If this is correct, we have that $$P(\overline{X}\geq 4)=1-P(\overline{X}< 4)=1-\Phi \left (\frac{4-3.5}{\frac{1.7}{\sqrt{100}}}\right )\approx 1-\Phi (2.94)=1-0.9984=0.0016$$
Is everything correct?
At 3) we define a new RV $Z:=n\cdot \overline{X}_n$ which describes the sum of lengths of $n$ songs.
For $n=200$ we have that $E(Z)=E(200\overline{X}_{200})=200\cdot E(\overline{X}_{200})=200\cdot 3.5=700$ and $V(Z)=V(200\overline{X}_{200})=200^2\cdot V(\overline{X}_{200})=200^2\cdot \frac{2.89}{200}=578$
Then $$P(Z\geq 700)=\Phi \left (\frac{700-700}{\sqrt{578}}\right )=\Phi (0)=0.5$$
Is this correct?
The answer to 1) is simply No.
Without any information on the distribution of the individual song lengths, we cannot determine the probability that the length of a randomly selected song exceeds a given limit. Knowing the average length and standard deviation is not enough.
For parts 2) and 3), the Central Limit Theorem applies.