Question

probability distributions
according to this simulation, how many calls can the library expect in one hour?
probability distribution
event | probability | cumulative probability | assigned numbers
0 | 0.05 | 0.05 | 01-05
1 | 0.15 | 0.20 | 06-20
2 | 0.25 | 0.45 | 21-45
3 | 0.30 | 0.75 | 46-75
4 | 0.20 | 0.95 | 76-95
5 | 0.05 | 1.00 | 96-100
intervals | 1 | 2 | 3 | 4 | 5 | 6
total

Explanation:

Step1: Define expected value formula

The expected value $E(X)$ is calculated as:
$$E(X) = \sum [x_i \cdot P(x_i)]$$
where $x_i$ is the number of calls, and $P(x_i)$ is its probability.

Step2: Substitute values into formula

$$\begin{align*} E(X) &= (0 \cdot 0.05) + (1 \cdot 0.15) + (2 \cdot 0.25) + (3 \cdot 0.30) + (4 \cdot 0.20) + (5 \cdot 0.05) \end{align*}$$

Step3: Calculate each term

$$\begin{align*} &0 \cdot 0.05 = 0\\ &1 \cdot 0.15 = 0.15\\ &2 \cdot 0.25 = 0.5\\ &3 \cdot 0.30 = 0.9\\ &4 \cdot 0.20 = 0.8\\ &5 \cdot 0.05 = 0.25 \end{align*}$$

Step4: Sum all terms

$$E(X) = 0 + 0.15 + 0.5 + 0.9 + 0.8 + 0.25 = 2.9$$

Answer:

2.9