The number of complaints which a bank manager receives per day follows a Poisson distribution with parameter `m = 4`. Find the probability that the manager receives only two complaints on a given day. Use `e^(-4) = 0.0183`

Answer

To find the probability that the bank manager receives only two complaints on a given day, we will use the formula for the Poisson distribution. The formula is given by:

\[ P(X = x) = \frac{e^{-m} \cdot m^x}{x!} \]

where:
- \( P(X = x) \) is the probability of receiving \( x \) complaints,
- \( m \) is the average number of complaints (in this case, \( m = 4 \)),
- \( x \) is the number of complaints we want to find the probability for (in this case, \( x = 2 \)),
- \( e \) is the base of the natural logarithm.

### Step-by-step Solution:

1. **Identify the parameters**:
- Here, \( m = 4 \) (the average number of complaints).
- We want to find the probability for \( x = 2 \) (the number of complaints).

2. **Substitute the values into the formula**:
\[ P(X = 2) = \frac{e^{-4} \cdot 4^2}{2!} \]

3. **Calculate \( e^{-4} \)**:
- We are given that \( e^{-4} = 0.0183 \).

4. **Calculate \( 4^2 \)**:
\[ 4^2 = 16 \]

5. **Calculate \( 2! \)**:
\[ 2! = 2 \times 1 = 2 \]

6. **Substitute these values back into the formula**:
\[ P(X = 2) = \frac{0.0183 \cdot 16}{2} \]

7. **Perform the multiplication**:
\[ 0.0183 \cdot 16 = 0.2928 \]

8. **Divide by \( 2 \)**:
\[ P(X = 2) = \frac{0.2928}{2} = 0.1464 \]

### Final Answer:
The probability that the bank manager receives only two complaints on a given day is \( P(X = 2) = 0.1464 \).