My Problem
Ben goes into Tinder every night before bedtime. Ben does this every night for 365 days (1 year). The time that Ben spends every night in Tinder is uniformally distributed between 20-30 minutes. What is the probability that Ben spends more than 150 hours during one year?
The solution
0.99
My attempt
I tried to solve the problem by first calculating the average number of hours he would spend during a year (average = 0.5*(30*365 + 20*365) = 9125 min = 152.0833 hours).
Then I use Poission distribution, X ∈ Po(152)
Pr(X > 150) = 0,54
My problem
I am clearly getting the wrong answer, what am I doing wrong?
I am thankful for any help.
We use the central limit theorem (since the sample size is large). Let $X_i\sim U(20,30)$ be i.i.d. random variables representing the number of minutes spent on Tinder on day $i$. Then each $X_i$ has expectation $\mathbb{E}[X_i]=\frac{1}{2}(20+30)=25$ and variance $\operatorname{Var}[X_i]=\frac{1}{12}(30-20)^2=25/3$.
The distribution of the sample mean $\overline{X}=\frac{X_1+X_2+\dots+X_{365}}{365}$ can be approximated by a normal distribution $Y\sim N(\mu,\sigma^2)$ with mean $\mu=25$ and standard deviation $\sigma=\frac{\sqrt{25/3}}{\sqrt{365}}=\sqrt{5/219}$. Hence the probability we want to find is $$P(Y\geq 150\cdot 60/365)\approx 0.9883.$$
Sidenote: If you want an exact value for the probability, then you can use that the sum of i.i.d. $U(0,1)$ random variables follow an Irwin-Hall distribution, for which the CDF is known. You can standardize your $X_i$ via a linear transformation.