You own an ice-cream shop. Every day when you open the shop, there is an $\text{Exp}(\frac{1}{10})$ random waiting time, in minutes, before the first customer arrives.

1. When the waiting time for the first customer exceeds 20 minutes, it is a bad day. Approximate the probability that you will have between 50 and 60 bad days in a year (365 days).
2. When the waiting time for the first customer is shorter than 3 seconds (i.e., 0.05 minutes), it is a great day. Approximate the probability that you will have exactly 3 great days in a year (365 days).

**My Attempt**

1. The probability of waiting over 20 minutes is
$$\mathbb{P}(X\geq20)=\int^\infty\_{20}0.1e^{-0.1x}dx=[e^{-0.1x}]^\infty\_{20}=e^{-2}\approx0.135$$
Then since $np(1-p)=365(0.135)(1-0.135)=42.7>10$, we can use the normal distribution to apprimate the probability. The expectation for an exponential distribution is $\frac{1}{\lambda}=\frac{1}{1/10}=10$. Thus
$$\mathbb{P}(50\leq x\leq60)=\mathbb{P}(a\leq\frac{x-\mathbb{E}[x]}{\sqrt{\text{Var}(x)}}\leq b)$$
$$=\mathbb{P}(\frac{50-10}{\sqrt{\text{Var}(x)}}\leq\frac{x-\mathbb{E}[X]}{\sqrt{\text{Var}(x)}}\leq\frac{60-10}{\sqrt{\text{Var}(x)}})\approx \Phi(\frac{50}{\sqrt{\text{Var}(x)}})-\Phi(\frac{40}{\sqrt{\text{Var}(x)}})$$
The part that I am having trouble with is how to find the variance for the exponential distribution. Can I get some pointers?

**Edit**
I think I know how to find the variance now. I got Var[X]=E[$X^2$]-E[X]=200-10=190.
So
$$\Phi(\frac{50}{\sqrt{190}})-\Phi(\frac{40}{\sqrt{190}})\approx0.00176$$

My concern now is that this probability is way too small. I don't think that it is reasonable. Can I get pointers on where I went wrong? Thank you

2. For this one, since 3 seconds is so small, we can consider this event rare, so we can use poisson approximation. First we find the probability of great day
$$\mathbb{P}(X\leq0.05)=1-[e^{-0.1}x]^\infty\_{0.05}\approx0.00498$$
We know that the expectation is equal to the parameter, $\lambda$, of the poisson distribution. Then we can find use the formula
$$e^{-\lambda}\frac{\lambda^k}{k!}$$

Is this the right approach?